How to prove that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere? Let $f:(a,b)\rightarrow\mathbb{R}$.
The statement to prove is that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere.
So if it wouldn't be true then we would have $\forall x \in (a,b):\int_{(a,x)}f_+d\lambda=\int_{(a,x)}f_-d\lambda\neq0$, so every interval contains positive and negative values of the function. It is possible to construct such an $f$ so my guess would be that this function is not Lebesgue measurable and we can't take this integral.
Thanks in advance!

Suppose for some $x \in A$ where $\lambda(A) > 0$ and $f(x) \neq 0$, but $\int_{(a,x)}fd\lambda = 0 $ (Assumption for contradiction)
Claim : 
$$
\int_{0}^{\infty} \lambda [ g \geq t]m(dt) = \int g \;d \lambda
$$
where $m$ is lebesgue measure. (You can prove it by Fubini theorem)
Then, for $I$ characteristic function, and $g = fI_{(a,x)}$
$$
0 = \int_{(a,x)}fd\lambda = \int gd\lambda = \int_{0}^{\infty} \lambda (g \geq t)m(dt) = 0
$$
Then you can see clearly this is contradiction by the above integral cannot be zero.
