$\textbf{Problem} $ Let $m$ be the Lebesgue measure on $\mathbb{R}$. Let $f_n: \mathbb{R} \rightarrow [0,\infty)$ be sequence of Lebesgue measurable functions. Show that there is a sequence $c_n$ of positive numbers such that \begin{align*} \sum_{n=1}^\infty \frac{f_n(x)}{c_n} \quad \textrm{converges m-a.e. x.} \end{align*} (Hint. For a measurable function $f : X \rightarrow [0,\infty)$, for any sequence $a_m \downarrow 0$, and for any $\epsilon>0$, let $A_m:=\{x:a_mf(x)>\epsilon\}$. Then $\cap_m A_m=\emptyset$.)

My attempt: Take $A_n= \{x: \frac{f_n(x)}{c_n}>1/2^n\}$ . By hint, $\cap_n A_n=\emptyset$. However, we don't know $A_n$ be deceasing sequence of subset..

Any help is appreciated....

Thank you!


Let $$ I(x) = \sum_{i=1}^{\infty} \frac{f_n(x)}{c_n} $$

Assume each $f_n$ is continuous. Then for each n, set $$ b_n = \max_{1 \leq i \leq n} \{ \sup_{x \in [-n,n]} f_i(x) \} $$ and set $c_n = 2^{n} b_n$. Then for each $x \in \mathbb{R}$, $ \exists N$ such that $\forall n \geq N$, $ f_n(x) / c_n < 2^{-n}$, and hence $I(x)$ converges everywhere.

Now we pass from continuous to measurable functions using Lusin's Theorem. So assume now that each $f_n$ is just a measurable function.

For each n, Lusin's theorem gives us a compact set $E_n \subset [-n,n]$ such that $f_i$ is continuous on $E_n$ for every $i = 1, ... , n$, and $m([-n,n] - E_n) < 2^{-n}$. Define $A_n = [-n,n] - E_n$. Then $$ \sum_1^{\infty} m(A_n) < \infty $$ so by Borel-Cantelli, $m(\limsup A_n) = 0$. If we now let

$$ b_n = \max_{1 \leq i \leq n} \{ \sup_{x \in \cup_{i=1}^n E_n} f_i(x) \} $$ and $c^n = 2^n b_n$, then for every $x \in \mathbb{R} - \limsup A_n$, $I(x)$ converges. $\blacksquare$


Since the interval $[-n,n]$ has finite Lebesgue measure, it follows from the continuity of the Lebesgue measure $m$ that

$$m \left( \left\{ x \in [-n,n]; r f_n(x) > \frac{1}{2^n} \right\} \right) \xrightarrow[]{r \to 0} 0$$

for each fixed $n \in \mathbb{N}$, and therefore we can choose $r_n>0$ such that

$$A_n := \left\{ x \in [-n,n]; r_n f_n(x) > \frac{1}{2^n} \right\}$$


$$m \left( A_n \right) \leq \frac{1}{2^n}.$$

By the monotone convergence theorem, this implies

$$\int_{\mathbb{R}} \left( \sum_{n \geq 1} 1_{A_n}(x) \right) \, m(dx) = \sum_{n \geq 1} m(A_n) < \infty,$$

and so

$$\sum_{n \geq 1} 1_{A_n}(x) < \infty$$

for Lebesgue almost all $x \in \mathbb{R}$, i.e. for almost all $x$ we can find $N=N(x)$ such that $1_{A_n}(x)=0$ for all $n \geq N$. By the very definition of $A_n$ this implies that the series

$$\sum_{n \geq 1} r_n f_n(x) $$

converges for almost all $x$, and setting $c_n := 1/r_n$ this proves the assertion.


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