Show that $f_n\to f$ in $L^1(\mu)$ iff $\sup\limits_{A \in \mathcal M} \left| \int_{A} f_n d\mu - \int_{A} f d\mu \right| \rightarrow 0$

Question

Let $(X, \mathcal M, \mu)$ be a measure space, $f_n$,$f \in L^1 (\mu)$. Show that $\int_X \mid f_n - f\mid d\mu \rightarrow 0$ as $n\rightarrow \infty$ if and only if $\sup_{A \in \mathcal M} \mid \int_{A} f_n d\mu - \int_{A} f d\mu \mid \rightarrow 0$ as $n\rightarrow \infty$.

It's easy to show one direction but how to show the other direction, i.e. how to show $\int_X \mid f_n - f\mid d\mu \rightarrow 0$ as $n\rightarrow \infty$ given the condition. Thanks!

Answer 1

I preassume that you meant $n\rightarrow\infty$ in stead of $n\rightarrow0$.

Firstly denote $g_{n}:=f_n-f$.

Secondly abbreviate $\mu\left(h\right):=\int_{X}hd\mu$ for integrable functions $h$.

You ask for a proof of $\mu\left(\left|g_{n}\right|\right)\rightarrow0$ on base of $s_{n}:=\sup\left\{ \left|\mu\left(g_{n}1_{A}\right)\right|\mid A\in\mathcal{M}\right\} \rightarrow0$.

Let $A_{n}:=\left\{ g_{n}>0\right\}\in\mathcal M$ and $B_{n}:=\left\{ g_{n}<0\right\}\in\mathcal M$.

Then $\mu\left(\left|g_{n}\right|\right)=\mu\left(g_{n}1_{A_{n}}\right)+\mu\left(-g_{n}1_{B_{n}}\right)=\left|\mu\left(g_{n}1_{A_{n}}\right)\right|+\left|\mu\left(g_{n}1_{B_{n}}\right)\right|\leq2s_{n}$.

Hence $s_n\rightarrow0$ implies that $\mu\left(\left|g_{n}\right|\right)\rightarrow0$