From the book I have been reading, it seems the following result is implicitly used:
If $f_{a}(x):=\sum_{n=0}^{\infty} s_{n}(a,x)$ converges uniformly with respect to (large) $a$, for example $\forall |x|<|a|$, and $\lim_{a \rightarrow \infty} s_{n}(a,x) = s_{n}(x)$ $\forall n$ then, \begin{eqnarray} \lim_{a \rightarrow \infty} \sum_{n=0}^{\infty} s_{n}(a,x) = \sum_{n=0}^{\infty} s_{n}(x). \end{eqnarray} I have not been able to find a reference for such a Theorem that justifies this kind of interchange with limit and sum and I could not think of a proof. Is there a Theorem known for this? Or, any idea how to prove the result?
Of course, the result does seem very reasonable. Especially in the example I am working with (I haven't written it to minimise details and keep focus on the main part of my problem).
Many thanks for all help.
The result is not true without extra assumptions.
If $X= \{ x\}$, then the result would state that we can interchange limits and integration whenever the limit of the integrals exists, which is simply not true.
However, even for arbitrary $X$, it is still not true. Let us define \begin{align*} s_n(a,x) = \begin{cases} 0 & a \not\in (n-1 , n] \\ 1 & a \in (n-1,n] \end{cases}. \end{align*} Then we have \begin{align*} f_a(x) = \sum_{n=1}^\infty s_n(a,x) = s_{\lceil a \rceil }(a,x) = 1 \end{align*} for all $a$. So in particular $f_a(x)$ converges uniformly to $1$. However \begin{align*} s_n(x) = \lim_{a \to \infty } s_n(a,x) = \lim_{ \substack{a \to \infty \\ a > n}} s_n(a,x) =0 \end{align*} for all $n \in \mathbb{N}$.
Thus we find \begin{align*} 1 = \lim_{a \to \infty} \sum_{n=1}^\infty s_n(a,x) \neq \sum_{n=1}^\infty s_n(x) = 0. \end{align*}