2
$\begingroup$

Let $\{f_n\}$ be a sequence of measurable functions defined on a measurable set $E$. Define $E_0$ to be the set of points $x$ in $E$ at which $\{f_n(x)\}$ converges. Is the set $E_0$ measurable.

I need to see if $A:=\{x \in E: f_n(x)\ \ \text{conerges}\}$ measurable or not. I only have the information above, this question actually gives me a headache because we just know that a set $E$ is measurable, so for the case when it has measure zero, then it is trivil that $E_0$ is measurable. But how about if $E$ has a positive outer measure??

I would appreciate any hint or help for that. Thank you,

$\endgroup$
2
$\begingroup$
  1. One knows that a sequence $(a_n)_n\subseteq\mathbb{C}$ converges if and only if $(a_n)_n$ is Cauchy.

  2. A sequence $(a_n)_n$ is Cauchy if and only if $\forall M\geq1$ there exists $N\geq1$ such that $|a_m-a_n|<1/M$ for $m,n \geq N$ (here it is important to take $M\in\mathbb{N}$ and not $\varepsilon>0$ because measurable sets only have a good behavior under countable unions and intersections).

  3. $$\{x\in E : f_n(x)\text{ converges }\} = \bigcap_{M=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{m=N}^{\infty} \bigcap_{n=N}^{\infty} \{x\in E : |f_m(x)-f_n(x)|<1/M\}$$

Can you finish from here?

The set $A=\{x\in E : |f_m(x)-f_n(x)|<1/M\}$ is measurable since $f_m-f_n$ is measurable and $A=(f_m-f_n)^{-1}((-1/M,1/M))$

$\endgroup$
  • $\begingroup$ Thankkk you got it. $\endgroup$ – Ahmed Sep 27 '18 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.