A sequence of measurable function and a set points such that this sequence converge

lebesgue-measuremeasure-theory

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,

Best Answer

  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 $A=(f_m-f_n)^{-1}((-1/M,1/M))$

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