Sufficient condition for having finite double limit of a double sequence. 
If $(s(n,m))$ is a double sequence such that
(i) the iterated limit $\lim_{m \to \infty} (\lim_{n \to \infty  }s(n,m))= a$, and
(ii) the limit $\lim _{n \to \infty}s(n,m)$ exists uniformly for every $m \in \mathbb N$,
then the double limit $\lim_{n,m \to \infty}s(n,m)=a$.

Can anyone please tell me how to prove this theorem?
 A: Your hypothesis (i) -- that $\lim_{m \to \infty} (\lim_{n \to \infty  }s(n,m))= a$ -- requires that the inner limit must exist for every $n$.
Even if we only assume that the single limit $\lim_{n \to \infty}s(n,m) = a_m$ exists for every $m$, and the limit $\lim_{m \to \infty} s(n,m) = b_n$ exists uniformly for every $n$.  It then follows that
$$\lim_{n,m \to \infty} s(n,m) = \lim_{m \to \infty} a_m = \lim_{n \to \infty} b_n$$
By the uniform convergence of $s(n,m) \to  b_n$, for any  $\epsilon > 0$ there exists a positive integer $Q(\epsilon)$ such that if $m \geqslant Q(\epsilon)$, then for all n,
$$\tag{*}|s(n,m) - b_n| < \epsilon $$
Taking any fixed $q \geqslant Q(\epsilon)$, we have
$$|b_j - b_k|\leqslant |b_j - s(q,j)| + |s(q,j) - s(q,k)| + |s(q,k) - b_k|$$
By (*) it follows that $|b_j - s(q,j)| < \epsilon$ and $|s(q,k) - b_k| < \epsilon$ for any values of $j$ and $k$.  Since $\lim_{m \to \infty} s(q,m) = a_q$ exists, it follows that with $q$ fixed $s(q,m)$ is a Cauchy sequence and there exists $M(\epsilon,q)$ such that if $j,k \geqslant M(\epsilon,q)$, then $|s(q,j) - s(q,k)|< \epsilon$.
Altogether, we have for all $j,k \geqslant M(\epsilon,q)$, $|b_j - b_k| < 3\epsilon$. Therefore, the sequence $(b_n)$ is a Cauchy sequence and there exists a number $b$ such that
$$\lim_{n \to \infty} \lim_{m \to \infty} s(n,m) = \lim_{n \to \infty}b_n = b$$
Again, for any $\epsilon > 0$, there exists a positive integer $N(\epsilon)$ such that if $n \geqslant N(\epsilon)$, then $|b_n -b | < \epsilon$.
Thus, if $n,m \geqslant \max(N(\epsilon),Q(\epsilon))$, then
$$|s(n,m) - a| \leqslant |s(n,m) - b_n| + |b_n - b| < 2\epsilon,$$
which proves that
$$\lim_{n,m \to \infty} s(n,m) = \lim_{n \to \infty} b_n = b$$
Given hypothesis (i) we must have $b = a$.
Finally, we can prove the existence of the other iterated limit and that
$$\tag{**}a = \lim_{m \to \infty} (\lim_{n \to \infty  }s(n,m))= \lim_{n,m \to \infty} s(n,m) = b$$
For every $\epsilon > 0$, we have for all sufficiently large $n,m $,
$$|s(n,m) - b| < \epsilon$$
Thus, for sufficiently large $m$,
$$| \lim_{n \to \infty}s(n,m) - b | = \lim_{n \to \infty}|s(n,m) - b| \leqslant \epsilon,$$
which proves (**).
