Clarification in swapping limit and sum ($\lim_n \sum_{j \in S} g(j)$. I think I "proved" something that is incorrect, but I'm not sure why it is incorrect. Clarification would be appreciated.
Suppose that $(S,F,P)$ is a probability space where $S$ is countable. Let $g:S \rightarrow \mathbb{R}$ be a bounded measurable function, and let $a_{n,j} \geq 0$ be a convergent real sequence in $n \in \mathbb{N}$ for each $j \in S$, and suppose that $\sum_{j \in S} a_{n,j}=1$ for each $n$.
I want to interchange the following (for each $n$ the sum on the LHS is finite, and the limit on the RHS exists and is finite),
$$\lim_{n \rightarrow \infty}\sum_{j \in S} g(j) a_{n,j}=\sum_{j \in S} \lim_{n \rightarrow \infty} g(j)a_{n,j},$$
and my thought was to split the above into,
$$\lim_n\sum_{j \in S, g(j)>0} g(j)a_{n,j}-\lim_n \sum_{j \in S, g(j) \leq 0} g(j) a_{n,j},$$
and then apply the monotone convergence theorem to each sum where the summands are all non-negative functions. Then I could recollect terms and get what I want. But this seems like I'm overlooking something; this looks like it suggests I can always interchange limits with sums, which I know isn't generally true. Is there something I'm missing? I hope I didn't forget to include any important information, I didn't want to overload the question with irrelevant details.
Thanks.
 A: Counterexample:
As $S$ is countable there exists a bijection $\pi:\mathbb{N}\to S$, so I construct specific candidates for $a_{n,j}$ and $g(j)$ indexed by $j\in \mathbb{N}$ instead of $j\in S$.
For each $n\in\mathbb{N}$ let $a_{n,n}=1$ and $a_{n,j}=0$ for $j\not =n$. Thus we have that $a_{n,j}\geq 0$ is a convergence series in $n\in\mathbb{N}$ for each $j\in  \mathbb{N}$, specifically it holds that $a_{n,j}\to 0$ as $n\to\infty$ for each $j\in\mathbb{N}$. Furthermore, we have that $\sum_{j=1}^\infty a_{n,j}=1$ for each $n\in\mathbb{N}$. 
Now let $g:\mathbb{N}\to\mathbb{R}$ be a bounded and measurable function given by $g(j)=1$ for all $j\in\mathbb{N}$. 
Now realize that
$$
\lim_{n\to\infty} \sum_{j=1}^\infty g(j)a_{n,j} = \lim_{n\to \infty} \sum_{j=1}^\infty a_{n,j} = \lim_{n\to \infty} 1 = 1,$$
and
$$
\sum_{j=1}^\infty \lim_{n\to \infty} g(j)a_{n,j} = \sum_{j=1}^\infty \lim_{n\to \infty} a_{n,j} =\sum_{j=1}^\infty 0 =0.
$$
What goes wrong? You can't use monotone convergence theorem, as the conditions on $g$ and $a$ does not entail that $g(j)a_{n,j}$ is monotonically increasing in $n$ for each $j\in\mathbb{N}$. You can furthermore not use dominated convergence theorem, as the conditions on $g$ and $a$ does not entail the existence of a summable sequence $(b_j)$ dominating $|g(j)a_{n,j}|$ for all $n\in\mathbb{N}$, as the smallest such sequence in the above example would be $b_j=1$ for all $j\in\mathbb{N}$, which is not summable.
