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 $d < b$.