$\lim\limits_{n\to\infty}\int_\mathbb R |f_n-f|d\lambda=0 \leftrightarrow \lim \inf_{n\in\mathbb N}f_n\leq f$

Question

I am wondering if that statement is true:
If $(f_n)_{n\in\mathbb N} \in \mathcal L^1(\mathbb R)$ so that $\lim\limits_{n\to\infty}\displaystyle\int_\mathbb R |f_n-f|d\lambda=0$, then $\lim \inf_{n\in\mathbb N}f_n\leq f$ almost everywhere

Answer 1

Let $g = \liminf_n f_n$, $\epsilon > 0$ and $A_\epsilon = \left\{x: g(x) - f(x) > \epsilon\right\}$

\begin{align} \epsilon \lambda\left(A_\epsilon\right) &\le \int_{A_\epsilon} \left(g - f\right)\mathrm d\lambda\\ &\le \int_{A_\epsilon} (f_n - f)\mathrm d\lambda\\ &\le \int_{\mathbb R} \left|f_n - f\right| \mathrm d \lambda \underset {n\to \infty}\to 0 \end{align}

This proves that $$\lambda \left(A_\epsilon\right) = 0$$

Since $$\left\{x: g(x) - f(x) > 0\right\} = \bigcup_{n\in \mathbb N_{\ge 1}} A_{\frac1n},$$

your statement follows immediately.