I was proving this theorem of Measure theory -
"Suppose $f$ is integrable and $\int_{a}^{x} f = 0 \forall x \in [a,b]$ then $f = 0$ almost everywhere."
Here is what I tried in the proof here -
Let us consider set $E =\{ x | f(x) > 0\}$ is of non-zero measure.
$E$ is measurable as $f^{-1}(0,\infty)$ is measurable , applying one of the characterizations of measurable sets , states that $\forall \epsilon > 0$ there is a closed set $F \subset E$ suchthat $m(E|F) < \epsilon$ and $m(F) \neq 0$.
so $[a,b]/F$ is open let us denote it by $O$.and it can be expressed as a countable disjoint union of open sets so $O =\cup_{n \geq 1} (a_{n},b_{n})$.
Now $\int_{a}^{b}f = 0$ as it will be true for $x = b$
$\int_{F}f + \int_{O}f = 0$ $\implies \int_{O}f \neq 0$
So,$\int_{\cup_{(a_{n},b_{n})}}f = \sum_{n \geq 1}\int_{(a_{n},b_{n})}f \neq0 $
Implying there exist $N$ such that $\int_{(a_{N},b_{N})}f \neq 0$ $\int_{a}^{b_{N}}f - \int_{a}^{a_{N}}f \neq 0$
$\implies \int_{a}^{b_{N}}f \neq 0 $ or $\int_{a}^{a_{N}}f \neq 0$ Which is a contradiction as to our integral $\int_{a}^{x}f = 0 \forall x \in [a,b]$ .
Now we conclude due to the contradiction that $E$ is a set of zero measure and I think that similar can be done for $f < 0$ ,
now how do I proceed in order to show that $f = 0$ almost everywhere
Any help is great!
,[almost everywhere part is proved above I think by proving that set of points where $f\neq 0$ is of measure zero.]