Suppose you have a double sequence $\displaystyle a_{nm}$.  What are sufficient conditions for you to be able to say that $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}$?  Bonus points for necessary and sufficient conditions.

For an example of a sequence where this is not the case, consider $\displaystyle a_{nm}=\left(\frac{1}{n}\right)^{\frac{1}{m}}$.  $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}}=\lim_{n\to \infty}{\left(\frac{1}{n}\right)^0}=\lim_{n\to \infty}{1}=1$, but $\displaystyle \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}=\lim_{m\to \infty}{0^{\frac{1}{m}}}=\lim_{m\to \infty}{0}=0$.

If you want to avoid hypotheses that involve uniform convergence, you can always cheat and use the counting measure on {0, 1, 2,...} and then use either the Monotone or Dominated Convergence Theorem from integration theory.

For instance, using the Monotone Convergence Theorem, we get the following (perhaps silly) sufficient criterion:

Proposition: If $a_{mn}$ is monotonically increasing in $m$, and is such that $c_{mn} = a_{mn} - a_{m-1,n}$ is monotonically increasing in $n$, then $\lim\limits_{m \to \infty} \lim\limits_{n \to \infty} a_{mn} = \lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{mn}$.

Proof: Our two hypotheses really just amount to saying that each $c_{mn} \geq 0$ and that the $c_{mn}$ are monotonically increasing in $n$.  So we can use the Monotone Convergence Theorem with respect to the counting measure:
$$\lim_{n\to \infty} \int c_{mn} = \int \lim_{n\to \infty} c_{mn}$$
But really, these integrals are sums, so:
$$\lim_{n\to \infty} \sum_{m=1}^\infty c_{mn} = \sum_{m=1}^\infty \lim_{n\to \infty} c_{mn}$$
Since infinite sums are just limits of partial sums, we have:
$$\lim_{n\to \infty} \lim_{M \to \infty} \sum_{m=1}^M c_{mn} = \lim_{M\to\infty}\lim_{n\to \infty} \sum_{m=1}^M c_{mn}$$
By our construction of $c_{mn}$, the left side is $\lim\limits_{n\to\infty} \lim\limits_{M\to\infty} a_{Mn}$, and the right side is $\lim\limits_{M\to\infty} \lim\limits_{n\to\infty}a_{Mn}$. $\lozenge$

Edit: In the above, our index ranges are $m,n \geq 1$, and we make the convention that $a_{0n} = 0$.

Using the Dominated Convergence Theorem, we could probably get something slightly more useful.
