The question:

Given a sequence of measurable functions $(f_n):\Omega\rightarrow \mathbb{R}$. Prove the following set $A$ is measurable: $$A=\{\omega:f_n(\omega)\in[a,b]\text{ for finitely many }n\}$$ $$B=\{\omega:f_n(\omega)\in[a,b]\text{ for infinitely many }n\}$$

I found another similar post but I don't think it was helpful.

My thoughts:

For $A$, $f_n(\omega)\in[a,b]$ for finitely many $n$ means: $$\forall \,N < \infty\,\,\exists \,\,n\leq N \text{ such that } f_n(\omega)\in [a,b]$$translate that to set language is:$$\bigcup_{N\in\mathbb{N}}\bigcap_{n\leq N} \{\omega:f_n(\omega) \in [a,b]\}$$However, this does not seem right, cause if for some $n \leq N$, set $\{\omega: f_n(\omega)\in[a,b]\}$ is empty, then the intersection over all $n \leq N$ would be empty, and that is not necessarily true.

For $B$, $f_n(\omega)\in[a,b]$ for infinitely many $n$ means: $$\exists \,N < \infty\,\,\forall \,\,n\geq N \text{ such that } f_n(\omega)\in [a,b]$$ in set it is: $$\bigcap_{N\in\mathbb{N}}\bigcup_{n\geq N} \{\omega:f_n(\omega) \in [a,b]\}$$

Are my thoughts correct? I apologize if I made any stupid mistakes. Since I don't have a solid background on sets or logic, any help would be very much appreciated!

  • 2
    $\begingroup$ Your characterization of $A$ is wrong; the statement you've written says 'the set of $n$ with $f_n(\omega)\in[a,b]$ is non-empty', but that's different from finite. What you want to say is that there's some $N_0$ such that for all $n\gt N_0$, $f_n(\omega)\not\in[a,b]$; the set of $n$ with $f_n(\omega)\in[a,b]$ is bounded. $\endgroup$ Aug 4, 2021 at 16:49
  • $\begingroup$ Thank you, I think it is working with $\{f_n(\omega)\in [a,b]\}^C$? But it feels like if we do $n>N_0,f_n(\omega)\notin [a,b]$, that says all but finitely many $n$, $f_n(\omega) \notin [a,b]$? Is it good enough? $\endgroup$
    – Felicks
    Aug 4, 2021 at 16:56
  • 1
    $\begingroup$ Notice that you are basically asking whether or not $\limsup f_{n}$ and $\liminf f_{n}$ are measurable. This is true: as shown in the answers below, it follows because there is a compatible definition of "limsup" and "liminf" of sets. See here for a formal definition of "limsup" and "liminf" of sets: en.wikipedia.org/wiki/Set-theoretic_limit $\endgroup$
    – user711689
    Aug 4, 2021 at 17:47

2 Answers 2



For some $N$, for all $n$, if $n\geq N$, then $f_{n}(\omega)\notin[a,b]$.

$f_{n}(\omega)\notin[a,b]$ is equivalent to $f_{n}(\omega)\in(-\infty,a)\cup(b,\infty)$.

The set becomes \begin{align*} \bigcup_{N=1}^{\infty}\bigcap_{n\geq N}f_{n}^{-1}(-\infty,a)\cup f_{n}^{-1}(b,\infty). \end{align*}


Hints: For each $n$, let $C_{n}=\{\omega\mid f_{n}(\omega)\in[a,b]\}$, which is measurable.

Note that $A=\cup_{n\in\mathbb{N}}\cap_{k\geq n}C_{k}^{c}$ and $B=\cap_{n\in\mathbb{N}}\cup_{k\geq n}C_{k}$. (Here, I assume that the word "for finitely many $n$" includes the case "for no $n$".)


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