3.1

Part a.) We first need to show that $K_\delta(x)$ satisfies $(i),(ii),$ and $(iii)$ listed at the top of page 109. Given $\delta > 0$, and $\phi$ is integrable s.t. $\int_{\mathbb{R}^d} \phi = 1$, we have that: $$\int_{\mathbb{R}^d} K_\delta (x)\hspace{0.1cm}dx = \int_{\mathbb{R}^d} \frac{1}{\delta^d} \phi(x / \delta) \hspace{0.1cm}dx$$ ...and by the dilation property of $L^1$ functions: $$\int_{\mathbb{R}^d} \frac{1}{\delta^d} \phi(x / \delta) \hspace{0.1cm}dx = \int_{\mathbb{R}^d} \frac{\delta^d}{\delta^d} \phi(x) \hspace{0.1cm}dx = \int_{\mathbb{R}^d} \phi(x) \hspace{0.1cm}dx = 1$$ Which satisfies $(i)$. Next, notice: $$\int_{\mathbb{R}^d} |K_\delta (x)| \hspace{0.1cm}dx = \int_{\mathbb{R}^d} |\phi(x)| \hspace{0.1cm}dx = || \phi ||_{L^1} < \infty$$ ...which satisfies $(ii)$. Finally, observe that: $$\int_{B_\mu} |K_\delta (x)| \hspace{0.1cm}dx = \int_{B_\mu} |K_\delta (x)| \hspace{0.1cm}dx = \int_{\mathbb{R}^d} \frac{1}{\delta^d} |\phi(x / \delta)| \chi_{B_\mu}(x) dx$$ ...again by the dilation property of $L^1$ functions: $$\int_{\mathbb{R}^d} \frac{1}{\delta^d} |\phi(x / \delta)| \chi_{B_\mu}(x) dx = \int_{\mathbb{R}^d} |\phi(x)| \chi_{B_\mu}(\delta x) dx = \int_{B_{\mu / \delta}} |\phi(x)| dx$$ Therefore, given that for any $\mu > 0$, we have $B_{\mu / \delta} \to \mathbb{R}^d$ as $\delta \to 0$, it follows directly by, say the Dominated Convergence Theorem that: $$\lim_{\delta \to 0} \int_{B_\mu^c} |K_\delta (x)| \hspace{0.1cm}dx = 0$$ ...which was $(iii)$.

Part b.) With the added assumptions that $|\phi| \leq M$ where $M > 0$ and $\phi$ is supported on a compact set $S \subset \mathbb{R^d}$, we need to show that $K_\delta (x)$ is an approximation to the identity. (Properties $(ii')$ and $(iii')$) Certainly: $$|K_\delta (x)| = |\frac{1}{\delta^d} \phi(x / \delta)| \leq \frac{1}{\delta^d}M$$ ...satisfying condition $(ii')$. Next, since $S$ is compact, let $\overline{B_r}$ be a ball of radius $r = \max \lbrace \max(S), 1 \rbrace$. $$|K_\delta (x)| \leq \frac{M}{\delta^d} \chi_S(x/\delta) \leq \frac{M}{\delta^d} \chi_{\overline{B_r}}(x/\delta)$$ Now, for $|x| > \delta r$, we have: $$\frac{M \delta}{|x|^{d+1}} \geq \frac{M}{\delta^d} \chi_{\overline{B_r}}(x/\delta) = 0$$ For $0 < |x| \leq \delta r$ we have: $$\frac{M \delta}{|x|^{d+1}} \geq \frac{M \delta}{|r\delta|^{d+1}} \geq \frac{M}{\delta^d}$$ ...satisfying condition $(iii')$.

Part c.) First, since $\int_{\mathbb{R}^d} K_\delta (y) dy = 1$, observe that: $$f(x) = \int_{\mathbb{R}^d}f(x) K_\delta (y) dy$$ So it now follows directly that: $$||(f*K_\delta) - f||_{L^1} = \int_{\mathbb{R}^d} \Bigg| \int_{\mathbb{R}^d} f(x - y)K_{\delta}(y)dy - f(x) \Bigg| \hspace{0.1cm}dx = \ldots$$ $$\ldots = \int_{\mathbb{R}^d} \Bigg| \int_{\mathbb{R}^d} \big(f(x - y) - f(x)\big)K_{\delta}(y)dy \Bigg| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy \hspace{0.1cm}dx$$ Now, by Fubini's Theorem and the triangle inequality that $\forall r > 0$: $$\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy \hspace{0.1cm}dx = \ldots$$ $$\int_{\mathbb{R}^d} \Bigg( \int_{B_r(0)} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy + \ldots$$ $$\ldots + \int_{B_r^c(0)} \big|f(x - y) - f(x)\big||K_{\delta}(y)|dy \hspace{0.1cm}\Bigg)dx \leq \ldots$$ $$\ldots \leq ||f(x-y) - f(x)||_{L^1} \int_{B_r(0)}|K_{\delta}(y)|dy + \ldots$$ $$\ldots + 2||f||_{L^1}\int_{B_r^c(0)}|K_{\delta}(y)|dy$$ Now, since $f$ is $L^1$, and from property $(ii)$ of good kernals (P.109), $\exists A > 0$ such that $||K_\delta||_{L^1} < A \hspace{0.1cm} \forall \delta$, we know $\forall \epsilon > 0$ there exists an $r > 0$ small enough such that: $$y \in B_r(0) \hspace{0.25cm} \Rightarrow \hspace{0.25cm} ||f(x-y) - f(x)|| < \frac{\epsilon}{2A}$$ And, from property $(iii)$ of good kernals (P.109), we have that $\forall \epsilon > 0 \hspace{0.25cm} \exists \delta > 0$ small enough such that: $$\int_{B_r^c(0)}|K_{\delta}(y)|dy < \frac{\epsilon}{4||f||_{L^1}}$$ Putting everything together, we finally see that if we choose both $\delta, r > 0$ small enough, that: $$||f(x-y) - f(x)||_{L^1} \int_{B_r(0)}|K_{\delta}(y)|dy + \ldots$$ $$\ldots + 2||f||_{L^1}\int_{B_r^c(0)}|K_{\delta}(y)|dy < \ldots$$ $$\ldots < \frac{\epsilon}{2A}\int_{B_r(0)}|K_{\delta}(y)|dy + 2||f||_{L^1} \frac{\epsilon}{4||f||_{L^1}}<\ldots$$ $$\ldots < \frac{\epsilon}{2A} A + \frac{\epsilon}{2} = \epsilon$$ ...as desired.

2.24

Part a.) Given the equation: $$(f*g)(x) = \int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm}dy$$ If we assume $f$ is integrable, and $\exists M \geq 0$ such that $|g| \leq M \hspace{0.1cm} \forall x \in \mathbb{R}^d$, we have: $$\big|(f*g)(x) - (f*g)(z)\big| = \ldots$$ $$\ldots = \Bigg|\int_{\mathbb{R}^d} \bigg(f(x-y) - f(z - y)\bigg)g(y) \hspace{0.1cm}dy\Bigg| \leq \ldots$$ $$\ldots \leq M\int_{\mathbb{R}^d} \bigg|f(x-y) - f(z - y)\bigg| \hspace{0.1cm}dy =\ldots$$ $$\ldots = M\int_{\mathbb{R}^d} \bigg|f(-y) - f((z - x) - y)\bigg| \hspace{0.1cm}dy$$ Thus since $f$ is $L^1(\mathbb{R}^d)$, by Proposition 2.5, $\forall \epsilon > 0,\hspace{0.25cm} \exists \delta$ such that $||z - x|| < \delta \hspace{0.25cm} \Rightarrow ||f(y) - f(y-(z-x))||_{L^1} < \epsilon$. Thus, since $$\big|(f*g)(x) - (f*g)(z)\big| \leq ||f(y) - f(y-(z-x))||_{L^1}$$ The convolution $(f*g)(x)$ must be uniformly continuous.

Part b.) If $f$ and $g$ are both $L^1(\mathbb{R}^d)$, we proved in exercise 21 part d that $(f*g)(x)$ is also $L^1(\mathbb{R}^d)$. Thus, since $(f*g)(x)$ is uniformly continuous, and integrable, we have (by exercise 6 part b) that: $$\lim_{|x| \to \infty} (f*g)(x) = 0$$ ...as desired.

2.23

Assume to the contrary that there does exist an $I \in L^1(\mathbb{R}^d)$ such that: $$(f*I) = f \hspace{0.25cm} \forall f \in L^1(\mathbb{R}^d)$$ It follows from the latter parts of exercise 21 that: $$\hat{f}(\xi) = \widehat{(f*I)}(\xi) = \hat{f}(\xi)\hat{I}(\xi)$$ Thus, since $\hat{f}(\xi)$ need not be zero, we have that $\hat{I}(\xi) = 1 \hspace{0.25cm} \forall \xi$. I.e. $\lim_{\xi \to \infty} \hat{I}(\xi) = 1$. This contradicts the Riemann-Lebesgue Lemma. Therefore, $I \notin L^1(\mathbb{R}^d)$.

