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).

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.

1.17

The goal of this exercise is to show that if we choose $c_n$'s as the hint prescribes, the set: $$E = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} \lim_{n \to \infty} \Big| \frac{f_n(x)}{c_n} \Big| > 0 \Big\rbrace $$ has measure zero. Define: $$ S_c = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} |f_n(x)| > \frac{c}{n} \Big\rbrace$$ Notice that $m(S_1) \leq 1$. Also, notice $S_{c+1} \subset S_{c} \hspace{0.25cm} \forall c$, and $\bigcap_{c=1}^{\infty}S_c = \emptyset$. Thus, by Corollary 3.3, $\lim_{n \to \infty} m(S_n) = 0$.

With that out of the way, notice that for the sets: $$E_n = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} \Big| \frac{f_n(x)}{c_n} \Big| > \frac{1}{n} \Big\rbrace $$ ...we can choose $c_n$ large enough such that for every $n$, $m(E_n) \leq \frac{1}{2^n}$. Certainly, it follows that ($\dagger$): $$ \sum_{n=1}^{\infty} m(E_n) \leq \sum_{n=1}^{\infty} \frac{1}{2^n} = 1 < \infty $$ Now, observe that $E = \sup{ \lbrace E_n \rbrace }$, i.e. (recalling problem 1.16), $E = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k}E_n$. Thus, applying the Borel-Cantelli Lemma, we have that $$ (\dagger) \Rightarrow m(E) = 0$$

1.16

This exercise is asking us to prove the Borel-Cantelli Lemma. In the measure theory settings, it states:

Suppose $\lbrace E_n \rbrace_{n=1}^{\infty}$ is a countable family of subsets of $\mathbb{R}^d$, and ($\dagger$) $$\sum_{k=1}^{\infty} m(E_k) < \infty$$ The set $E = \bigcap_{n=1}^{\infty} \bigcup_{k \geq n} E_k$ has measure zero.

Part a.) We first need to show $E$ is a measurable set. Recall that the measurable sets form a $\sigma$-algebra, $\mathcal{M}$, which is closed under countable unions and intersections.

Since each $E_k \in \mathcal{M}$, certainly $A_n = (\bigcup_{k \geq n} E_k) \in \mathcal{M} \hspace{0.25cm} \forall n$. It follows that since $E = \bigcap_{n=1}^{\infty} A_n$, that $E \in \mathcal{M}$.

Part b.) Assume to the contrary that $m(E) = \delta > 0$. Notice that if we define: $$S_N = \bigcap_{k=1}^N \bigcup_{n \geq k} E_n = \bigcup_{n \geq N} E_n $$ Then, certainly, $S_N \searrow E$, and $\forall N$, $$\delta \leq m(S_N) = m(\bigcup_{n \geq N} E_n) \leq \sum_{n=N}^{\infty} m(E_n)$$ Which certainly contradicts $\dagger$, completing the proof.