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Subset of the preimage of a semicontinuous real function is Borel

A real function $f$ on the line is upper semi-continuous at $x$, if for each $\epsilon > 0$, there exists $\delta > 0$ such that $|x-y|<\delta$ implies that $f(y) < f(x) + \epsilon$. Check that if $f$ is everywhere upper semi-continuous, then it is measurable.

I could not do this question.

Thanks and regards.

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marked as duplicate by t.b., Rasmus, Asaf Karagila, J. M., Srivatsan Nov 18 '11 at 20:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 7 down vote accepted

I'm going to show that the set $U = \{x \,:\,f(x) \lt t\}$ is open for each $t \in \mathbb{R}$ (so $f$ is indeed measurable):

We have $x \in U$ if and only if $f(x) \lt t$. Fix $x \in U$. Take $\varepsilon = t-f(x)$. Your definition of upper semi-continuity yields a $\delta$ such that $|x-y| \lt \delta$ implies $f(y) \lt f(x) + \varepsilon = t$, so $|x-y| \lt \delta$ implies $y \in U$.

See also this related question.

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Because I'm lazy... Seriously: the very first lemma one proves in measure theory is that it suffices to check that $\{x : f(x) \mathrel{\Box} t\}$ is measurable where $\Box$ can be any one of $\leq, \lt, \geq, \gt$ and I feel that it is safe to assume that OP knows that the Borel $\sigma$-algebra is generated by the open sets from the other questions. – t.b. Nov 18 '11 at 19:32
Yes, true. But that doesn't mean he remembers it... anyway, will remove my comment. – Rudy the Reindeer Nov 18 '11 at 19:38

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