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We are given a space X with $\sigma$-algebra (of subsets of X) $\mathbb{F}$ and a sequence of measurable (w.r.t $\mathbb{F}$) functions $f_n: X \rightarrow \mathbb{R}$ for $n \in \mathbb{N}$.
Now consider a set $$H = \{x \in X : \text{the sequence} \{f_n(x)\}^\infty_{n=1} \text{is unbounded from above and bounded from below} \}$$ .
We want to show that $H \in \mathbb{F}$.


My attempt or rather intuition:
$H$ can be written as intersection of
$$A = \{x\in X: \text{the sequence is bounded from below} \}$$

and $X-B$, where $$B = \{x\in X: \text{the sequence is bounded from above} \}$$
Now since $A$ and $X-B$ are "sort of"* inverse images of $[m,\infty]$ and $[-\infty,M]$ respectively ($m$ is the lower bound and $M$ is the upper bound) of functions $f_n$ which are measurable then H is "constructed" from sets that are in $\mathbb{F}$ using "legal" set operations that keep them inside $\mathbb{F}$.
* I wrote sort of because they are rather countable unions or intersections of inverse images (?)

Is this intuition correct? Could you add precision to this reasoning?

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Using your idea of writing $H = A \cup (X-B)$, one can proceed as follows: to show that $A$ is measurable, write $$ A = \bigcup_{k \in \mathbb{N}} \{ x: f_n(x) \geq -k \text{ for all } n \in \mathbb{N} \} = \bigcup_{k \in \mathbb{N} } \bigcap_{ n \in \mathbb{N}} \{x: f_n(x) \geq -k \}.$$ Since each function $f_n$ is measurable, each set $\{x: f_n(x) \geq -k \}$ is measurable. Taking the intersection over $n \in \mathbb{N}$ and the union over $k \in \mathbb{N}$ preserves measurability. A similar argument shows that your set $B$ is measurable.

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A different approach: Both $\limsup f_n, \liminf f_n$ are measurable functions from $X$ into $[-\infty,\infty].$ The set of interest equals

$$\{x\in X: \limsup f_n (x) = \infty\}\cap \{x\in X: \liminf f_n (x) > -\infty\},$$

which is a measurable set.

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  • $\begingroup$ Thanks a lot! That's way simpler... $\endgroup$ – Wojtek Maślakiewicz Jan 20 at 20:57

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