2.22

This exercise is asking us to prove the Riemann-Lebesgue Lemma. Exactly as the hint prescribes, first observe that $\xi \cdot \xi' = \frac{1}{2}$, and then by the translation invariance of the Lebesgue integral: $$\hat{f}(\xi) = \int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i (x - \xi') \cdot \xi} \hspace{0.1cm}dx = \ldots$$ $$= \int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i x \cdot \xi} e^{-2\pi i \xi \cdot \xi'} \hspace{0.1cm}dx = -\int_{\mathbb{R}^d} f(x - \xi')e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx$$ So we can certainly rewrite $\hat{f}(\xi)$ as: $$\hat{f}(\xi) = \frac{1}{2} \int_{\mathbb{R}^d}\big(f(x)- f(x - \xi')\big)e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx$$ Now, observe that: $$\lim_{|\xi| \to \infty} |\hat{f}(\xi)| = \lim_{|\xi| \to \infty} \Bigg|\frac{1}{2} \int_{\mathbb{R}^d}\big(f(x)- f(x - \xi')\big)e^{-2\pi i x \cdot \xi}\hspace{0.1cm}dx\Bigg| = \dagger$$ ...and thus, by the triangle inequality, and since $\xi' \to 0$ if $|\xi| \to \infty$, it's clear that: $$\dagger \leq \lim_{\xi' \to 0} \frac{1}{2} \int_{\mathbb{R}^d}\big|f(x)- f(x - \xi')\big| \hspace{0.1cm}dx = 0$$ ...from Proposition 2.5 (p. 74).

2.21

Part a.) Since the product of two measurable functions is measurable, it suffices to show that $f(x-y)$ and $g(y)\chi_{\mathbb{R}^d(x)}$ are each measurable in $\mathbb{R}^{2d}$.

Conveniently, since $f$ is measurable on $\mathbb{R}^d$, it follows directly from Proposition 3.9 (p. 86) that $f(x-y)$ is measurable on $\mathbb{R}^{2d}$. Also, since $g$ is measurable on $\mathbb{R}^d$, it follows directly from Corollary 3.7 (P. 85) that $g(y)\chi_{\mathbb{R}^d(x)}$ is measurable on $\mathbb{R}^{2d}$.

Part b.) Since we know $f(x-y)g(y)$ is measurable, by Tonelli's Theorem we have: $$\int_{\mathbb{R}^{2d}} |f(x-y)g(y)| \hspace{0.1cm}d(x,y) \hspace{0.25cm}=\hspace{0.25cm} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y)g(y)| \hspace{0.1cm}dx\hspace{0.1cm}dy$$ ...and from the translation invariance of integration we get: $$\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y)g(y)| \hspace{0.1cm}dx\hspace{0.1cm}dy = \int_{\mathbb{R}^d} |g(y)| \int_{\mathbb{R}^d} |f(x-y)| \hspace{0.1cm}dx\hspace{0.1cm}dy = \ldots$$ $$\ldots = ||f||_{L^1(\mathbb{R}^d)} \int_{\mathbb{R}^d} |g(y)| dy = ||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)} < \infty$$ ...since both $f$ and $g$ are $L^1$.

Part c.) Since $f(x-y)g(y)$ was just shown to be integrable, it follows directly from Fubini's Theorem that for almost every $x \in \mathbb{R}^d$, : $$\int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy < \infty$$ I.e., the convolution: $$(f*g)(x) = \int_{\mathbb{R}^d} f(x-y)g(y)\hspace{0.1cm}dy$$ ...is well-defined for a.e. $x \in \mathbb{R}^d$.

Part d.) Observe that: $$\int_{\mathbb{R}^d} |(f*g)(x)\hspace{0.1cm}| dx = \int_{\mathbb{R}^d} \Bigg| \int_{\mathbb{R}^d}f(x-y)g(y)\hspace{0.1cm}dy \hspace{0.1cm}\Bigg| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{\mathbb{R}^d} \int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy \hspace{0.1cm}dx$$ ...which, by part b, we see: $$||(f*g)||_{L^1(\mathbb{R}^d)} = \int_{\mathbb{R}^d} |(f*g)(x)|\hspace{0.1cm} dx \leq \int_{\mathbb{R}^d}|f(x-y)g(y)|\hspace{0.1cm}dy \hspace{0.1cm}dx = \ldots$$ $$\ldots = ||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)}$$ Now, if $f$ and $g$ are positive functions, $|f(x-y)g(y)|=f(x-y)g(y)$, so equality of $||(f*g)||_{L^1(\mathbb{R}^d)}$ and $||f||_{L^1(\mathbb{R}^d)}||g||_{L^1(\mathbb{R}^d)}$ follows (again) directly from part b.

Part e.) Let's first check that $\hat{f}(\xi)$ is bounded. Recall that $|e^{i \theta}| = 1 \hspace{0.25cm} \forall \theta \in \mathbb{R}$. Then, observe: $$|\hat{f}(\xi)| = \Bigg| \int_{\mathbb{R}^d} f(x) e^{-2\pi i x \xi} \hspace{0.1cm} dx \Bigg| \leq \ldots$$ $$\ldots \leq \int_{\mathbb{R}^d} |f(x)||e^{-2\pi i x \xi}| \hspace{0.1cm}dx = \int_{\mathbb{R}^d} |f(x)|\hspace{0.1cm}dx = ||f||_{L^1(\mathbb{R}^d)}$$ Thus, $\hat{f}(\xi)$ is bounded.

Now, let's see if $\hat{f}(\xi)$ is continuous. We begin by observing: $$|\hat{f}(\xi) - \hat{f}(\mu)| = \Bigg| \int_{\mathbb{R}^d} f(x) \big(e^{-2\pi i x \cdot \xi} - e^{-2\pi i x \cdot \mu}\big) \hspace{0.1cm} dx \Bigg| \leq \ldots$$ $$\ldots \leq \int_{\mathbb{R}^d}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx$$ Note that since $f$ is $L^1(\mathbb{R}^d$, for any $\epsilon > 0$ we have that there exists an $R > 0$ such that: $$\int_{B_R^c} |f(x)| \hspace{0.1cm} dx \leq \frac{\epsilon}{4}$$ (Where $B_R$ is a ball of radius $R$ centered the origin.)

Now, since $\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \leq 2$, we see: $$\int_{\mathbb{R}^d}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx + 2\int_{B_R^c}|f(x)| \hspace{0.1cm}dx$$ From here, require: $$||\xi - \mu|| < \delta = \frac{\epsilon}{8 \pi R ||f||_{L^1(\mathbb{R}^d)}}$$ Now, it should be clear from the Cauchy Schwartz inequality that on $B_R$: $$|x \cdot (\xi - \mu)| \leq R \delta = \frac{\epsilon}{8 \pi ||f||_{L^1(\mathbb{R}^d)}}$$ Therefore, plugging it all in, we finally see: $$\int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx \leq \ldots$$ $$\leq \int_{B_R}|f(x)| \Big[\big| \cos(2\pi x \cdot (\xi - \mu)) - 1 \big| + \big|\sin(2\pi x \cdot (\xi - \mu))\big|\Big] \hspace{0.1cm}dx$$ $$\leq \int_{B_R}|f(x)| \Big[\big| \cos\big(\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}}\big) - 1 \big| + \big|\sin\big(\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}}\big) \big| \Big]$$ $$\leq \int_{B_R}|f(x)| \Big[\frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}} + \frac{\epsilon}{4 ||f||_{L^1(\mathbb{R}^d)}} \Big] \hspace{0.1cm}dx \hspace{0.25cm} \leq \ldots$$ $$\ldots \leq \frac{\epsilon ||f||_{L^1(\mathbb{R}^d)}}{2 ||f||_{L^1(\mathbb{R}^d)}} = \frac{\epsilon}{2}$$ Thus, we've just shown, for a sufficiently large $R > 0$: $$||\xi - \mu|| < \delta = \frac{\epsilon}{8 \pi R ||f||_{L^1(\mathbb{R}^d)}} \hspace{0.25cm} \Rightarrow \hspace{0.25cm} |\hat{f}(\xi) - \hat{f}(\mu)| \leq \ldots$$ $$\ldots \leq \int_{B_R}|f(x)|\big|e^{-2\pi i x \cdot (\xi - \mu)} - 1 \big| \hspace{0.1cm}dx + 2\int_{B_R^c}|f(x)| \hspace{0.1cm}dx \hspace{0.25cm} \leq$$ $$\ldots \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ ...as desired.

