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.

Added later: See also this related thread for an argument using a different definition of (upper) semicontinuity.


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