Argument $\int_{-\infty}^{t}\int_{-\infty}^{x_2}f_{X_1,X_2}dx_1dx_2=\int_{-\infty}^{t}\int_{-\infty}^{x_1}f_{X_1,X_2}dx_2dx_1$

Question

Right now I have two integrals $$ \int_{-\infty}^{t} \int_{-\infty}^{x_2} f_{X_1,X_2}(x_1,x_2) \, dx_1 dx_2 = \int_{-\infty}^{t} \int_{-\infty}^{x_1} f_{X_1,X_2}(x_1,x_2) \, dx_2 dx_1 $$ with the property that $f_{X_1,X_2}(x_1,x_2)=f_{X_1,X_2}(x_2,x_1)$ for all possible $(x_1,x_2)$. This intuitively make sense if I just draw the area bounded by these two integrals, but I am writing a proof right now and really don't know how can I formally explain it. Some hint would be really appreciated!

Answer 1

By swapping the names of the variables $(x_1, x_2) \leftrightarrow (x_2, x_1)$ in the first expression, you almost get the second equation. But the arguments of the function are now swapped. Fortunately, you have the hypothesis that your function is symmetric in those arguments, so $$ \def\cb#1{\color{blue}{#1}} \def\cg#1{\color{green}{#1}} \begin{align*} \cb{\int_{-\infty}^{t}} \cg{\int_{-\infty}^{x_2}} f_{X_1,X_2}(\cg{x_1}, \cb{x_2}) \, \cg{dx_1} \cb{dx_2} &= \cg{\int_{-\infty}^{t}} \cb{\int_{-\infty}^{x_1}} f_{X_1,X_2}(\cb{x_2}, \cg{x_1}) \, \cb{dx_2} \cg{dx_1} \\ &= \cg{\int_{-\infty}^{t}} \cb{\int_{-\infty}^{x_1}} f_{X_1,X_2}(\cg{x_1}, \cb{x_2}) \, \cb{dx_2} \cg{dx_1}. \end{align*} $$