Finally we want to show: $$\widehat{(f*g)}(\xi) = \hat{f}(\xi)\hat{g}(\xi)$$ Proceed by directly applying Fubini's Theorem: $$\widehat{(f*g)}(\xi) = \int_{\mathbb{R}^d} \Bigg[\int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm}dy \Bigg] e^{-2\pi i \xi x} \hspace{0.1cm} dx = \ldots$$ $$\ldots = \int_{\mathbb{R}^d} \Bigg[\int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm} e^{-2\pi i \xi (x - y + y)} dy \Bigg] \hspace{0.1cm} dx$$ $$\ldots = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \Big(f(x-y) e^{-2\pi i \xi (x-y)} \Big) \Big(g(y) e^{-2\pi i \xi y}\Big) \hspace{0.1cm}dy \hspace{0.1cm} dx$$ $$\ldots = \int_{\mathbb{R}^d} \Big(g(y) e^{-2\pi i \xi y}\Big) \int_{\mathbb{R}^d} \Big(f(x-y) e^{-2\pi i \xi (x-y)} \Big) \hspace{0.1cm}dx \hspace{0.1cm} dy$$ $$\ldots = \hat{f}(\xi)\int_{\mathbb{R}^d} g(y) e^{-2\pi i \xi y} \hspace{0.1cm} dy = \hat{f}(\xi)\hat{g}(\xi)$$ ...as desired.

2.20

As the hint prescribes, define: $$\mathcal{C} = \lbrace E \subset \mathbb{R}^d \hspace{0.25cm} | \hspace{0.25cm} E^y \in \mathcal{B}(\mathbb{R}) \rbrace$$
First, note that $\mathcal{C}$ is trivially non-empty. Next, does $E \in \mathcal{C} \hspace{0.25cm} \Rightarrow E^c \in \mathcal{C}$?

Yes. Notice $(E^c)^y = (E^y)^c$ (with respect to the slice, not $\mathbb{R}^2$). Thus, since $E^y$ is Borel, $E^c \in \mathcal{C}$.

Is $\mathcal{C}$ closed under countable unions and intersections? Yes, using identical reasoning from above. Thus, $\mathcal{C}$ is a $\sigma$-algebra.

Does $\mathcal{C}$ contain the open sets?

Indeed let $\mathcal{O}$ be an open set in $\mathbb{R}^d$. Notice that for any $x$ lying on any slice $\mathcal{O}^y$ $\exists r > 0$ such that $B_r(x) \subset \mathcal{O}$ Certainly, $B_r(x)^y \subset \mathcal{O}^y \hspace{0.25cm} \Rightarrow \mathcal{O}^y$ is open, and therefore Borel. Thus, any open set in $\mathbb{R}^2$ is in $\mathcal{C}$.

Now, since $\mathcal{B}(\mathbb{R}^2)$ is the smallest $\sigma$-algebra that contains the open sets in $\mathbb{R}^2$, we have that $\mathcal{B}(\mathbb{R}^2) \subset \mathcal{C}$. Thus, every slice of a Borel set $E$ must be Borel, as well.

2.19

First, noting the definition of our set $E_\alpha$: $$E_\alpha = \big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \big| \hspace{0.25cm} |f(x)| > \alpha \big\rbrace$$ Observe that $E_\alpha$ is certainly measurable. Now, consider the integration: $$\int_0^\infty m(E_\alpha) \hspace{0.1cm} d\alpha = \int_0^\infty \int_{E_\alpha} dx \hspace{0.1cm} d\alpha$$ Since constant functions over measurable sets are measurable, we can apply Tonelli's Theorem to see: $$\int_0^\infty \int_{E_\alpha} dx \hspace{0.1cm} d\alpha = \int_{E_\alpha} \int_0^\infty d\alpha \hspace{0.1cm} dx = \int_{\mathbb{R}^d} \int_0^\infty \chi_{[0,|f(x)|)} d\alpha \hspace{0.1cm} dx$$ Next, notice from the definition of $E_\alpha$ that: $$\int_{\mathbb{R}^d} \int_0^\infty \chi_{[0,|f(x)|)} d\alpha \hspace{0.1cm} dx = \int_{\mathbb{R}^d} m\big([0,|f(x)|)\big) \hspace{0.1cm} dx = \int_{\mathbb{R}^d} |f(x)| dx$$ ...as desired.

2.18

Let $Q = [0,1]$. Since $|f(x) - f(y)| \in L^1(Q \times Q)$, and $|f(x)| < \infty \hspace{0.25cm} \forall x \in Q$, it follows directly from Fubini's Theorem that for almost every $y \in Q$: $$\int_Q |f(x)-f(y)| \hspace{0.1cm}dx = \int_Q |f(x)- C| \hspace{0.1cm}dx < \infty$$ Where $C$ is a fixed real number in the range of $f$. Thus, it's easy to see: $$\int_Q |f(x)| \hspace{0.1cm} dx \leq \int_Q |f(x) - C| + |C| \hspace{0.1cm} dx = \ldots$$ $$\ldots = |C| + \int_Q |f(x) - C| \hspace{0.1cm} dx < \infty$$ Thus, $f \in L^1(Q)$, as desired.

2.17

Part a.) By construction, it's straight-forward to see that since for any $x \geq 0 \hspace{0.25cm} \exists n \in \mathbb{N}$ such that $x \in [n,n+1)$, that $f_x(y) = a_n - a_n = 0$ for all $x \geq 0$. Thus, certainly: $$\forall x, \int_{\mathbb{R}} f_x(y) dy = 0$$ ...and therefore: $$\int_\mathbb{R} \int_\mathbb{R} f_x(y) dy \hspace{0.1cm} dx = 0$$ Part b.) By construction again, you'll observe that since every $y \geq 0$ is between an $n \in \mathbb{Z}$ and $n+1$, that: $$\int_{\mathbb{R}} f^y(x) \chi_{[n,n+1)}(y) dx = a_n - a_{n-1}$$ (Aside from the $n=0$ case, where the integral simply evaluates to $a_0$.) Thus, since: $$a_n = \sum_{k \leq n} b_k$$ We can see: $$a_n - a_{n-1} = b_n$$ ...and finally, since these were constructed to be piecewise-constant functions: $$\int_{\mathbb{R}} \int_{\mathbb{R}} f^y(x)dx \hspace{0.1cm} dy = \sum_{n=0}^\infty \int_{[n,n+1)} \Big(\int_{\mathbb{R}} f^y(x)dx\Big) dy \hspace{0.1cm} = \ldots$$ $$\ldots = a_0 + \sum_{n=1}^\infty (a_n - a_{n-1}) \hspace{0.25cm} = \hspace{0.25cm} \sum_{n=0}^\infty b_n = s < \infty$$
Part c.) Lastly, since $b_n > 0 \hspace{0.25cm} \forall n$, observe that $a_n > a_0 = b_0 > 0 \hspace{0.25cm} \forall n$. Thus, $$\int_{\mathbb{R}} \int_{\mathbb{R}} |f(x,y)| dx \hspace{0.1cm} dy \hspace{0.25cm} > \hspace{0.25cm} a_0 + 2\sum_{n=1}^\infty a_0 \hspace{0.25cm} = \hspace{0.25cm} \infty$$

2.16

First, recall from the invariance properties of $L^1(\mathbb{R})$ functions that for any $f$ that's non-negative and $L^1(\mathbb{R})$: $$\int_\mathbb{R} f(-x)dx = \int_\mathbb{R} f(x)dx$$ In addition, for any $\delta > 0$, $$\int_\mathbb{R} f(\delta x)dx = \frac{1}{\delta} \int_\mathbb{R} f(x)dx$$ Now, let's combine these conditions. Assume $f$ is still non-negative and $L^1(\mathbb{R})$, and $\delta > 0$, then observe: $$\int_\mathbb{R} f(-\delta x)dx = \int_\mathbb{R} f(\delta x)dx = \frac{1}{\delta} \int_\mathbb{R} f(x)dx$$ Thus, it should be clear that for any non-negative $f \in L^1(\mathbb{R})$, and $\delta \neq 0$: $$\int_\mathbb{R} f(\delta x) dx = \frac{1}{|\delta|} \int_\mathbb{R} f(x) dx$$ ...which can clearly be extended to any $f \in L^1(\mathbb{R})$ by linearity. ($\dagger$)

Now, suppose $f$ is integrable on $\mathbb{R}^d$, and $\delta = (\delta_1, \ldots, \delta_d)$ is a $d-$tuple of non-zero real numbers such that: $$f^\delta (x) = f(\delta x) = f(\delta_1 x_1, \delta_2 x_2, \ldots, \delta_d x_d)$$ We need to show $f^\delta (x)$ is integrable such that: $$\int_{\mathbb{R}^d} f^\delta (x) dx = |\delta_1|^{-1} \cdot \cdot \cdot |\delta_d|^{-1} \int_{\mathbb{R}^d} f(x) dx$$ Proceed by first noting that since $f \in L^1(\mathbb{R}^d)$, it follows directly from Fubini's Theorem and $\dagger$ that given $\delta = (\delta_1, 1, \ldots, 1)$ that: $$|\delta_1|^{-1} \int_{\mathbb{R}^d} f(x) \hspace{0.1cm} dx \hspace{.25cm} =\hspace{.25cm} |\delta_1|^{-1} \int_\mathbb{{R}^{d-1}} \int_{\mathbb{R}} f(x_1,y) \hspace{0.1cm} dx \hspace{0.1cm} dy = \ldots$$ $$\ldots = \int_{\mathbb{R}^{d-1}} \int_{\mathbb{R}} f(\delta_1 x_1, y) \hspace{0.1cm} dx \hspace{0.1cm} dy \hspace{.25cm}=\hspace{.25cm} \int_{\mathbb{R}^d} f(\delta x) dx$$ ...and since we can simply continue in this fashion for any $n \leq d$ (by a simple induction argument) we observe for $f \in L^1(\mathbb{R}^d)$, and any fixed, nowhere-zero, $d$-tuple $\delta$: $$\int_{\mathbb{R}^d} |f^\delta (x)|dx = \int_{\mathbb{R}^d} |f(\delta_1 x_1, \ldots, \delta_d x_d)| dx = \ldots$$ $$\ldots = |\delta_1|^{-1} \cdot \cdot \cdot |\delta_d|^{-1} \int_{\mathbb{R}^d} |f(x)| dx < \infty$$ We see that $f^\delta$ must be $L^1(\mathbb{R}^d)$.

