# 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.

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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.

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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.

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Sorry, I should have mentioned that $f$ goes to $\mathbb{R}$ and not $[0,\infty)$. I have already edited it in question. – haemhweg Dec 5 '13 at 9:13
My claim hold for general $f$ haemhweg – Block Jeong Dec 5 '13 at 9:14
Didn't you mean $\lambda[f\geq t]$ in first integral? And then if $f = -1$ we always have $\lambda[f\geq t] = 0$ for all $t \geq 0$, so we would get $\int_0^\infty -1 d\lambda$ = 0. – haemhweg Dec 5 '13 at 9:21
Oh..! you are right... I am not sure but what about decomposing $f$ to two positive function $f^+$ and $f^-$? – Block Jeong Dec 5 '13 at 9:36