$f_n$ measureable $\implies$ $m\left(\bigcap_{k=0}^\infty F_k\right)=0$

Question

I have that $f_n$ is measurable on a finite measure space.

Define $F_k=\{\omega:|f_n|>k \}$

$F_k$ are measurable and have the property $F_1 \supseteq F_2\supseteq\cdots$

Can I then claim that $m\left(\bigcap_{n=1}^\infty F_n\right) = 0$?

Answer 1

If $f$ is real valued, $\cap F_k$ is empty. so its measure is zero.

Answer 2

If $f$ maps to $\mathbb{R}\cup \{ \pm\infty\}$, the set is still measurable as this intersection is the preimage of the point $$ +\infty $$ which is measurable, since it is closed and thus Borel.

In order for the statement that this set be measure zero, you need to restrict your attention to functions which attain only real (and not extended real) values.

Answer 3

Assuming you meant $$F_k = \{\omega \;:\; |f_{k}(\omega)| > k\},$$ then this is true, even if $f$ is not measurable (if $f: A \to \mathbb{R}$). We need only note that for any fixed point $a$, $f(a)$ will be less than $k$ once $k$ gets large enough, so it will be excluded from $F_k$ for large enough $k$. So $$ \bigcap_{k=1}^\infty F_k = \varnothing. $$