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

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

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

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 $$m(x:\lim_{n\to\infty}|f_{n}(x)-f(x)|\geq\epsilon)=0.$$
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 $dx$.