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?

Question

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!

Answer 1

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.

Answer 2

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.