# $(f_n)$ Borel measurable implies $\sup_n f_n$ and $\inf_n f_n$ Borel measurable

Suppose $(f_n)$ is a sequence of Borel measurable functions. Show that both $\sup_nf_n$ and $\inf_nf_n$ are Borel measurable.

Attempt:

Suppose $(f_n)$ is a sequence of Borel measurable functions. That is, $f^{-1}_n((\alpha,\infty])$ is a Borel set for all $\alpha\in\overline{\mathbb{R}}$. (where $\overline{\mathbb{R}}=[-\infty,\infty]$, the extended real numbers)

Then $\forall\alpha\in\overline{\mathbb{R}}, \{\sup_nf_n>\alpha\}=\cup_{n=1}^{\infty}\{f_n>\alpha\}=\cup f^{-1}_n((\alpha,\infty])$.

Similarly for the infimum, I get that $\{\inf_n f_n\geq \alpha\}=\cap_{n=1}^{\infty}\{f_n\geq \alpha\}=\cap_{n=1}^{\infty}f^{-1}_n((\alpha,\infty])$

If I was working in $\mathbb{R}=(-\infty,\infty)$ instead of $\overline{\mathbb{R}}=[-\infty,\infty]$, then I could conclude that both are Borel sets right away since they were unions and intersections of open sets, but since my intervals are half closed, I'm not sure how to conclude that these intervals are Borel sets. Can someone provide a hint? Thanks.

## 1 Answer

Even though it is a half closed interval, $(\alpha,\infty]$ is an open set; one way to see this is that its complement, $[-\infty,\alpha]$, is closed. Thus, you are done.

• Of course. Thanks. – Sujaan Kunalan Mar 2 '15 at 5:38