Use Fubini's Theorem to verify the integration by parts formula, and justify that the hypotheses of Fubini's Theorem are satisfied. Let $f$ and $g$ be Lebesgue integrable real-valued functions on $[0,1]$, and define
$$F(x) = \int_0^x f(t)dt, \quad \text{and} \quad G(x) = \int_0^x g(t) dt.$$
Use Fubini's Theorem to verify the integration by parts formula
$$\int_0^1 F(x)g(x)dx = F(1)G(1) - \int_0^1 f(x)G(x)dx.$$
Justify that the hypotheses of Fubini's Theorem are satisfied.
This one seems straightforward to us, but for some reason we can't get anything nice to appear.
Here are our attempts for the left-hand side and the right-hand side respectively after the hint:
$$\begin{aligned} \int_0^1 F(x)g(x)dx &amp;= \int_0^1 \int_0^x f(t)dtg(x)dx\\
&amp;= \int_0^1\int_0^1 1_{[0,x]}f(t)g(x)dtdx\\
&amp;= \int_0^1 f(t) \int_0^1 1_{[0,x]}g(x)dxdt. \end{aligned}$$
$$\begin{aligned} F(1)G(1) - \int_0^1 f(x)G(x)dx &amp;= F(1)G(1) - \int_0^1 f(x)\int_0^x g(t)dtdx\\
&amp;= F(1)G(1) - \int_0^1\int_0^1 1_{[0,x]}f(x)g(t)dtdx\\
&amp;= F(1)G(1) - \int_0^1g(t)\int_0^1 1_{[0,x]}f(x)dxdt.\end{aligned}$$
The first few steps just involves rewriting the terms as they are defined and moving constants inside or outside of an integral as needed.
$$
\begin{align*}F(1)G(1)&amp;=\Big(\int_0^1f(t)dt\Big)\Big(\int_0^1 g(x)dx\Big)= \int_0^1g(x)\Big(\int_0^1 f(t)dt\Big)dx\\
&amp;= \int_0^1g(x)\Big(\int_0^x f(t)dt\Big)dx + \int_0^1g(x)\Big(\int_x^1 f(t)dt\Big)dx\\
&amp;= \int_0^1 g(x)F(x)dx + \int_0^1g(x)\Big(\int_x^1 f(t)dt\Big)dx\end{align*}$$
Now we use Fubini's Theorem to show that the second integral takes the appropriate form. Well
$$\begin{align*}\int_0^1 g(x)\Big(\int_x^1 f(t)dt\Big)dx &amp;= \int_0^1g(x)\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)f(t)dt dx\\
&amp;= \int_0^1\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)f(t)dt dx\\
&amp;= \int_0^1\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)f(t)dxdt\\
&amp;= \int_0^1f(t)\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)dx dt \\ 
&amp;= \int_0^1f(t)\Big(\int_0^1 1_{[0\leq x\leq t]}(x)g(x)dx\Big) dt \\ 
&amp;= \int_0^1f(t)\Big(\int_0^t g(x)dx\Big) dt \\ 
&amp;= \int_0^1 f(t)G(t)dt\end{align*}$$
Hence after a change of variables we have that
$$F(1)G(1) = \int_0^1 g(x)F(x)dx + \int_0^1 f(x)G(x)dx $$
$$f(x)=u(x)v(x)$$
$$\frac{df}{dx}=\frac{du}{dx}v+\frac{dv}{dx}u$$
$$df=v.du+u.dv$$
$$\int df=\int v.du+\int u.dv$$
$$\int u.dv=f-\int v.du$$
$$\int u.dv=uv-\int v.du$$
If you don't have to use Fubini's theorem