3.3

Part a.) Under the given conditions, we know that $\forall \epsilon > 0$ there exists $r > 0$ such that: $$\frac{m(E \cap [-r,r])}{m([-r,r])} \geq (1-\epsilon)$$ Now, simply observe (by the translation / rotation invariance of the Lebesgue Measure): $$m(-E \cap [-r,r]) = m(E \cap [-r,r]) \geq (1-\epsilon)2r$$ Thus, for $\epsilon < \frac{1}{2}$: $$0 < (1-2\epsilon)2r \leq m(-E \cap E \cap [-r,r])$$ ...and thus, since these sets have positive measure, we can always find a sequence of $x_n's$ that satisfy the condition.

Part b.) It suffices to simply consider the $cE$ case, where $c > 1$. Since: $$\frac{m(cE \cap [-r,r])}{m([-r,r])} = \frac{m(E \cap [-\frac{r}{c}, \frac{r}{c}])}{m([-\frac{r}{c},\frac{r}{c}])}$$ We have: $$m(cE \cap [-r,r]) = \frac{m([-r,r])}{m([-\frac{r}{c},\frac{r}{c}])}m\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big) = \ldots$$ $$\ldots = cm\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big)$$ Thus, like in part a, we have that since $\forall \epsilon > 0$ there exists an $r_0 > 0$ such that $\forall r \in (0,r_0)$: $$m(E \cap [-r,r]) > (1 - \epsilon)2r$$ Therefore, since $\frac{r}{c} < r$, we have: $$m(cE \cap [-r,r]) = cm\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big) > c(1 - \epsilon)2\frac{r}{c} = \ldots$$ $$\ldots = (1 - \epsilon)2r > 0$$ ...when $\epsilon < \frac{1}{2}$. Thus, under these settings, the estimate: $$m(cE \cap E \cap [-r,r]) > (1 - 2\epsilon)2r$$ ...holds, $\implies$ we can choose a sequence of $x_n's$ satisfying the desired condition.

To see that this condition holds for $0 < c < 1$, observe that if we simply define $cE = F$, that: $\frac{1}{c}F = E$, and the statement holds per the previous argument.

Now, since we know it works for $c > 0$, and $c = -1$ (from Part a.), simply combine the arguments to see that this clearly works for $c \in \mathbb{R} \backslash \lbrace 0 \rbrace$.

3.2

Recall the statement from exercise 3.1c, however, instead of: $$\int_\mathbb{R}^d K_\delta (y) = 1$$ Write: ($\dagger$) For some particular $C \in \mathbb{R}$: $$\int_\mathbb{R}^d K_\delta (y) = C$$ The new statement should read: $$ \dagger \hspace{0.25cm} \Rightarrow (f*K_\delta)(x) \to Cf(x) \hspace{0.25cm} \text{as} \hspace{0.25cm} \delta \to 0$$ The argument follows identically to how the $C=1$ case is shown for approximations to the identity. Now, simply consider $C = 0$, and we're done.

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 $f \in L^1$.