If $\int_\mathbb{R} |f_n| \le \frac{1}{n^2}$, then $f_n \to 0 \lambda$-a.e.

Question

Assume that $f_n : \mathbb{R}\to \mathbb{R}$ is a sequence of Lebesgue measurable functions such that $\int_\mathbb{R} |f_n|d\lambda \le \frac{1}{n^2}$. Show that $f_n \to 0 \lambda$-a.e.

My attempt :

Choose $\delta>0$ and denote the set $E^\delta_n = \{x\in \mathbb{R} : |f_n|\ge \delta\}$ $$\frac{1}{n^2}\ge\int_\mathbb{R} |f_n|d\lambda \ge \int _{E^\delta_n}|f_n|d\lambda \ge \delta \lambda\Big(\{x\in \mathbb{R} : |f_n|\ge \delta\}\Big)$$ taking limit as $n\to\infty$; $$\lim_{n\to\infty}\lambda(E^\delta_n) =\lim_{n\to\infty}\lambda\Big(\{x\in \mathbb{R} : |f_n|\ge \delta\}\Big)=0$$ so $f_n \to 0 \lambda$-a.e

Answer 1

Another way, but somehow it is more complicated:

Let $S$ be the set defined by

\begin{align*} S=\bigcap_{N}\bigcup_{m}\bigcap_{n\geq m}\left\{x\in\mathbb{R}:|f_{n}(x)|<\dfrac{1}{N}\right\}. \end{align*}

This set collects those points $x$ such that $f_{n}(x)\rightarrow 0$. We are to show that $\lambda(S^{c})=0$, this means that $f_{n}(x)\rightarrow 0$ $\lambda$-a.e.

The set $S^{c}$ can be written as \begin{align*} S^{c}=\bigcup_{N}\bigcap_{m}\bigcup_{n\geq m}\left\{x\in\mathbb{R}:|f_{n}(x)|\geq\dfrac{1}{N}\right\} \end{align*}

Fix $N$ and $m$, then for all $n\geq m$, we have \begin{align*} \dfrac{1}{N}\lambda\left(\left\{x\in\mathbb{R}：|f_{n}(x)|\geq\dfrac{1}{N}\right\}\right)\leq\int_{\mathbb{R}}|f_{n}|d\lambda\leq\dfrac{1}{n^{2}}, \end{align*} so \begin{align*} \dfrac{1}{N}\lambda\left(\bigcup_{n\geq m}\left\{x\in\mathbb{R}：|f_{n}(x)|\geq\dfrac{1}{N}\right\}\right)\leq\dfrac{1}{N}\sum_{n\geq m}\lambda\left(\left\{x\in\mathbb{R}：|f_{n}(x)|\geq\dfrac{1}{N}\right\}\right)\leq\sum_{n\geq m}\dfrac{1}{n^{2}}, \end{align*} and hence \begin{align*} \lambda\left(\bigcap_{m}\bigcup_{n\geq m}\left\{x\in\mathbb{R}：|f_{n}(x)|\geq\dfrac{1}{N}\right\}\right)\leq\lambda\left(\bigcup_{n\geq m}\left\{x\in\mathbb{R}：|f_{n}(x)|\geq\dfrac{1}{N}\right\}\right)\leq N\sum_{n\geq m}\dfrac{1}{n^{2}}. \end{align*} Since this is valid for all $m=1,2,...$ and we know that $\displaystyle\sum_{n\geq m}\dfrac{1}{n^{2}}\rightarrow 0$ as $m\rightarrow\infty$, it follows that \begin{align*} \lambda\left(\bigcap_{m}\bigcup_{n\geq m}\left\{x\in\mathbb{R}：|f_{n}(x)|\geq\dfrac{1}{N}\right\}\right)=0. \end{align*} Note that this is also valid for all $N=1,2,...$, it follows that \begin{align*} \lambda(S^{c})\leq\sum_{N}\lambda\left(\bigcap_{m}\bigcup_{n\geq m}\left\{x\in\mathbb{R}：|f_{n}(x)|\geq\dfrac{1}{N}\right\}\right)=0, \end{align*} we are done.

Answer 2

Another way, by Monotone Convergence Theorem or Tonelli: \begin{align*} \int_{\mathbb{R}}\sum_{n}|f_{n}|=\sum_{n}\int_{\mathbb{R}}|f_{n}|\leq\sum_{n}\dfrac{1}{n^{2}}<\infty, \end{align*} so for a.e. $x$, \begin{align*} \sum_{n}|f_{n}(x)|<\infty, \end{align*} and hence $f_{n}(x)\rightarrow 0$.