2.15

It is clear by direct calculation (or from Problem 10, Part II) that our function:

$f(x) = \Bigg\lbrace \begin{array}{cc} x^{-1 / 2} & x \in (0,1) & \\ 0 & \text{otherwise} \\ \end{array}$

...is $L^1(\mathbb{R})$.

The goal of the exercise is to prove that for some fixed enumeration of the rationals $\lbrace q_n \rbrace_{n=1}^\infty$, $$F(x) = \sum_{k=1}^\infty 2^{-k} f(x - q_k)$$ ...is also $L^1(\mathbb{R})$. Proceed by simply defining the sequence of functions: $$F_n(x) = \sum_{k=1}^n 2^{-k} f(x - q_k)$$ ...and first observing that $F_n(x)$ is $L^1(\mathbb{R}) \hspace{0.25cm} \forall n$: $$\int_\mathbb{R} F_n(x) dx = \int_\mathbb{R} \sum_{k=1}^n 2^{-k} f(x - q_k) dx = \dots$$ $$\ldots = \sum_{k=1}^n 2^{-k} \int_\mathbb{R} f(x - q_k)dx = C \sum_{k=1}^n 2^{-k}$$ Where $C = \int_\mathbb{R} f(x)dx$ (by translation invariance). Thus, since $\lbrace F_n \rbrace_{n=1}^\infty$ is a collection of non-negative measurable functions such that $F_n \leq F_{n+1} \hspace{0.25cm} \forall n$, and $F_n \nearrow F$ monotonically, we have by the Monotone Convergence Theorem, $$\int_\mathbb{R} |F(x)| dx = \int_\mathbb{R} F(x)dx = \lim_{n \to \infty} \int_\mathbb{R} F_n(x)dx = \ldots$$ $$\ldots = C \sum_{k=1}^\infty 2^{-k} = C < \infty$$ Thus, $F \in L^1(\mathbb{R})$. Next, observe that even if some function $G = F$ almost everywhere, on any fixed interval $I \subset \mathbb{R}, G|_I = F|_I$ almost everywhere. Thus, the sets: $$\mathcal{G}_I^n = \lbrace x \in G|_I \hspace{0.25cm} \big| \hspace{0.25cm} G(x) > n \rbrace \hspace{0.25cm} \text{and} \hspace{0.25cm} \mathcal{F}_I^n = \lbrace x \in F|_I \hspace{0.25cm} \big| \hspace{0.25cm} F(x) > n \rbrace$$ Have the same (positive $\forall n$) measure. Thus, $G$ must be unbounded on any interval.

2.12

First, define the sets: $$S_n^k = \Big[\frac{k-1}{2^n}, \frac{k}{2^n}\Big]$$ From these intervals, define the sequence of functions: $$f_n^k = \chi_{S_n^k}(x) + \chi_{S_n^k}(-x)$$ Where for each $n$ to progress $+1$, $k$ must sweep from $1$ to $2^n + 1$.

Certainly, for every $n$, $$\int_{\mathbb{R}} f_n^k = \frac{1}{2^{n-1}}$$ Since the collection of functions $\lbrace f_n^k \rbrace$ is countable, with a well-defined sequence, let us just enumerate them with the single index $m$. We have: $$\int_{\mathbb{R}} |f_m - 0| = \int_{\mathbb{R}} f_m \to 0$$ However, for any $x \in \mathbb{R}$, it's clear that $\lim_{m\to \infty} f_m(x)$ does not exist. To expand this to $\mathbb{R}^d$, just replace the intervals with a closed balls centered at the origin subtracting open balls with the same dyadic rational-difference... (onion-layers).

2.11

Consider first the set: $$A = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < 0 \rbrace$$ It's clear that: $$A = \bigcup_{n=1}^\infty \lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < - \frac{1}{n} \rbrace$$ Assume to the contrary that $m(A) > 0$. We have: $$m(A) \leq \sum_{n=1}^\infty m(\lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < - \frac{1}{n} \rbrace)$$ Since $m(A) > 0$, for at least one $n$, $m(\lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < - \frac{1}{n} \rbrace)$ > 0. Call this set $E$. We have: $$\int_{E} f \leq \int_E -\frac{1}{n} = -\frac{1}{n} m(E) < 0$$ ...a contradiction. It's obvious that if you switch that conditions around such that for any measurable set $S \subset \mathbb{R}^d$, $$\int_S f \leq 0$$ that the same line of reasoning will result in $f \leq 0$ almost everywhere. Combining these conditions two conditions, i.e., for any measurable set $S \subset \mathbb{R}^d$, $$\int_S f = 0$$ will directly result in $0 \leq f \leq 0$ almost everywhere. I.e. $f = 0$ almost everywhere.

2.10

Given $f$ measurable on $\mathbb{R}^d$, and the sets: $$E_{2^k} = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) > 2^k \rbrace$$ $$F_k = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} 2^k < f(x) \leq 2^{k+1} \rbrace$$ The exercise is asking us to prove:

$$f \in L^1 \iff \sum_{k=-\infty}^\infty 2^k m(F_k) < \infty \iff \sum_{k=-\infty}^\infty 2^k m(E_{2^k}) < \infty$$ Denote these conditions as $a, b,$ and $c$ respectively. Proof of this statement needs to be broken up into four parts:

Part 1: $a \Rightarrow b$. Simply observe that: $$\sum_{k=-\infty}^\infty 2^k m(F_k) \hspace{0.25cm} =\hspace{0.25cm} \sum_{k=-\infty}^\infty \int_{F_k} 2^k dx \hspace{0.25cm} \leq \ldots$$ $$\ldots \leq \hspace{0.25cm} \sum_{k=-\infty}^\infty \int_{F_k} f(x) dx \hspace{0.25cm} = \hspace{0.25cm} \int_{\mathbb{R}^d} f(x)dx < \infty$$ Part 2: $b \Rightarrow a$. Similarly: $$\frac{1}{2} \int_{\mathbb{R}^d} f(x)dx \hspace{0.25cm} \leq \hspace{0.25cm} \frac{1}{2} \sum_{k=-\infty}^\infty \int_{F_k} f(x) dx \leq \ldots$$ $$\ldots \leq \hspace{0.25cm} \frac{1}{2} \sum_{k=-\infty}^\infty \int_{F_k} 2^{k+1} dx = \sum_{k=-\infty}^\infty 2^{k} m(F_k) < \infty$$ Part 3: $c \Rightarrow b$. Certainly, $F_k \subset E_{2^k}$. Thus: $$\sum_{k=-\infty}^\infty 2^{k} m(F_k) \leq \sum_{k=-\infty}^\infty 2^{k} m(E_{2^k}) < \infty$$ Part 4: $b \Rightarrow c$. First observe that $E_{2^k} = \bigcup_{n=k}^\infty F_k$. Indeed, $$\sum_{k=-\infty}^\infty 2^{k} m(E_{2^k}) \leq \sum_{k=-\infty}^\infty \sum_{n=k}^\infty 2^{k} m(F_{n}) = \ldots$$ ...(by Fubini's Theorem) $$\ldots = \sum_{n=-\infty}^\infty \sum_{k=-\infty}^n 2^{k} m(F_{n}) = \sum_{n=-\infty}^\infty \Bigg(2^n m(F_n) \sum_{k=-\infty}^n 2^{k-n} \Bigg) = \ldots$$ $$\ldots = \sum_{n=-\infty}^\infty 2^{n+1} m(F_n) < \infty$$ Thus, $a \iff b \iff c$, as desired.


The second half of the exercise wants us to investigate the following functions:

$f(x) = \Bigg\lbrace \begin{array}{cc} |x|^{-a} & |x| \leq 1 & \\ 0 & \text{otherwise} \\ \end{array}$

and

$g(x) = \Bigg\lbrace \begin{array}{cc} |x|^{-b} & |x| > 1 & \\ 0 & \text{otherwise} \\ \end{array}$

Where $B$ is the closed unit ball centered at the origin.

