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?


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.


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.

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

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