I am doing Exercise 1.6.51 in Terence Tao's introduction to measure theory.
Let $F : [a, b]\to\mathbb R$ be an absolutely continuous function, and let $\phi: [a, b]\to\mathbb R$ be a continuously differentiable function supported in a compact subset of $(a, b)$. Show that $\int_{[a,b]}F'\phi(x) \,dx = -\int_{[a,b]} F\phi' (x)\, dx$.
I made a "proof", as follows, that only requires $\phi$ to be absolutely continuous, instead of being continuously differentiable (which is stronger than absolute continuity in this case). I believe there's something wrong, but I cannot find it out. I would be grateful if someone can help.
Below, by integral I mean Lebesgue integral.
Since $F$ and $\phi$ are absolutely continuous on $[a,b]$, we know that $F$ and $\phi$ are integrable, and that $F'$ and $\phi'$ are integrable as functions defined almost everywhere on $[a,b]$.
So both $F'\phi$ and $F\phi'$ are integrable. Then, we need only show $$\int_{[a,b]}(F'\phi+F\phi')(x)\,dx=0.$$
At any $x$ where $F$ and $\phi$ are differentiable, we know that $F\phi$ is differentiable with $(F\phi)'(x)=(F'\phi+F\phi')(x)$. And almost every point in $[a,b]$ is such a point since $F$ and $\phi$ are absolutely continuous. So we need only show $$\int_{[a,b]}(F\phi)'(x)\,dx=0.$$
Since $F$ and $\phi$ are absolutely continuous on $[a,b]$, so is $F\phi$. So the second fundamental theorem gives $$\int_{[a,b]}(F\phi)'(x)\,dx=(F\phi)(b)-(F\phi)(a),$$ which is indeed $0$ since the support of $\phi$ is a subset of $(a,b)$.