Prove that $\int_{y=0}^{\infty} \int_{x:g(x) > y} f(x) \,dx \,dy = \int_{x:g(x)>0} \int_{y=0} ^{y=g(x)}\,dy$ Prove that $\int_{y=0}^\infty \int_{x:g(x) > y}  f(x) \,dx \,dy = \int_{x:g(x)>0} \int_{y=0}^{y=g(x)}\,dy ~f(x) \,dx$ where $f,g$ are continuous functions
Attempt: Let $I$ = $\int_{y=0}^\infty \int_{x:g(x) > y}  f(x) \,dx \,dy $. Now, we need to interchange the order of integration. The inner integral
$$\int_{x:g(x) > y}  f(x) \,dx$$
provides us a clue with its limits confined to $\{x:g(x) > y)\} ~$ i.e $~\{x: y<g(x) \} \implies $limits of $y$ will vary from $0$ to $g(x)$. Since the outer integral is over positive $y$ and $g(x) > y \implies g(x) > 0 ~\forall~x$ in our question.
Hence, the outer limit will be over $x : g(x) > 0$.

Although I have tried to prove the question in the above manner, this does not seem rigorous and solid proof in any way. Could someone please guide me towards more convincing proof? Will my argument serve as a proof?

 A: I'm not sure if this is any help to you, but in a similar situation I was told to use Indicator-functions, so what you are doing would roughly look like:
$$\int_{y=0}^{\infty} \int_{\{g(x)>y\}} f(x) \,dx \,dy = \int_{\mathbb{R}}\int_{\mathbb{R}} \mathbb{1}_{\{0<y\}}\cdot \mathbb{1}_{\{y < g(x)\}} \cdot f(x) \,dx \,dy \\= \int_{\mathbb{R}}\int_{\mathbb{R}} \mathbb{1}_{\{0<y<g(x)\}}\cdot \mathbb{1}_{\{0 < g(x)\}} \cdot f(x) \,dx \,dy $$
The thing you have check is that the product of the indicator functions before and after the second equality are actually the same function. Now if you change the order of integration, if you can, then you can pull out the second indicator function from the now inner integral since it doesn't depend on y, so you get:
$$\int_{\mathbb{R}} \mathbb{1}_{\{0 < g(x)\}} \cdot f(x) \int_{\mathbb{R}} \mathbb{1}_{\{0< y < g(x)\}} \cdot\,dy \,dx = \int_{\{g(x)>0\}}f(x) \int_{y = 0}^{g(x)}\, dy \,dx$$
I'm not sure though if this is any more rigorous than what you did, since it boils down to the same logic.
