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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.

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  • $\begingroup$ there are a lot of ways to show this, depending on the theoretic background. You can show that $(f_j)\to f$ point-wise a.e. implies that $(f_j)\to f$ almost uniformly when the space have finite measure. And from here that $(f_j)\to f$ almost uniformly implies that $(f_j)\to f$ in measure $\endgroup$ – Masacroso Dec 4 '18 at 22:40
  • $\begingroup$ is almost uniform like saying uniform almost everywhere? or is it a slightly weaker notion of uniform convergence? $\endgroup$ – Hossien Sahebjame Dec 4 '18 at 23:04
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    $\begingroup$ a sequence of measurable functions converges almost uniformly in the measure space $(X,\mu)$ if, for any chosen $\epsilon>0$, there is some measurable $A_\epsilon\subset X$ such that $(f_n)\to f$ converges uniformly in $A_\epsilon$ and $\mu(A_\epsilon^\complement)<\epsilon$. However this doesn't mean that $(f_n)\to f$ uniformly a.e.! $\endgroup$ – Masacroso Dec 4 '18 at 23:11

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