For the first function, notice that after solving for $|x|$, the sets $E_{2^k}$, where $k \geq 0$ are: $$E_{2^k} = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} |x| < 2^{-\frac{k}{a}} \rbrace$$ I.e. for positive $k$, $m(E_{2^k}) = v_d (2^{-\frac{k}{a}})^d$. Notice also that for $k < 0$, $m(E_{2^k}) = v_d$. Thus, $$\sum_{k=-\infty}^\infty 2^k m(E_{2^k}) = \sum_{k=-\infty}^0 v_d 2^k + \sum_{k=1}^\infty 2^k v_d (2^{-\frac{k}{a}})^d = \ldots$$ $$\ldots = 2v_d + \sum_{k=1}^\infty v_d 2^{k(1-\frac{d}{a})}$$ ...which will only converge if $d > a$. Thus, if $d > a$, $f(x)$ is $L^1$.

For the second function, for $-\infty < k < 0$: $$E_{2^k} = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} |x| < 2^{-\frac{k}{b}} \rbrace$$ Similar to before, notice that for $k \geq 0, m(E_{2^k}) = 0.$ Thus,: $$\sum_{k=-\infty}^\infty 2^k m(E_{2^k}) = \sum_{k=-\infty}^{-1} 2^k v_d (2^{-\frac{k}{b}})^d = \sum_{k=-\infty}^{-1} v_d 2^{k(1 -\frac{d}{b})}$$ ...which will clearly only converge if $d < b$. Thus, $g(x)$ is $L^1$ if $d < b$.

2.9

This exercise is asking us to prove Tchebychev's Inequality, stated as:

Suppose $f \geq 0$, and $f$ integrable. If $\alpha > 0$ and the set $E_\alpha$ is defined as: $$E_\alpha = \lbrace x | f(x) > \alpha \rbrace$$ Then: $$m(E_\alpha) \leq \frac{1}{\alpha} \int f$$

Simply re-write the definition of our sets $E_\alpha$: $$E_\alpha = \lbrace x | \frac{f(x)}{\alpha} > 1 \rbrace$$ Next, observe that: $$m(E_\alpha) = \int_{E_\alpha} dx \hspace{0.25cm} \leq \int_{E_\alpha} \frac{f(x)}{\alpha} dx \hspace{0.25cm} \leq \frac{1}{\alpha} \int f(x) dx$$ ...as desired.

2.8

Recall from Proposition 1.12 (ii) on P.65:

Suppose $f$ is integrable on $\mathbb{R}^d$. Then $\forall \epsilon > 0$ there is a $\delta$ > 0 such that: $$m(E) < \delta \hspace{0.25cm} \Rightarrow \int_E |f| < \epsilon$$

Now, simply make the observation that: $$|F(x) - F(y)| = \Big| \int_{-\infty}^x f(t) dt - \int_{-\infty}^y f(t) dt \Big| \leq \ldots$$ $$\ldots \leq \Big| \int_x^y f(t) dt\Big| \leq \int_x^y |f(t)|dt$$ Since we already have that $f$ is integrable, certainly, $\forall \epsilon > 0 \hspace{0.25cm} \exists \delta > 0$ such that: $$|x-y| < \delta \Rightarrow \int_x^y |f(t)|dt < \epsilon$$ Thus, we have $F(x)$ must be uniformly continuous.

2.7

First, assume $f$ is almost-everywhere finite. Next, partition $\mathbb{R}^d$ into almost-disjoint closed unit cubes $\lbrace Q_k \rbrace_{k=1}^\infty$. Similarly to how the exercise defines $\Gamma$, define: $$\Gamma_k = \lbrace (x,y) \in Q_k \times \mathbb{R} \hspace{0.25cm} \big| \hspace{0.25cm} y = f(x) \rbrace$$ Next, define the sets: $$F_{k,n}^i = \lbrace x \in Q_k \hspace{0.25cm} \big| \hspace{0.25cm} \frac{i}{2^n} \leq f(x) < \frac{i+1}{2^n} \rbrace$$ Then, define: $$E_{k,n}^i = F_{k,n}^i \times \Big[\frac{i}{2^n}, \frac{i+1}{2^n}\Big)$$ And, finally: $$E_{k,n} = \bigcup_{i = -\infty}^{\infty} E_{k,n}^i$$ All of the above sets clearly inherit their measurability from $f$. Notice the following: $$\Gamma_k \subset E_{k,n} \hspace{0.25cm} \forall n \in \mathbb{N}$$ and ($\dagger$), $$E_{k,n+1} \subset E_{k,n} \hspace{0.25cm} \forall n \in \mathbb{N}$$ Now, observe that $$m^{d+1}(E_{k,n})\hspace{0.25cm} \leq \hspace{0.25cm}\sum_{i=-\infty}^\infty m^{d+1}(E_{k,n}^i) \leq \ldots$$ $$\ldots \leq \sum_{i=-\infty}^\infty m^{d}(F_{k,n}^i) \cdot m\Big(\big[\frac{i}{2^n}, \frac{i+1}{2^n}\big)\Big) = \ldots$$ $$\ldots = \frac{1}{2^{n-1}}\sum_{i=-\infty}^\infty m^{d}(F_{k,n}^i) \hspace{0.25cm} \leq \hspace{0.25cm} \frac{1}{2^{n-1}} \cdot m^d(Q_k) = \frac{1}{2^{n-1}}$$ Observe, by $\dagger$, that since the sets $E_{k,n}$ are collapsing, with finite measure, that $\Gamma_k$ must be measurable since: $$m^*(\Gamma_k) \leq \lim_{n \to \infty} m(E_{k,n}) = \lim_{n \to \infty} \frac{1}{2^{n-1}} = 0$$ Thus,each $\Gamma_k$ is measurable, which implies $\Gamma$ is measurable, since $\mathcal{M}$ is closed under countable unions. Lastly, we just need to observe that $$m^{d+1}(\Gamma) \leq \sum_{k = 1}^{\infty} m^{d+1}(\Gamma_k) = 0$$ ...which is what we set out to show.

2.6

Part a.) Define:

$g_n(x) = \Bigg\lbrace \begin{array}{cc} (n^2 2^n x + n) & x \in [-\frac{1}{n2^n},0) & \\ (-n^2 2^n x + n) & x \in [0,\frac{1}{n2^n}] \\ \end{array}$

Certainly, $\int_{\mathbb{R}} g_n = \frac{1}{2^n}$, for every $n$. Now, just define: $$f_m(x) = \sum_{n=1}^m g_n(x - n)$$ Now we can define $f(x) = \lim_{m \to \infty} f_m(x)$. Since each $f_m$ is positive, and $f_m \nearrow f$, monotonicly, we have --by the Monotone Convergence Theorem that $f$ must be measurable, and: $$\lim_{m \to \infty} \int_{\mathbb{R}} f_m(x) = \int_{\mathbb{R}} f$$ By construction, we can directly calculate $\int_{\mathbb{R}} f$ = $\sum_{n=1}^\infty \frac{1}{2^n} = 1$. Thus, $f$ is $L^1$. Lastly, observe that the $\limsup_{x \to \infty} f(x) = \infty$, since $f(n) = n$ for all $n \in \mathbb{N}$.

Part b.) Assume to the contrary that there exists a function $f$ such that $f$ is $L^1$, uniformly continuous on $\mathbb{R}$, and $f(x) \nrightarrow 0$ as $x \to \infty$.

($\dagger$) Certainly, $\limsup_{x \to \infty} |f(x)| = \alpha > 0$.

Since $f$ is uniformly continuous, we know $f$ must be finite on any compact set. Define the sequence $\lbrace x_n \rbrace_{n=1}^\infty$ such that $x_n \in [n, n+1]$ and $|f(x_n)| = \max_{x \in [n, n+1]}{(|f(x)|)}$.

By $\dagger$, we know $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ $\hspace{0.25cm} |f(x_n)| > \frac{\alpha}{2}$. Fix a particular $\epsilon$ such that $0 < \epsilon < \frac{\alpha}{4}$. Since $f$ is uniformly continuous, we know that for this particular $\epsilon$, $\exists \delta > 0$ such that $|x - y| < \delta \hspace{0.25cm} \Rightarrow |f(x) - f(y)| < \epsilon$.

Now, consider again our sequence $\lbrace x_n \rbrace_{n=1}^\infty$. For our particular epsilon, redefine the $\delta$ determined by $\epsilon$ such that $\delta = \min\lbrace\delta(\epsilon), \frac{1}{2} \rbrace$.

We have, by the reverse triangle inequality, that for every $n \geq N$: $$\big||f(x_n)| - |f(x_n + \delta)|\big| \leq |f(x_n) - f(x_n + \delta)| \leq \epsilon < \frac{\alpha}{4}$$ Define the sets: $$E_n = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} x_{2n} \leq x \leq x_{2n} + \delta \rbrace$$ I skipped every other $x_n$ to ensure $\lbrace E_n \rbrace$ will be pair-wise disjoint.

