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.

2.21

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.

2.18

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

2.10

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 $d < b$.

2.4

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.