Let $X$ be a set, $\mathcal{A}$ a sigma-algebra on $X$ and $\mathcal{B}$ the Borel algebra on $\mathbb{R}$.
Let $\lbrace f_n:(X,\mathcal{A})\to (\mathbb{R},\mathcal{B})\rbrace_{n\in\mathbb{Z_+}}$ be a sequence of measurable functions, such that a $K>0$ exists with $|f_n(x)|<K$ for all $n \in \mathbb{Z_+}, x\in X$.
How to show, that $f:(X, \mathcal{A})\to(\mathbb{R},\mathcal{B}), x \mapsto \limsup\limits_{n\to\infty}f_n(x)$ is also measurable?
I used the following steps:
$\lbrace \sup\limits_{n\in \mathbb{N}}f_n < K\rbrace=\bigcap\limits_{n=1}^{\infty} \lbrace f_n<K\rbrace$ so the supremum is measurable.
Also, $\limsup\limits_{n\to\infty}f_n$=$\inf\limits_{n\in\mathbb{N}}$$\sup\limits_{l\geq n}f_l$ so the limit superior is measurable.
Can it be done like that or is there another way to show it right?
Best Answer
Notice that $$\left\{x:\sup_{n\in \Bbb N}f_n(x)> \alpha\right\}=\bigcup_{n=1}^{\infty}\left\{x:f_n(x)>\alpha\right\}$$ for each $\alpha\in \Bbb R$. Hence, $\displaystyle\sup_{n\in \Bbb N} f_n$ is a measurable function.
Now, since $$\inf_{n\in \Bbb N} g_n=-\sup_{n\in \Bbb N} (-g_n),$$ we can say that $\displaystyle\inf_{n\in \Bbb N} g_n$ is also measurable whenever $g_n$ are also measurable.
Finally, $$\limsup_{n\rightarrow \infty}f_n=\inf_{n\in \Bbb N}\left(\sup_{i≥n}f_i\right).$$ So that $\displaystyle\limsup_{n\rightarrow \infty}\ f_n$ is also measurable.