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Suppose $E$ is measurable subset of $\Bbb R$, $(f_n)$ is sequence of measurable functions from $E$ to $[-\infty, \infty] $ , $f$ is function from $E$ to $[-\infty , \infty]$.

If $\lim\limits_{n \to \infty} f_n=f$ (Almost everywhere) then $\lim\limits_{n \to \infty} f_n=f$ ( in measure on$E$)

Is this Statement True?

$\lim\limits_{n \to \infty} f_n=f$ (in measure on $E$) i.e. $\forall \epsilon \gt 0$ ,$\exists N \in \Bbb N s.t. \forall n \ge N :$ $m^*(\{x \in E |$ $ |f_n(x)-f(x)|\ge \epsilon\}) \lt \epsilon$

$\lim\limits_{n \to \infty} f_n=f$ (Almost everywhere) i.e. $m^*(\{ x \in E | \lim\limits_{n \to \infty} f_n\not =f\}=0$

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    $\begingroup$ It's not necessarily true if $E$ has infinite measure. $\endgroup$ – David Mitra Dec 1 '14 at 20:20
  • $\begingroup$ @DavidMitra in fact this question is (True\False). Do you have any Counterexample? $\endgroup$ – agustin Dec 1 '14 at 20:23
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    $\begingroup$ $E=\Bbb R$, $f_n=\chi_{[n,n+1]}$. (You would also have to assume $f_n$ and $f$ to have finite values a.e. for the statement to hold.) $\endgroup$ – David Mitra Dec 1 '14 at 20:25
  • $\begingroup$ @DavidMitra It's Ok, that's work. $\endgroup$ – agustin Dec 1 '14 at 20:32
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HINT:

You need to show $$\lim_{n\to\infty}m(x:|f_{n}(x)-f(x)|\geq\epsilon)=0$$ for all $\epsilon>0$. From pointwise a.e. convergence, we have $$m(x:\lim_{n\to\infty}|f_{n}(x)-f(x)|\geq\epsilon)=0.$$

So you need to find a way to justify bringing the limit out of the measure (which is valid; commuting the limits in the reverse direction is not necessarily though, as shown by the usual counter-examples).

EDIT:

For some reason I thought the OP was working on a probability space (or a space with finite measure). If the $m(E)=+\infty$, the claim is false as exhibited by a moving bump function out to horizontal infinity where $E=\mathbb{R}$ with Lebesgue measure $dx$.

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