Show that $\int_a^b f(u) \int_a^u f(v) dv du=\int_a^b f(u) \int_u^b f(v) dv du$

For $ u\in\left[a,b\right] $ define a function $ f_{u}:\left[a,b\right]\rightarrow\mathbb{R} $ as following : $ f_{u}\left(v\right)=\left\lbrace\begin{aligned}f\left(u\right)f\left(v\right),\ \ \ \ &\textrm{If }a\leq v\leq u\\ 0\ \ \ \ \ \ \ \ ,\ \ \ \ &\textrm{If }u\leq v\leq b\end{aligned}\right. \cdot $

Fubini's theorem allows us to write the following : $$ \int_{a}^{b}{\int_{a}^{b}{f_{u}\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int_{a}^{b}{\int_{a}^{b}{f_{u}\left(v\right)\mathrm{d}u}\,\mathrm{d}v} $$

Since the left side can be written as : $ \int\limits_{a}^{b}{\int\limits_{a}^{b}{f_{u}\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int\limits_{a}^{b}{\int\limits_{a}^{u}{f\left(u\right)f\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int\limits_{a}^{b}{f\left(u\right)\int\limits_{a}^{u}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u} \cdot $

And the right side can be written as : $ \int\limits_{a}^{b}{\int\limits_{a}^{b}{f_{u}\left(v\right)\mathrm{d}u}\,\mathrm{d}v}=\int\limits_{a}^{b}{\int\limits_{v}^{b}{f\left(u\right)f\left(v\right)\mathrm{d}u}\,\mathrm{d}v}=\int\limits_{a}^{b}{f\left(v\right)\int\limits_{v}^{b}{f\left(u\right)\mathrm{d}u}\,\mathrm{d}v} \cdot $

We have : $$ \int_{a}^{b}{f\left(u\right)\int_{a}^{u}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int_{a}^{b}{f\left(v\right)\int_{v}^{b}{f\left(u\right)\mathrm{d}u}\,\mathrm{d}v} $$

We can substitute $ \left(u,v\right) $ to $ \left(v,u\right) $ in the right side to get : $$ \int_{a}^{b}{f\left(u\right)\int_{a}^{u}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int_{a}^{b}{f\left(u\right)\int_{u}^{b}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u} $$