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?


$\{x:\ sup_{n\in \Bbb N}f_n\ (x)> \alpha\}=\bigcup_{n=1}^{\infty}\{x:\ f_n(x)>\alpha\}$ for each $\alpha\in \Bbb R$. Hence $sup\ f_n$ is a measurable function.

Now since $inf_{n\in \Bbb N}\ g_n=-sup_{n\in \Bbb N}\ (-g_n) $ we can say $inf_{n\in \Bbb N}\ g_n$ is also measurable whenever $g_n$ are also measurable.

Finally $lim\ sup_{n\rightarrow \infty}\ f_n=inf_{n\in \Bbb N}(sup_{i≥n}\ f_i)$. So that $lim\ sup_{n\rightarrow \infty}\ f_n$ is also measurable.

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