In Conway's functional analysis book (Ed. 2) on page 306, $Tf=if'$ is defined on $D(T)=\{f\in AC[0,1]: f'\in L^2(0,1),~f(0)=f(1)=0\}$. In some part of the proof he has $f\in L^2(0,1)$ and he obtained $f(x)=\int_0^xg(t)dt$ (where $g\in L^2(0,1)$) only almost everywhere. Then, he says "we may assume that $f(x)=\int_0^xg(t)dt$ for all $x$". Could you please explain this "passage" to me?
Many thanks.
Math
$f\in AC[0,1]\subset L^2(0,1)$ but $f$ and $\int_0^x g(t)dt$ are equal up to a measurable set of measure 0 so they must be in the same equivalence class of $L^2(0,1)$ then we can choose $f$ simply as $\int_0^x g(t)dt$