$\int_{0}^{x} f(z)dz =0$ 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.

Complement of a set of measure $0$ is dense. That is elementary and you should provide some context to your question. — geetha290krm, Apr 19 '23 at 08:36
$\int_0^x f(z) dz$ is a continuous function that is zero almost everywhere, hence it is identically zero. — PierreCarre, Apr 19 '23 at 08:36

score 6 · Accepted Answer · answered Apr 19 '23 at 08:41

6

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.

answered Apr 19 '23 at 08:41

PierreCarre

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

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