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!

Consider arbitrary $x \in (a,b)$. Then for every suitably small $\epsilon > 0$, 
Let, $$ I_\epsilon : = \int_{((x-\epsilon),(x + \epsilon))} f d\lambda = \int_{(a,x+\epsilon)} f d\lambda - \int_{(a,x - \epsilon)} f d\lambda = 0 - 0 = 0. $$
Also, $$ \lim_{\epsilon \rightarrow 0} \frac{I_{\epsilon}}{2\epsilon} = \lim_{\epsilon \rightarrow 0} 0 = f(x) \: (\lambda) \mbox{ - a.e. by Lebesgue differentiation theorem }. $$
Hence we are done.   
