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Let $(f_n)$ be a sequence of measurable functions defined on a measurable set $E$. Is the set $A=\{x\in E|\lim_{n \to\infty} f_n(x)\ exists\ (finitely)\}$ measurable?

The following is my answer:

Define $g:A\to\mathbb{R}$ as $g(x)=\limsup f_n(x)-\liminf f_n(x) \ \forall x\in A$. Then $g(x)=0 \ \forall x\in A$ since $x\in A\implies \limsup f_n(x)=\limsup f_n(x)=\lim f_n(x)$ and $g$ is measurable as $\limsup f_n,\ \liminf f_n$ are measurable functions. Thus $g^{-1}((-1,1))=A$ is measurable as $(-1,1)$ is open.

Can somebody please tell me if this proof is alright? Thank you.

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  • $\begingroup$ Yes, and you just need to consider $g^{-1}(\{0\})$. As $g(\cdot)$ is a measurable function, preimage of $\{0\}$ under $g(\cdot)$ is a measurable set. For an alternative proof, see below. $\endgroup$ – TBTD Apr 26 '17 at 17:03
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To be precise, let's call our measurable space $(X, \mathcal{B})$.

Your logic is not right - all you get is that $A$ is measurable with respect to the induced $\sigma$-algebra $\mathcal{B}_A = \{A \cap B : B \in \mathcal{B}\}$. But this is trivially true, and doesn't imply $A \in \mathcal{B}$.

By this same logic, I could take a non-measurable set $V \notin \mathcal{B}$ and define the function $g : V \to \mathbb{R}$ to be the constant function $0$, which is "clearly a measurable function". Then $g^{-1}((-1,1)) = V$ but this does not prove $V$ is measurable. The issue is that $g$ is measurable with respect to $\mathcal{B}_V$ only; it does not even make sense to speak of it being $\mathcal{B}$-measurable, since that notion applies only to functions whose domain is $X$.

To get an argument that works, you have to define your function $g$ on all of $X$ (or at least on some set already known to be in $\mathcal{B}$). You can still use something like the function $g = \limsup f_n - \liminf f_n$, and consider $g^{-1}(\{0\})$, but you will have to define $g$ in an appropriate way when the limsup and/or liminf are infinite.

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  • $\begingroup$ How would you define such a function? Please help. $\endgroup$ – Janitha357 May 4 '17 at 15:26
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As an alternative, note that the complement of this set is $$ A^c = \left\{x \in E : \limsup_{n\to\infty}f_n(x) > \liminf_{n\to\infty}f_n(x)\right\}. $$

This means, for every $x \in A^c$, $\exists r\in\mathbb{Q}$ such that $$ \limsup_{n\to\infty}f_n(x) > r>\liminf_{n\to\infty}f_n(x). $$ Now, make the following observation: $$ A^c=\bigcup_{r\in\mathbb{Q}}\underbrace{\left\{x\in E:\limsup_{n\to\infty}f_n(x)>r\right\}}_{\triangleq V_n}\cap\underbrace{\left\{x\in E:\liminf_{n\to\infty}f_n(x)<r\right\}}_{\triangleq W_n}. $$ Since each of the functions $\limsup_{n\to\infty}f_n(x)$ and $\liminf_{n\to\infty}f_n(x)$ are measurable, so does $V_n$ and $W_n$. Finally, since countable union of measurable sets ($V_n \cap W_n$ in our case) is measurable, so does $A^c$, hence $A$.

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  • $\begingroup$ Your proof is beautiful. Thank you very much :) $\endgroup$ – Janitha357 Apr 26 '17 at 17:13

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