Double integral variable substitution 
Let $(a,b)\in (0,1)$ and let $T>0$. Consider the following integral:
$$
\iint_{\Gamma }f(x)g(x+t)\,dt\,dx,
$$
where
$
\Gamma =\left\{ (t,x)\in (0,T)\times (0,1):t+x\in (0,1)\right\} .
$

Let $s=x+t$, then $s\in (0,1)$ and $s-t\in (0,1)$, $t\in (0,T)$.
Am I right?
$$
\iint_{\Gamma }f(x)g(x+t)\,dt\,dx={\int_{0}^{1}\int_{0}^{\min(1-x,T)}}f(x)g(t+x)\,dt\,dx={\int_{0}^{1}\int_{0}^{\min(s,T)}}f(s-t)g(s)\,dt\,ds,
$$
Thank you.

The substitution is fine. Here is a variation with an intermediate step.

We obtain
\begin{align*}
\color{blue}{\int_{x=0}^1}&\color{blue}{\int_{t=0}^{\min(1-x,T)}f(x)g(t+x)\,dt\,dx}\\
&=\int_{t=0}^{\min(T,1)}\int_{x=0}^{1-t}f(x)g(t+x)\,dx\,dt\tag{1}\\
&=\int_{t=0}^{\min{(T,1)}}\int_{s=t}^{1}f(s-t)g(s)\,ds\,dt\tag{2}\\
&\,\,\color{blue}{=\int_{s=0}^{1}\int_{t=0}^{\min{(s,T)}}f(s-t)g(s)\,dt\,ds}\tag{3}
\end{align*}
in accordance with OPs calculation.

Comment:

*

*In (1) we change the order of integration.


*In (2) we substitute $s=x+t, ds=dx$ and set upper and lower limit of the integral accordingly.


*In (3) we change the order of integration once more.
