Variable substitution in double integral 
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?
$$
\iint_{\Gamma }f(x)g(x+t)\,dt\,dx={\int_{0}^{1}\int_{0}^{\min(s,T)}}f(s-t)g(s)\,dt\,ds,
$$
Thank you.
 A: I think you're right!
Your domain $\Gamma$ is defined by the inequalities:
$$\left\lbrace\begin{matrix} 0 & < & x & < & 1 \\ 0 & < & t & < & T \\ 0 & < & x+t & < & 1 \end{matrix}\right. $$
You can translate these into inequalities with $(s,t)$, they then read:
$$\left\lbrace \begin{matrix} 0 & < & s-t & < & 1 \\ 0 & < & t & < & T \\ 0 & < & s & < & 1 \end{matrix} \right.$$
Now the first inequality can be decomposed as $t < s$ and $s<1+t$. Assuming the last two inequalities hold, $s<1+t$ is always true so it can be discarded without loosing any information. Formally:
$$\left\lbrace \begin{matrix} 0 & < & s-t & < & 1 \\ 0 & < & t & < & T \\ 0 & < & s & < & 1 \end{matrix} \right. \Longleftrightarrow \left\lbrace \begin{matrix} 
 & & t & < & s \\ 0 & < & t & < & T \\ 0 & < & s & < & 1 \end{matrix}\right.$$
The last system can be rewritten as:
$$ \left\lbrace \begin{matrix} 0 & < & t & < & \min (s,T) \\ 0 & < & s & < & 1 \end{matrix}\right.$$
and this gives you your new domain, as we have shown:
$$ \left\lbrace\begin{matrix} 0 & < & t & < & \min (1-x,T) \\ 0 & < & x & < & 1 \end{matrix} \right. \Longleftrightarrow \left\lbrace \begin{matrix} 0 & < & t & < & \min (s,T) \\ 0 & < & s & < & 1 \end{matrix}\right.$$
