Let $(X, M, \mu)$ be a measure space, and $f_n:X \rightarrow \mathbb R$ sequence of measurable functions.
How can I show that the set of $x$ that for them $f_n(x)$ has a subsequence that converges to $0$ is a measurable set?
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- 1$\begingroup$ I guess you meant fn:X→R $\endgroup$Amit Keinan– Amit Keinan2021-01-22 14:32:59 +00:00Commented Jan 22, 2021 at 14:32
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2 Answers
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Note that there is a subsequence of $\left(f_n(x)\right)_n$ converging to $0$ if and only if $\forall k\in\mathbb{N}\exists n\in\mathbb{N}$ s.t. $\left\vert f_n(x)\right\vert<\frac{1}{k}$. Hence, \begin{align*} &\left\{x\in X:\left(f_n(x)\right)_n\text{ has a subsequence converging to }0\right\}\\ =&\left\{x\in X:\forall k\in\mathbb{N}\exists n\in\mathbb{N}\text{ s.t. }\left\vert f_n(x)\right\vert<\frac{1}{k}\right\}\\ =&\bigcap_{k\in\mathbb{N}}\bigcup_{n\in\mathbb{N}}\left\{x\in X:\left\vert f_n(x)\right\vert<\frac{1}{k}\right\} \end{align*} Since $f$ is measurable, it is a countable union and intersection of measurable sets and thus measurable itself.
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I do not have sufficient reps to comment yet, so I can only write an answer (sorry!). I think the answer above @junjios is missing something. The definition of convergence is that for every $k\in\mathbb{N}$, there is some $N\in\mathbb{N}$ such that FOR ALL $n\geq N$, we have $|f_n(x)|<\frac{1}{k}$. The intersection of union above only captures one instance of an $n$, and does not account for $n+1,n+2,\cdots$. I think the set you want to look at is $$\bigcap_{k=1}^\infty\bigcup_{m=1}^\infty\bigcap_{n=m}^\infty\left\{x\in X:|f_n(x)|<\frac{1}{k}\right\}.$$ The rest of the argument follows.