...and now we can finally observe that: $$\int_\mathbb{R} |f| \geq \int_{(\bigcup_{n=N}^\infty E_n)} \frac{\alpha}{4} = \sum_{n = N}^\infty \frac{\delta\alpha}{4} = \infty$$ ...contradicting the assumption that $f \in L^1$.

2.5

Part a.) By the definition of the infimum, we have $\forall \epsilon > 0 \hspace{0.25cm} \exists z \in F$ such that $|x - z| < \delta(x) + \epsilon$. Thus, $$\delta(y) \leq |y - z| \leq |y - x| + |x - z| \leq |y - x| + \delta(x) + \epsilon$$ Thus, since $\epsilon$ can be arbitrarily small, we have: $$\delta(y) - \delta(x) \leq |y-x|$$ Now, repeat this argument for $y$, and we have: $$-|y-x| \leq \delta(y) - \delta(x) \leq |y-x| \hspace{0.25cm} \Rightarrow |\delta(y) - \delta(x)| \leq |y-x|$$

Part b.) First observe that if $y \in F$, then $\delta(y) = 0$. Then, certainly: $$I(x) = \int_F \frac{\delta(y)}{|x-y|^2} dy + \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy = \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy$$ Next, notice that since $F^c$ is an open set, for any $x \in F^c$ we know $\exists r > 0$ such that $B_{r}(x) \subset F^c$, and for any $y \in B_{r}(x)$, $\exists M > 0$ such that $\delta(y) \geq M$. (We can always do this by simply choosing $r$ small enough to fit the ball in $F^c$, then using the ball of radius $\frac{r}{2}$.) Therefore, we have: $$\int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \geq M \int_{B_{r}(x)} \frac{1}{|x-y|^2} dy$$ Now, just observe that we can just center $x$ at the origin, and our calculation becomes: $$M\int_{B_{r}(x)} \frac{1}{|x-y|^2} dy = \lim_{b \to 0} \int_b^\infty \frac{1}{y^2}dy = \infty$$ ...as desired.

Part c.) As the hint prescribes, if we can show $I(x)$ is $L^1(F)$ then we have that $I$ must be almost everywhere finite. Indeed, first notice (from part b) that: $$\int_F I(x) dx = \int_{F} \int_{\mathbb{R}} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx = \int_{F} \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx$$ Before I can use Tonelli's Theorem to swap the limits of integration, I need to first check that the function: $$h(x,y) = \chi_{(F \times F^c)}\frac{\delta(y)}{|x-y|^2}$$ ...is measurable. Certainly, $\chi_{(F \times F^c)}$ is measurable since $F$ and $F^c$ are Borel. Next notice that the sets: $$\lbrace y \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} \delta(y) > \alpha \rbrace$$ are measurable, since they're just the union of $F$ and countably-many closed intervals. Lastly, the function $\frac{1}{|x-y|^2}$ is obviously measurable since it's continuous almost everywhere. Thus, since the product of measurable functions are measurable, we have $h(x,y)$ is measurable.

Thus, it follows from Tonelli's theorem that: $$\int_{F} \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx = \int_{F^c} \int_{F} \frac{\delta(y)}{|x-y|^2} dx \hspace{0.1cm} dy = \ldots$$ $$\ldots = \int_{F^c} \delta(y) \int_{F} \frac{1}{|x-y|^2} dx \hspace{0.1cm} dy$$ Now, this part I found to be kind of tricky (at first). Observe that: $$F \subset \lbrace x \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} |x - y| \geq \delta(y) \rbrace$$ Then we have: $$\int_{F} \frac{1}{|x-y|^2} dx \leq 2\int_{\delta(y)}^\infty \frac{1}{x^2} dx = \frac{2}{\delta(y)}$$ Finally, we have: $$\int_{F^c} \delta(y) \int_{F} \frac{1}{|x-y|^2} dx \hspace{0.1cm} dy \leq 2\int_{F^c} \delta(y) \int_{\delta(y)}^\infty \frac{1}{x^2} dx \hspace{0.1cm} dy \leq 2 m(F^c)$$ ...and since $m(F^c) < \infty$ we have $I$ must be $L^1 (F)$.

2.4

It suffices to consider $f$ to be a non-negative $L^1$ function. It's not at first apparent that $g(x)$ is an $L^1$ function. First, define the set: $$T(b) = \lbrace (x,t) \in \mathbb{R} \times \mathbb{R} \hspace{0.25cm} \big|\hspace{0.25cm} 0 < x \leq t, \hspace{0.25cm} x \leq t \leq b \rbrace$$ (It's just the triangle formed by taking the square $[0,b] \times [0,b]$, bisecting it with the line $x = t$, and taking the top half.) Next, define the function: $$F(x,t) = \frac{f(t)}{t} \chi_{T(b)}$$ Certainly, $F$ is measurable, since $\chi_{T(b)}$, $f$, and $\frac{1}{t}$ are measurable functions. Next, observe that:

$$\int_{[0,b]} g(x) dx = \int_{[0,b]} \int_{[x,b]} \frac{f(t)}{t}dtdx = \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{f(t)}{t} \chi_{T(b)} dt dx$$ From here, Tonelli's Theorem gives, that since $F$ is measurable: $$\int_{[0,b]} \int_{[x,b]} \frac{f(t)}{t}dtdx = \int_{[0,b]} \int_{[0,t]} \frac{f(t)}{t} dx dt = \int_{[0,b]} f(t) dt$$ Thus, since $f \in L^1$, we have: $$\int_{[0,b]} g(x) dx = \int_{[0,b]} f(t) dt < \infty \hspace{0.25cm} \Rightarrow g \in L^1$$ ...as desired.

2.3

Given that $f$ is $2\pi$-periodic, i.e. $f(x) = f(x + 2n\pi)$ for any $n \in \mathbb{Z}$, just as the hint prescribes, observe that, for some $k \in \mathbb{Z}$, the interval: $$I = (a,b] \subset (k\pi, (k+4)\pi]$$ Define $c = (k+2)\pi$. Certainly, if $c \neq a$, then $c \in (a,b]$. Now, $$\int_{(a,b]} f(x)dx = \int_{(a,c]} f(x)dx + \int_{(c,b]} f(x)dx$$ Since $f(x)|_{(c,b]} = f(x - 2\pi)|_{(k\pi,a]}$: $$\int_{(c,b]} f(x)dx = \int_{(k\pi, a]} f(x)dx$$ So we can certainly re-write: $$\int_{(a,b]} f(x)dx = \int_{(k\pi, c]} f(x)dx = \int_{(k\pi,(k+2)\pi]}f(x)dx$$ Finally, since we can certainly break up the integral: $$\int_{(k\pi,(k+2)\pi]}f(x)dx = \int_{(k\pi,(k+1)\pi]}f(x)dx + \int_{((k+1)\pi,(k+2)\pi]}f(x)dx$$ ... by $2\pi$-periodicity, again, it's apparent that: $$\int_{(a,b]} f(x)dx = \int_{(k\pi,(k+2)\pi]}f(x)dx = \int_{((k+1)\pi,(k+3)\pi]}f(x)dx$$ Thus, we can say this equality holds for any integer $k$, specifically, $k= -1$. Therefore: $$\int_{(a,b]} f(x)dx = \int_{(-\pi, \pi]}f(x)dx$$

2.2

Since continuous functions with compact support are dense in $L^1(\mathbb{R}^d)$, we know that $\forall \epsilon > 0$ there exists a $g$, continuous with compact support such that $|| f - g ||_{L^1} < \frac{\epsilon}{3}$. It follows that since: $$f(\delta x) - f(x) = f(\delta x) - g(\delta x) + g(\delta x) -g(x) + g(x) -f(x)$$ We have, by the triangle inequality: $$||f(\delta x) - f(x) ||_{L^1} \leq ||g(\delta x) - g(x) ||_{L^1} + ||f(\delta x) - g(\delta x) ||_{L^1} + ||f(x) - g(x) ||_{L^1}$$ ...and thus, since $|| f - g ||_{L^1} < \frac{\epsilon}{3}$, we have: $$||f(\delta x) - f(x) ||_{L^1} \leq ||g(\delta x) - g(x) ||_{L^1} + ||f(\delta x) - g(\delta x) ||_{L^1} + \frac{\epsilon}{3}$$ Glance ahead to exercise 2.16 and note the dilation property of $L^1$ functions: $$\int_{\mathbb{R}^d} h(x) \hspace{0.1cm}dx = |\delta|^d \int_{\mathbb{R}^d} h(\delta x) \hspace{0.1cm}dx$$ It's clear from this property that: $$||f(\delta x) - g(\delta x) ||_{L^1} = \frac{1}{|\delta|^d} ||f - g||_{L^1} \leq \frac{\epsilon}{3|\delta|^d} \leq \frac{\epsilon}{3}$$ Now, given that $g(x)$ is continuous, with compact support, we have that $g(x)$ and $g(\delta x)$ are uniformly continuous. Thus, there exists a bound $M$ for $g$. Next, see that since the sets $E$ and $\delta E$ are compact, the sets $E \Delta \delta E$, and $E \cap \delta E$ must be compact as well. It follows directly that: $$\int_{E \cup \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx = \ldots$$ $$\ldots = \int_{E \cap \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx + \int_{E \Delta \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx \leq \ldots$$ $$\ldots \leq \int_{E \cap \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx + 2M m(E \Delta \delta E)$$

