# Subset of the preimage of a semicontinuous real function is Borel

I'm in a jam with this problem:

Let $f: \mathbb{R} \to [-\infty,\infty]$ be lower semicontinuous, and let $A = \{ x:f(x)\ge a \}$. Is $A$ necessarily a Borel set in $\mathbb{R}$?

I've actually managed to prove that if $A$ has no excluded points, then it is Borel, and it's easy to use that to show that it's also Borel if there are only countably many such points, but I wasn't able to do so, so I guess a hint in this direction would be appreciated. I do feel, however, that there's a much less roundabout way to resolve this problem, so if someone could gently hint me to the right direction I would be most grateful.

Observe that $\{x\,:\,f(x) \geq a\} = \bigcap_{n=1}^{\infty} \left\{x \,:\,f(x) \gt a-\frac{1}{n}\right\}$, and the latter is a countable intersection of open sets by definition of lower semicontinuity.