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).