Show that every upper semi-continuous real function is measurable

Question

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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.

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Answer 1

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.