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.

2.18

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.

2.10

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 $d < b$.

2.6

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.

1.18

The exercise didn't specify anything about $f = \infty$ on sets with positive measure. I'll assume $f$ is a.e. finite. From here, it, of course, suffices to consider only non-negative measurable functions. For any such $f$, we need to find a sequence of continuous functions $\lbrace f_n \rbrace_{n = 1}^{\infty}$ such that the set: $$E = \Big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \bigg| \hspace{0.25cm} \lim_{n \to \infty} |f(x) - f_n(x)| > 0 \Big\rbrace$$ has measure zero. First, define the $d-$dimensional cubes: $$Q_n = [-n,n]\times \ldots \times[-n,n]$$ And the functions: $$g_n(x) = \min{\lbrace n,f(x) \rbrace}\chi_{Q_n}$$ It follows from Lusin's Theorem that for any of our $g_n$'s, we can find a set $F_n$ such that $g_n \chi_{F_n}$ is continuous, and $m(Q_n \backslash F_n)\leq \frac{1}{2^{n}}$. Now, we can simply define $f_n = g_n \chi_{F_n}$.

Lastly, define: $E_n = Q_n \backslash F_n$. Notice that: $$\sum_{n = 1}^{\infty} m(E_n) \leq \sum_{n=1}^{\infty} \frac{1}{2^n} \leq 1$$ Now, of course, we simply need to show that $E$ is $\sup{\lbrace E_n \rbrace}$. This actually isn't all that hard. First, notice: $$\bigcup_{n \geq k} E_n = \Big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \bigg| \hspace{0.25cm} |f(x) - f_n(x)| > 0, \hspace{0.25cm} n \geq k \Big\rbrace$$ Naturally, we arrive that the result that $E = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k} E_n$, and thus, by the Borel-Cantelli Lemma, $m(E) = 0$, as desired.