The Lebesgue integrable function $f$ on $[a,b]$ satisfies the following condition:
$\int_a^xf d\mu=0$ for any $x \in [a,b]$
I want to prove that $f=0$ a.e., so I tried $f=f_+-f_-$ , where $f_+=\max \{f,0\} $ and $f_-=-\min\{f,0\}$, and I tried to prove that the integration on $[a,b]$ of $f_+$ and $f_-$ is both zero, but it doesn't work. Is there any other approach to the proof? I already know that $\int_x^yf d\mu=0$ for any $[x,y] \subset [a,b]$.
p.s. I'm just a beginner of real analysis, so I don't know anything but basic measure theory and lebesgue integration.
Denote $F(x) = \int_a^x f(y) \textrm{d}\mu$, then we have $F(x) = 0$ for all $x \in [a,b]$ and so $F'= 0$.
So by Lebesgue Differentiation Theorem we have a.s. $f = F' = 0$.