measurability of function that differs from continuous function

Question

If $f: \mathbb{R} \to \mathbb{R}$ is a function, and for all $\epsilon > 0$, $g_\epsilon: \mathbb{R} \to \mathbb{R}$ is a continuous function such that the set: $$ E_\epsilon = \{ x \in \mathbb{R} : f(x) \neq g_\epsilon(x)\} $$ is Lebesgue measurable with $m(E_\epsilon) < \epsilon$, show that $f$ is measurable.

I started by writing $E^c_\epsilon = \{ x : f-g = 0\} = (f-g)^{-1}(\{ 0\})$, which implies that $f-g$ is a measurable function, but then I have to show that $f$ is measurable from there which doesn't seem straight forward.

Answer 1

Two ingredients for the proof:

• A continuous function is measurable. If we change a function on sets with zero Lebesgue measure, it is still measurable, so a function that is continuous almost everywhere (apart from a set with measure zero) is measurable.
• Assume that there would be a set of positive Lebesgue measure on which $f$ is not continuous. By choosing $\varepsilon$ smaller than that measure, you will get a contradiction.

By the way, what you wrote about $f-g$ seems to be wrong. You must check that every preimage of a measurable set is measurable, this does not follow that easily here.