Question about integral identity $\int_V d A g(A)= \int_U d\alpha \int_V dA f(\alpha,A)g(A)$ when $\int_U d \alpha f(\alpha,A)=1$ If$$\int_U d \alpha f(\alpha,A)=1$$ for $\forall A\in\mathbb{R}$
Does it mean 
$$\int_V d A  g(A)= \int_U d\alpha \int_V dA f(\alpha,A)g(A)$$
It there any counterexample? Or in which requirement does above statement hold? This kind of trick are always meet in physics and I'm really confused about it.
 A: 1. Interchanging the order of integration - and generally interchanging two limiting operators - is not always justified. Here is a tweak of a famous example: Let
$$ f(\alpha, A) = \frac{(A^2+1)(A^2 - \alpha^2)}{(A^2 + \alpha^2)^2}. $$
Then for any $A \in (0, 1)$ we have
$$ \int_{0}^{1} d\alpha \, f(\alpha, A) = 1 $$
and thus
$$ \int_{0}^{1} dA \int_{0}^{1} d\alpha \, f(\alpha, A) = 1. $$
On the other hand, it is not hard to check that
$$ \int_{0}^{1}d\alpha \int_{0}^{1} dA \, f(\alpha, A)
= \int_{0}^{1} d\alpha \left( \frac{2\alpha^2}{1+\alpha^2} - 2\alpha \operatorname{arccot}(\alpha) \right) = 1 - \frac{\pi}{2}. $$
2. Of course, interchanging the order of integrations is such an important idea that many theorems are developed to guarantee this under certain conditions. The Fubini-Tonelli theorem is probably the most famous of this kind. Ignoring some technical details (especially about measurability), they are stated as follows:

Tonelli's Theorem. If $f$ is a non-negative function on $U\times V$, then the following equality holds unconditionally:
$$ \int_{U}dx \int_{V} dy \, f(x,y)
= \int_{V}dy \int_{U} dx \, f(x,y)
= \iint_{U\times V} d(x,y) \, f(x, y).$$

Here, the first two are iterated integrals and the last one is a double integral. And

Fubini's Theorem. If $\int_{U\times V} d(x,y) \, |f(x,y)| < \infty$, then
$$ \int_{U}dx \int_{V} dy \, f(x,y)
= \int_{V}dy \int_{U} dx \, f(x,y)
= \iint_{U\times V} d(x,y) \, f(x, y). $$

Notice that we can check instead $\int_{U}dx \int_{V} dy \, |f(x,y)| < \infty$ or $\int_{V}dy \int_{U} dx \, |f(x,y)| < \infty$, thanks to the Tonelli's theorem.
