Prove there exists a measurable subset $E' \subset E$ such that $f$ is almost everywhere of one sign on $E'$.

Question

Let $f \in L\left(\mathbb{R}\right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E \subset \mathbb{R}$ such that $m(E) > 0$ and $f(x) \neq 0$ for all $x \in E$. Prove there exists a measurable subset $E' \subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.

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My (weak) attempt:

Let $E_\alpha = \{E \setminus A: m(E) = m(E \setminus A)\}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.

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I think all I need is a hint to start it.

Answer 1

Hint: take $P = \{f > 0\}$ and $N= \{f < 0\}$. Now,

$$ E = (E \cap P) \sqcup (E \cap N). $$

Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?