Interchange intergration and summation From Fernique theorem we know that if $\mu$ is a gaussian measure in a separable Banach space X we  will have that 
$\int_{X}e^{a\left \| x \right \|^{2}}\mu(dx)<\infty$
We can also see
$\int_{X}\sum_{n=0}^{\infty}a^{n}\frac{\left \| x \right \|^{2n}}{n!}\mu(dx)<\infty$
I saw on the internet that we can interchange this sum with the intergral because 
1)$f_{n}=a^{n}\frac{\left \| x \right \|^{2n}}{n!} \geq 0$
2)This intergral is finite so we can interchange
I would like to know if the previous assumpions are correct.
Also can someone give me an example were we use dominated convergence in order to show that we can interchange this integration and summation ??
 A: *

*The line $f_n=a^n\frac{||x||^{2n}}{n!}\geq 0$ hints toward a classical application of the monotone convergence theorem: suppose $f_n$, $n\in\mathbb{N}$, are nonnegative (measurable) functions. Then, $$\int_X\sum_{n=1}^\infty f_n(x)\, \mu(dx)=\sum_{n=1}^\infty \int_X f_n(x)\, \mu(dx),$$ regardless of whether the integrals or their sum actually converges to a finite value. Implicitly, the equality means that if one side is $+\infty$, then the other is as well (note that the Lebesgue integral of any nonnegative measurable function is well defined as an element of $[0,+\infty]$).
The proof of this equality is as follows: setting $s_N:=\sum_{n=1}^N f_n$, we obtain a measurable, increasing sequence of functions (in the sense that $s_N\leq s_{N+1}$), since the functions $f_n$ are all nonnegative. Now, we apply the monotone convergence theorem to this sequence and conclude $$\int_X\sum_{n=1}^\infty f_n(x)\, \mu(dx)=\int_X\lim_{N\to\infty}s_N(x)\, \mu(dx)\stackrel{(1)}{=}\lim_{N\to\infty}\int_X s_N(x)\, \mu(dx)\stackrel{(2)}{=}\sum_{n=1}^\infty \int_X f_n(x)\, \mu(dx),$$ where (1) is the monotone convergence theorem and (2) follows from the linearity of the integral.

*The line about the finiteness of the integral hints toward the dominated convergence theorem. Because of the special structure of the functions $f_n$, we can obtain the equality (1) in the above computation via the dominated convergence theorem instead of the monotone convergence theorem, thus proving the same claim: we want to interchange limit and integration for the $s_N$, so we need to find an integrable bound for $|s_N|$. We have $|s_N(x)|=s_N(x)\leq e^{a||x||^2}$, which is integrable by the theorem you quoted and of course independent of $N$. Hence, we can apply the dominated convergence theorem to conclude (1). The rest of the computation then proceeds in exactly the same way. 

