Let $f,f_k$ for $k \in \mathbb{N}$ be measurable finite a.e. on measurable set $E$. Suppose $f_k \rightarrow f$ point wise a.e., then $f_k$ converges to $f$ in measure on $E$.
This is a basic result and the proof follows from basic definitions I believe, I want to know why we need finite valued and finite measure, is it for the obvious reasons that infinite measure does not converge and finite valued so that the limit as $k$ tends to $\infty$ makes sense?
because converging pointwise a.e. means that for $x \in E$ with $E$ having positive finite measure, one has
lim$_{k \rightarrow \infty}$ $f_k(x) = f(x)$
so
lim$_{k \rightarrow \infty}$ $f_k(x) – f(x)$ = $0$,
and we wish to show for any $\epsilon > 0$ that
$m(\{x : \vert f_k(x) – f(x) \vert > \epsilon \}) \rightarrow 0$ as $k \rightarrow \infty$
and since $x$ is taken out of some set of finite positive measure, we obtain (and I am not sure this works)
$m(\{x : \vert 0 \vert > \epsilon \}) \rightarrow 0$ as $k \rightarrow \infty$
I get confused here, So sorry for being naive in measure theory, I apologize and thank you in advance! if there is any simple small subtle silly errors I am making, any intuitive assumptions I am incorrectly making please let me know, I am very eager to strengthen my measure theory skills.
Best Answer
Let $f_n \to f$ pointwise a.e. on a set $E$ of finite measure. Then $f_n \to f$ in measure.
Let $\epsilon>0$ be given and put
$$A_n:=\{x \in E : \vert f_n(x) - f(x) \vert > \epsilon\}.$$
Using the DCT, put $f_n:=\chi_{A_n}$, then $f_n \leq f_{n+1}$ and $f_n \to f$ pointwise a.e. where $f \equiv 0$ and if $g=\chi_E$, since $m(E)<\infty$, $g$ is integrable. Then $\vert f_n(x) \vert \leq g(x)$ for all $x$ thus
$$\lim_{n \to \infty}\int f_n = \int \lim_{n \to \infty} f_n =\int 0 = 0.$$
and we are done.