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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?

My try:

Let $f(x)=\lim _{n\to \infty} f_n(x)$

$E_0=\{x\in E:f_n(x) \mbox{converges}\}$

To show that $E_0$ is measurable consider the set $\{x:f(x)>c\}$ for any $c\in \Bbb R$.

If $f_n(x)$ converges for some $a\in E_0$ then $f(a)=\lim _{n\to \infty} f_n(a)$ then $|f_n(a)-f(a)|<\frac{1}{n}\forall n\in \Bbb N$

So $f(a)-\frac{1}{n}<f_n(a)<f(a)+\frac{1}{n};\forall n\in \Bbb N$

Then $\{x:f(x)>c\}=\cap_{{n\in \Bbb N}}[\{x:f_n(x)>f(a)-\frac{1}{n}\}\cap \{x:f_n(x)<f(a)+\frac{1}{n}\}]$ which is measurable as $f_n$ is measurable and measurable sets form a $\sigma-$ algebra.

Is the proof correct?

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  • How do you define $f$ on the complement of $E_0$? – John Dawkins May 07 '16 at 16:34
  • $A=\cap_{{n\in \Bbb N}}[{x:f_n(x)>f(a)-\frac{1}{n}}\cap {x:f_n(x)<f(a)+\frac{1}{n}}]$, $A$ is the is the set of $x$ where $f_n$ converges to $c$ where $c=f(a)$, it does not give you all $x$ such that the sequence converges – clark May 07 '16 at 16:38
  • I'd suggest thinking of $E_0$ as the set of $x$s for which ${f_n(x)}$ is a Cauchy sequence. – John Dawkins May 07 '16 at 16:45
  • if you know that $\limsup f_n$ and $\liminf f_n$ are measurable functions then $E_0=[ \limsup f_n=\liminf f_n]$$ – clark May 07 '16 at 16:47
  • Your try is not correct. E.g. convergence $f_n(a)\to f(a)$ does not guarantee that $|f_n(a)-f(a)|\leq\frac1{n}$ for each $n$. Have a look at this answer – drhab May 07 '16 at 17:05

1 Answers1

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Let $E_{c}$ be the set in question.

Then note that $E_{c} = \bigcap_{m \in \mathbb{N}}\bigcup_{N \in \mathbb{N}} \bigcap_{p \in \mathbb{N}} \{E: |f_{N+p} - f_{N}| < \frac{1}{m} \}$.

Daniel Akech Thiong
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