$\iint f(x)g(y)dxdy =\int f(x)dx \int g(y)dy $ . How could this be proven I tried this with Fubini, so on the LHS we can think of $g(y)$ as constant and take it out of the integral and in the following step we can think ok $f(x)$ as a constant such. In other word $\int g(y) \big[\int f(x)dx\big]dy=\big[\int f(x)dx\big]\int g(y)dy$.
But I do not know how to come from $\int f(x)dx \int g(y)dy \Rightarrow\iint f(x)g(y)dxdy $ pls help
Both expressions are the same. Note that
$$\iint f(x) g(y)\, dx\, dy=\int g(y)\left(\int f(x)\, dx\right)\, dy$$
because $g(y)$ is a constant for each $y$ in the inner integral. Now call $A:=\int f(x)\, dx$, then $A$ is a constant, assuming that the integral converges, and from above we knows that
$$\int g(y)A\, dy=A\int g(y)\, dy=\left(\int f(x)\, dx\right)\left(\int g(y)\, dy\right)$$
Revert the steps for the other direction of the proof (start by naming $A:=\int f(x)\, dx$ and put it inside of the other integral).