Now observe that for every $\delta \neq 1 \hspace{0.25cm} \exists \alpha$ such that $\delta x = x + \frac{x}{\alpha}$. Since $E \cap \delta E$ is compact, we can select the diameter, $r = \max_{x,y \in E \cap \delta E} |d(x,y)|$. Certainly now, $x$, and $x + \frac{x}{\alpha} \in B_{|r| / |\alpha|}(x)$. Thus, for any $\xi > 0$ we can have for a $\delta$ sufficiently close to 1, (i.e. $\alpha$ sufficiently large.) we have $$|x - \delta x| \leq \frac{|r|}{|\alpha|} < \xi$$ Thus, for any $\epsilon > 0$, there exists a $\delta$ sufficiently close to $1$ such that: $$\int_{E \cap \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx < \frac{\epsilon}{6}$$ and, since $m(E \Delta 2E) < \infty$ (by compactness) and $E \Delta \delta E \searrow \emptyset$ we have from corollary 3.3 (in chapter 1) that for some $\delta$ sufficiently close to $1$, $m(E \Delta \delta E) < \frac{\epsilon}{12}$. Thus, choose $\delta$ close enough to $1$ such that both of these things happen, and we'll have: $$||g(x) - g(\delta x)||_{L^1} = \int_{E \cup \delta E} |g(x) - g(\delta x)| \hspace{0.1cm}dx < \frac{\epsilon}{6} + 2M\frac{\epsilon}{12M} = \frac{\epsilon}{3}$$ Finally, we have for $\delta$ sufficiently close to 1: $$||f(\delta x) - f(x) ||_{L^1} \leq ||g(\delta x) - g(x) ||_{L^1} + ||f(\delta x) - g(\delta x) ||_{L^1} + ||f(x) - g(x) ||_{L^1} \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$$

1.37

The exercise defines: $$\Gamma = \lbrace (x,y) \in \mathbb{R}^2 \hspace{0.25cm} \big| \hspace{0.25cm} y = f(x), x \in \mathbb{R} \rbrace$$ For any $k \in \mathbb{Z}$, let $\hspace{0.25cm} I_k = [k,k+1]$.

Since $f$ is continuous on $\mathbb{R}$, $f$ is actually uniformly continuous an any compact interval. This means that on any $I_k$, $\hspace{0.25cm} \forall \epsilon > 0, \hspace{0.25cm} \exists \delta > 0$ such that $|x - y| < \delta \Rightarrow \hspace{0.25cm} |f(x) - f(y)| < \epsilon$.

For any $\epsilon > 0$, choose $m$ large enough such that $$|x - y| \leq \frac{1}{2^{m + |k|+2}} \hspace{0.25cm} \Rightarrow \hspace{0.25cm} |f(x) - f(y)| < \epsilon$$

Next, let $\lbrace I_k^j \rbrace$ be a collection of compact intervals of length $\frac{1}{2^{m + |k|+2}}$ such that we can write: $$I_k = \bigcup_{j=1}^{2^m} I_k^j$$ Next, notice, if: $$\Gamma_k = \lbrace (x,y) \in \mathbb{R}^2 \hspace{0.25cm} \big| \hspace{0.25cm} y = f(x), x \in I_k \rbrace$$ Then: $$\Gamma_k \subset \bigcup_{j=1}^{2^m} \big(I_k^j \times f(I_k^j)\big)$$ And: $$m(\Gamma_k) \leq m\Big(\bigcup_{j=1}^{2^m} \big(I_k^j \times f(I_k^j)\big)\Big) \leq \sum_{j=1}^{2^m} m\big(I_k^j \times f(I_k^j)\big) \leq \ldots$$ $$\ldots \leq \sum_{j=1}^{2^m} \frac{\epsilon}{2^{m + |k|+2}} = \frac{2^m \epsilon}{2^{m + |k| + 2}} = \frac{\epsilon}{2^{|k| + 2}}$$ Lastly, notice that $\Gamma = \bigcup_{k=-\infty}^{\infty} \Gamma_k$. I.e. $$m(\Gamma) \leq \sum_{k=-\infty}^{\infty} m(\Gamma_k) \leq \sum_{k=-\infty}^{\infty} \frac{\epsilon}{2^{|k| + 2}} \leq \ldots$$ $$\ldots \leq 2\sum_{k=0}^{\infty} \frac{\epsilon}{2^{|k| + 2}} = 2\frac{\epsilon}{2} = \epsilon$$ Thus, since $\epsilon > 0$, arbitrarily, we have that $m(\Gamma) = 0$.

1.34

I'll prove such a function exists by actually constructing one.

Let $F:[0,1] \to [0,1]$ such that:

$F(x) = \Bigg\lbrace \begin{array}{cc} f(x) & x \in \mathcal{C}_\epsilon & \\ g(x) & x \in [0,1]\cap \mathcal{C}_\epsilon^c \\ \end{array}$

Our mapping $f(x):\mathcal{C}_\epsilon \to \mathcal{C}_\gamma$ is constructed by simply reassinging $x \in \mathcal{C}_\epsilon$'s binary expansion to $\mathcal{C}_\gamma$'s. To do this, simply observe that for any $x$ in $\mathcal{C}_\epsilon$,

$x = \sum_{n=1}^{\infty} \alpha_n \epsilon^n$ where $\hspace{0.25cm} \alpha_n \in \lbrace 0, 2 \rbrace$, and is unique to $x$.

Thus,

$f(x) = \sum_{n=1}^{\infty} \alpha_n \gamma^n$

Next, we need to construct $g(x)$. These functions are going to be continuous (linear) functions that map open disjoint intervals to corresponding open disjoint intervals.

First, lets define the set $\beta_{n,k}$ as the "binary representation of the integer $k$ with a string of length $n$". For example: $$\beta_{5,3} = (0,0,1,0,1)$$ Next, augment $\beta$ with a $1$ on the rightmost-side of the string. Call this set $\phi_{n,k}$. I.e. $$\phi_{5,3} = (0,0,1,0,1,1)$$ Now, let $\phi_{n,k}(i)$ be the $i^{th}$ element of the string, from the left.
(I.e. $\phi_{5,3}(3) = 1$.)

Moving forward, we'll use $\phi_{n,k}$ to define the sets $\mathcal{O}_{n,k}^\epsilon$ (and similarly for $\mathcal{O}_{n,k}^\gamma$) such that: $$\bigcup_{n = 0}^\infty \bigcup_{k = 0}^{2^n - 1} \mathcal{O}_{n,k}^\epsilon = \hspace{0.25cm} [0,1]\cap \mathcal{C}_\epsilon^c$$ Without further ado, $$\mathcal{O}_{n,k}^\epsilon = \Bigg( \sum_{i=1}^{n+1} \Big(\frac{(1-\epsilon)^{i-1}(1+\epsilon)}{2^i}\phi_{n,k}(i)\Big)-\epsilon \Big(\frac{(1-\epsilon)}{2}\Big)^n, \sum_{i=1}^{n+1} \Big(\frac{(1-\epsilon)^{i-1}(1+\epsilon)}{2^i}\phi_{n,k}(i)\Big) \Bigg)$$ While the expressions are regrettably unpleasant, notice that for each $n,k, \epsilon$, it successfully reproduces the correct $(k+1)^{th}$ open interval for the $(n+1)^{th}$ generation of an $\epsilon$-cantor set. For example, we know that the second interval in the third generation of the standard ($\epsilon = \frac{1}{3}$) Cantor set is $\big(\frac{7}{27}, \frac{8}{27}\big)$. We can easily check...

First, lets calculate: $$\sum_{i=1}^{4} \Big(\frac{(\frac{2}{3})^{i-1}(\frac{4}{3})}{2^i}\phi_{3,2}(i)\Big) = \frac{2}{9} + \frac{2}{27} = \frac{8}{27}$$ With that out of the way, it's easy to check now that: $$\mathcal{O}_{2,1}^{\frac{1}{3}} = \Big(\frac{7}{27}, \frac{8}{27} \Big)$$ So... it works! This was constructed so that the collection of all $\mathcal{O}_{n,k}^\epsilon$'s are pairwise-disjoint. From here, it's actually easy to define a linear function from $\mathcal{O}_{n,k}^\epsilon$ $\to$ $\mathcal{O}_{n,k}^\gamma$. The slope from one set to the other is (obviously) $\big(\frac{\gamma}{\epsilon}\big)^{n+1}$. The $y$-intercept $b_{n,k}$ is fairly straightforward to calculate... (just pick the middle points of the corresponding intervals, and evaluate.)

