I am aware that using the counting measure on Lebesgue you can get important results. I am wondering if there is a way to exchange the limits using Dominated Convergence Theorem.
Let $a_{n,m}$ be indexed from 0 to infinity. $|a_{n,m}|<M$ for all indexes. $\lim_{m\to \infty} a_{n,m}$ exists for all m.
Does DCT allow you do say $\lim_{n \to \infty} \lim_{m \to \infty} a_{n,m} = \lim_{m \to \infty} \lim_{n \to \infty} a_{n,m}$
One possibility is the following. We write $b_{n,m} := a_{n,m} - a_{n-1,m}$ (with $a_{0,m} = 0$). Then, $a_{n,m} = \sum_{k=1}^n b_{k,m}$. Thus, $$ \lim_{m \to \infty}\lim_{n\to\infty} a_{n,m} = \lim_{m \to \infty}\lim_{n\to\infty} \sum_{k=1}^n b_{k,m} = \lim_{m \to \infty}\sum_{k = 1}^\infty b_{k,m}. $$ Now, you can try to apply DCT (under some restrictive assumptions on $b_{k,m}$). If this is possible, we can continue \begin{align} \lim_{m \to \infty}\lim_{n\to\infty} a_{n,m} &= \lim_{m \to \infty}\sum_{k = 1}^\infty b_{k,m} = \sum_{k = 1}^\infty\lim_{m \to \infty} b_{k,m} = \lim_{n\to\infty}\sum_{k=1}^n \lim_{m\to\infty}b_{k,m} = \lim_{n\to\infty} \lim_{m\to\infty} \sum_{k=1}^n b_{k,m} \\&= \lim_{n\to\infty} \lim_{m\to\infty} a_{n,m} . \end{align} To summarize: Yes, it is possible, but you need different assumptions on your sequence $a_{n,m}$.
No. Take, for instance,$$a_{nm}=\begin{cases}1&\text{ if }m\geqslant n\\0&\text{otherwise.}\end{cases}$$It's bounded, but$$\lim_{n\to\infty}\lim_{m\to\infty}a_{nm}=1\quad\text{and}\quad\lim_{m\to\infty}\lim_{n\to\infty}a_{nm}=0.$$
