$\int_{0}^{x} f(z)dz =0$ for all $x \in A$ where $\lambda([0,\infty) \setminus A) =0$, i.e. a.e., implies $f$ is zero a.e.

Question

This question is similar to the question appearing here. The difference in my question is that I don't allow the integral $\int_{0}^{x} f(z) dz =0$ everywhere, but only almost everywhere. $\lambda$ will denote the Lebesgue measure.

Q: Assume $f \in L^1(\lambda)$, $\int_{0}^{x} f(z)dz =0$ for all $x \in A$ where $\lambda([0,\infty) \setminus A) =0$, i.e. a.e., implies $f$ is zero a.e.?

If you know this is true from the literature you can just link me the reference.

Answer 1

Since $f$ is integrable, $F(x)=\int_0^x f(z) dz$ is continuous (see for instance the answers to this post). You are assuming that $F(x) = 0$ almost everywhere but, being continuous, it will be zero everywhere.