Discussion
Let \(f : [0, 1] \to \mathbb{R}\) be a function. Which one of the following is a sufficient condition for \(f\) to be Lebesgue measurable?
Answer: C
A sufficient condition is \(f\) is continuous almost everywhere on [0, 1].
If \(f\) is continuous outside a null set \(N\), then \(f\) restricted to \([0, 1] \setminus N\) is continuous, hence Borel measurable there. Every subset of \(N\) is Lebesgue measurable because Lebesgue measure is complete, so \(\{f \gt c\}\) is measurable for every \(c\).
The others fail. Take a non-measurable set \(V\) and \(f = 1\) on \(V\), \(-1\) off it: then \(|f| = 1\) is measurable and \(-1 \le f \le 1\) is squeezed between continuous functions, but \(f\) is not measurable. And \(f(x) = x\) on \(V\), \(-x\) off it, takes each value at most twice, so every level set is finite and measurable, yet \(f\) is not measurable.
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