Thus, we can finally fully define $g$ as follows: $$g(x) = \sum_{n=0}^{\infty} \sum_{k = 0}^{2^n - 1} \Big(\big(\frac{\gamma}{\epsilon}\big)^{n+1} x + b_{n,k}\Big)\chi_{\mathcal{O}_{n,k}^\epsilon}$$ Finally, we have our function $F(x) = f(x) + g(x)$, which satisfies all three properties outlined in the exercise.

1.33

Recall two things. First, that $m^*(E) = 0 \hspace{0.25cm} \Rightarrow E \in \mathcal{M}$. Second, recall from problem 32, part a, that any measurable subset to a non-measurable set must have measure zero. Thus, since $\mathcal{N}^c \cup \mathcal{N} = [0,1]$, it suffices to show that $m(\mathcal{N}^c) = 1$.

Assume to the contrary that for some $\epsilon \in (0,1)$, $\hspace{0.25cm} \mathcal{N}^c \subset U \subset [0,1]$, measurable, such that $m(U) = 1 - \epsilon$. It directly follows that $([0,1] \cap U^c) \subset \mathcal{N}$. This is a contradiction, because $m \big(([0,1] \cap U^c) \big) = \epsilon > 0$.

Therefore, $m^*(\mathcal{N}) + m^*(\mathcal{N}^c) > m(\mathcal{N} \cup \mathcal{N}^c)$.

1.28

Given $E \subset \mathbb{R}$, $m^*(E) > 0$, assume to the contrary that for any open interval $I$, $m^*(E \cap I) < \alpha m(I)$.

By the definition of the outer measure, we know that $\forall \epsilon > 0$ $\exists \lbrace Q_k \rbrace_{k=1}^\infty$ such that $E \subset \bigcup_{k=1}^\infty Q_k$, and: $$\sum_{k=1}^\infty |Q_k| \leq m^*(E) + \epsilon$$ Next, define $\lbrace \mathcal{O}_k \rbrace_{k=1}^\infty$ to be open intervals s.t. $Q_k \subset \mathcal{O}_k$ for all $k$, and $m(\mathcal{O}_k)) = m(Q_k) + \frac{\epsilon}{2^k}$. (Note, this can always be done by adding $\frac{\epsilon}{2^{k+1}}$ to each side of the interval $Q_k$.)

We constructed $\lbrace \mathcal{O}_k \rbrace_{k=1}^\infty$ with the following property in mind ($\dagger$): $$\sum_{k=1}^\infty m(\mathcal{O}_k) \leq \big(\sum_{k=1}^\infty m(Q_k)\big) + \epsilon \leq m^*(E) + 2\epsilon$$ Now, just observe that $E \subset \bigcup_{k=1}^\infty \mathcal{O}_k$, so, by ($\dagger$): $$m^*(E) \leq \sum_{k=1}^\infty m^*(E \cap \mathcal{O}_k) \leq \alpha \sum_{k=1}^\infty m(\mathcal{O}_k) \leq \alpha(m^*(E) + 2\epsilon)$$ So finally, we finally arrive at the result: $$m^*(E) \leq \alpha m^*(E)$$ Which, of course, gives that $m^*(E) = 0$... a contradiction.

1.27

Since $E_1$ and $E_2$ are compact sets in $\mathbb{R}^d$, certainly $E_2 \backslash E_1$ is a bounded and measurable. For any $t > 0$, The sets $(E_2 \backslash E_1) \cap \overline{B_{t}(0)}$ are also bounded and measurable. For simplicity's sake, from here on out, let's say:

$$S_t = (E_2 \backslash E_1) \cap \overline{B_{t}(0)}$$ Define $f(t) = m(S_t)$. If I can prove $f$ is a continuous function, we're done --(by the intermediate value theorem).

Side note: Notice the function $g(t) = \alpha t^d$ is continuous, but not uniformly continuous. ($\dagger$)

Let $0 \leq \tau < t$. Notice $|f(t) - f(\tau)| = |m(S_t) - m(S_\tau)|$.

Certainly, $S_\tau \subset S_t$, thus:.

$$|m(S_t) - m(S_\tau)| = m(S_t) - m(S_\tau) = m(S_t \backslash S_\tau)$$ Now, notice that $$(S_t \backslash S_\tau) \subset \overline{B_t(0)} \backslash \overline{B_\tau(0)}$$ so: $$m(S_t \backslash S_\tau) \leq m(\overline{B_t(0)} \backslash \overline{B_\tau(0)}) \leq \alpha(d)\big(t^d - \tau^d)$$ where $\alpha(d) = \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)}$, the volume of the $d$-dimensional unit ball. (Recall we proved the relation between $B_1$ and $B_t$ in problem 6 of this chapter.) Thus, by $\dagger$, $$|t - \tau| < \delta \Rightarrow \alpha(d)\big(t^d - \tau^d) \leq \epsilon(t)$$ Therefore, $f$ is a continuous function, and by the IVT, given any $c$ with $a < c < b$, we can find a $t^* > 0$ such that $m(E) = m\big(E_1 \cup S_{t^*}\big) = c$, where $E_1 \cup S_{t^*}$ is compact (by construction).

1.23

First, $\forall x \in \mathbb{R}$ and $\forall n \in \mathbb{N}$ define $x_{r,n} = \frac{r}{2^n}$ where $x \in \big(\frac{r}{2^n}, \frac{r+1}{2^n} \big]$.

Define $f_n(x,y) = f(x_{r,n},y)$. Are these $f_n$'s measurable? Observe that: $$\lbrace (x,y) \in \mathbb{R}^2 \big| f_n(x,y) > \alpha \rbrace$$ $$= \bigcup_{r = -\infty}^{\infty} \bigg(\frac{r}{2^n}, \frac{r+1}{2^n} \bigg] \times \lbrace y \in \mathbb{R} \big| f(x_{r,n},y) > \alpha \rbrace$$ Certainly, the intervals $\big(\frac{r}{2^n}, \frac{r+1}{2^n} \big]$ are measurable, and further, the sets $\lbrace y \in \mathbb{R} \big| f(x_{r,n},y) > \alpha \rbrace$ are measurable, given that $f(x,y)$ is continuous in the $y$-direction for any fixed $x$. Thus, their product is measurable. Lastly, since the measurable sets form a $\sigma$-algebra, we have that $f_n$ must be a measurable function $\forall n \hspace{0.25cm} \in \mathbb{N}$.

Next, if we can confirm that $f_n \to f$ as $n \to \infty$ on $\mathbb{R}^2$, we're done (given property 4 of measurable functions on p. 29.)

Fix an arbitrary $(x,y) \in \mathbb{R}^2$. Since $f$ is continuous in both directions, we have that since $x_{r,n} \to x$ as $n \to \infty$, that $f_n(x,y) = f(x_{r,n},y) \to f(x,y)$ as $n \to \infty$. Thus, for any $(x,y)$ in $\mathbb{R}^2$, $f_n$ converges to $f$ point-wise.

Thus, $f$ is measurable on $\mathbb{R}^2$.

1.26

It's fairly easy to see that since $B = A \cup (B \backslash A)$, that since both $A$ and $B$ are measurable, $m(B) = m(A) + m(B \backslash A)$. Thus, $m(B \backslash A) = 0$. Similarly, we observe $m^{*}(E \backslash A) = 0$, and thus $E \backslash A$ is measurable, with measure zero.

Certainly, $E = A \cup (E \backslash A)$. Thus, since $A$ is measurable, and the union of any two measurable sets is measurable, we have $E$ must be measurable.

1.24

First, break the integers into two countably infinite subsets: $$S = \lbrace n \in \mathbb{N} \hspace{0.25cm} | \hspace{0.25cm} n = 2^{m+1} \hspace{0.25cm} \forall m \in \mathbb{N} \rbrace$$ and $$\mathbb{N} \backslash S$$ Next, let $\lbrace a_n \rbrace_{n \in S}$ be an enumeration of $\mathbb{Q} \cap [0,1]^c$, and $\lbrace b_n \rbrace_{n \in \mathbb{N} \backslash S}$ be an enumeration of $\mathbb{Q} \cap [0,1]$.

Certainly, the set $\lbrace r_n \rbrace_{n \in \mathbb{N}} = \big(\lbrace a_n \rbrace_{n \in S}\big) \bigcup \big(\lbrace b_n \rbrace_{n \in \mathbb{N} \backslash S}\big)$ enumerates $\mathbb{Q}$.

Notice the set: $$A = \bigcup_{n=1}^{\infty} \Big( r_n - \frac{1}{n}, r_n + \frac{1}{n} \big) = T \cup V$$ where $$T = \bigcup_{n=1}^{\infty} \Big( a_n - \frac{1}{n}, a_n + \frac{1}{n} \big)$$ and $$V = \bigcup_{n=1}^{\infty} \Big( b_n - \frac{1}{n}, b_n + \frac{1}{n} \big)$$ Observe that $m(A) \leq m(T) + m(V)$, however, by construction, both $m(T) \leq 3$ and $m(V) \leq 2$. (Those numbers, 2 and 3 are both just "safe" over-estimates.) Thus, since $m(A)$ is finite, we have that $A^c$ must be non-empty.