Equivalence of these double integrals. I'm going over my old measure theory notes and saw something I wasn't quite sure about:
$$\int_{\mathbb R}\int_{\mathbb R} f(x-t)g(t)\ \mathsf dt\ \mathsf dx = \int_{\mathbb R} f(x)\mathsf dx \int_{\mathbb R} g(x)\mathsf dx. $$
I have "(change of variables)" written next to it, so I'm thinking it is really simple, but I am just not seeing it. I would actually prefer a hint to a full solution because I really should know this.
Edit: it turns out "change of variables" meant "Jacobian" and not "U-substitution" - so @Asvin's hint allowed me to write a solution (although if there is any error, please correct me). 
However, the context of the problem was that $f=\chi_A$, $g=\chi_B$ where $A$ and $B$ have finite Lebesgue measure, which justifies the use of Tonelli's theorem and so @Chu's solution is valid as well, as $0\leqslant \chi_A(x-t)\chi_B(t)\leqslant 1$ and so
$$\iint\limits_{\mathbb R^2} \chi_A(t-x)\chi_B(t)\ \mathsf d\mu(t\times x)\leqslant\mu(A)\mu(B)<\infty. $$
(in fact a simple computation yields that the value of the integral is in fact $\mu(A)\mu(B)$).
 A: By the Fubini's theorem, we can change the order of integration. We have that 
$\int_{\mathbb{R}}\int_{\mathbb{R}}f(x-t)g(t)dtdx=\int_{\mathbb{R}}\int_{\mathbb{R}}f(x-t)g(t)dxdt$
Now, let $u=x-t$. Then $du=dx$. By the change of variable theorem, 
$\int_{\mathbb{R}}f(x-t)dx = \int_{\mathbb{R}}f(u)du$. Substituting we have, 
$\int_{\mathbb{R}}\int_{\mathbb{R}}f(x-t)g(t)dxdt=\int_{\mathbb{R}}\int_{\mathbb{R}}f(u)g(t)dudt$. Since the variables $u$ and $t$ are independent, 
$\int_{\mathbb{R}}\int_{\mathbb{R}}f(u)g(t)dudt= \int_{\mathbb{R}}f(u)du\int_{\mathbb{R}}g(t)dt$. 
A: Define $\varphi:\mathbb R^2\to\mathbb R^2$ by $(x,t)\mapsto (x-t, t)=:(u,v)$, then the Jacobian is
$$|J| = \begin{vmatrix}\frac{\partial u}{\partial x}  & \frac{\partial u}{\partial t}\\ \frac{\partial v}{\partial x}  & \frac{\partial v}{\partial t}   \end{vmatrix} = \begin{vmatrix}1 & -1\\ 0 & 1\end{vmatrix} = 1. $$
It follows that
$$\int_{\mathbb R}\int_{\mathbb R}f(x-t)g(t)\ \mathsf dt\ \mathsf dx = \int_{\mathbb R}\int_{\mathbb R}|J|f(u)g(v)\ \mathsf d u\ \mathsf d v = \left(\int_{\mathbb R} f(x)\ \mathsf dx\right) \left(\int_{\mathbb R} g(x)\ \mathsf dx\right). $$
