Partial integration for lebesgue integrable functions I want to show the following:
Let be two Lebesgue integrable functions given: $f,g:[a,b] \rightarrow \mathbb R$. We define the functions:
$$F,G: [a,b] \rightarrow \mathbb R : F(x)=\int_{[a,x]}^ \! f(t) \, dt, G(x)=\int_{[a,x]}^ \! g(t) \, dt. $$
Show that the rule for the partial integration holds also in this case, i.e. It holds:
$$\int_{[a,b]}^ \! F(x)g(x) \, dx =F(b)G(b)-\int_{[a,b]}^ \! f(x)G(x) \, dx$$
Hint: Use the theorem of Fubini. 
I tried a bit , but I didn't come to any solution.
 A: Let $\phi(s,t) = f(s)g(t)$.
Let $X_+ = \{ (s,t) \in [a,b]^2 \mid t>s \}$, $X_- = \{ (s,t) \in [a,b]^2 \mid t<s \}$, we see that $m ([a,b]^2 \setminus (X_+ \cup X_-) ) = 0$.
Then $\int_{[a,b]^2} \phi = F(b)G(b) = \int_{X_+} \phi + \int_{X_-} \phi$.
Now compute the latter two integrals using Fubini.
A: Taking $F(x)$ to be $\int_{[a,x]} f(t)\,dt$ and $G(y)$ to be $\int_{[a,y]} g(s)\,ds$, we have:
\begin{align}
F(b)G(b) & = \int_{[a,b]} f(x)\,dx\cdot\int_{[a,b]} g(y)\,dy = \int_{[a,b]} \left( \int_{[a,b]} f(x)\,dx \right) g(y)\,dy \\[8pt]
& = \int_{[a,b]}\left(\int_{[a,b]}f(x)g(y)\,dx\right) \,dy \overset{\text{Fubini}}= \int_{[a,b]\times[a,b]} f(x)g(y)\, d(x,y) \\[8pt]
& = \int_J f(x)g(y)\, d(x,y) + \int_K f(x)g(y)\, d(x,y)
\end{align}
where $J=\{(x,y) : a \le y\le x\le b\}$ and $K=\{(x,y): a\le x<y\le b\}$.
Then
\begin{align}
\int_J f(x)g(y)\,d(x,y) & = \int_{[a,b]\times[a,b]} f(x)g(y) 1_J(x,y) \, d(x,y) \\[8pt]
& \overset{\text{Fubini}}= \int_{[a,b]} \left(\int_{[a,b]} g(y) 1_J(x,y)\,dy\right) f(x)\,dx \\[8pt]
& = \int_{[a,b]} \left(\int_{[a,x]} g(y)\,dy\right) f(x)\,dx = \int_{[a,b]}G(x)f(x)\,dx.
\end{align}
Procede similarly with $\displaystyle\int_K$ and you've got it.
