When can you switch the order of limits?

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.

• Interesting trick. I have never heard of the counting measure until now (though I am surprised, since it is a nice example of a non-Lebesgue measure). Actually, I thought I remembered in the measure theory that I learned that every countable set was measure zero for any measure, but I could be remembering it wrong. – asmeurer Dec 23 '10 at 5:04
• Perhaps countable sets have measure zero for every Borel measure? Is that right? – Jesse Madnick Dec 23 '10 at 5:07
• But yes, in any case, the counting measure assigns $+\infty$ to any infinite set. – Jesse Madnick Dec 23 '10 at 5:08
• I'm going to go ahead and mark this one as the answer, because it is the best of the ones given so far (it has the least strong condition). But be aware that if someone posts a better answer, I will switch it (you can do that, right?). – asmeurer Dec 27 '10 at 8:18
• But don't call it "cheating". Call it "an advanced method". – GEdgar Jun 14 '11 at 17:40

Here's some results.

Let us say that a sequence $a_{nm}$ of real numbers indexed by pairs of positive integers $(n,m)$ converges to $L$ if and only if for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L|\lt \epsilon$.

This is a fairly strong condition. We have:

Theorem. Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:

1. For each (sufficiently large) $n_0$, $\lim\limits_{m\to\infty}a_{n_0m}$ exists;
2. $\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}a_{nm} = L$.

Proof. If 2 holds, then we must have 1 (otherwise the expression in 2 does not even make sense). Now assume that 1 holds, and let $\lim\limits_{m\to\infty}a_{nm} = L_{n}$. We want to prove that $\lim\limits_{n\to\infty}L_n=L$.

Let $\epsilon\gt 0$. Then there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L| \lt \epsilon$. Let $M_N\gt N$ be such that for all $m\geq M_N$, $|a_{M_Nm}-L_{M_N}|\lt \epsilon$. Since $M_n\gt N$, we have $$|L_{M_N} - L| \leq |L_{M_N}-a_{M_Nm}| + |a_{M_Nm}-L| \leq 2\epsilon,$$ so this proves that $L_n\to L$ as $n\to\infty$. In particular, we have for the iterated limit $$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{nm} = \lim_{n\to\infty}L_n = L.\ \Box$$

A symmetric argument shows that

Theorem. Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:

1. For each (sufficiently large) $m_0$, $\lim\limits_{n\to\infty}a_{nm_0}$ exists;
2. $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}a_{nm} = L$.

However, $\lim\limits_{n,m\to\infty}a_{nm}=L$ does not imply the existence of the iterated limits; in fact, existence of the double limit and the existence of one iterated limit does not suffice to imply that the other iterated limits exists. Take $a_{nm}=\frac{(-1)^n}{m}$. Then $\lim\limits_{n,m\to\infty}a_{nm}=0$ (given $\epsilon\gt 0$, pick $N$ such that $\frac{1}{N}\lt\epsilon$), and the limit as $m\to\infty$ of $\frac{(-1)^n}{m}$ exists for each fixed $n$, but $\lim\limits_{n\to\infty}\frac{(-1)^n}{m}$ does not exist for any $m$. (By taking $a_{nm} = \frac{(-1)^n}{m} + \frac{(-1)^m}{n}$ you get one in which neither iterated limit exists.)

So, a sufficient conditions for the iterated limits to exist is:

Corollary. Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}=L$. Then the iterated limits $$\lim_{n\to\infty}\lim_{m\to\infty}a_{nm}\text{ and }\lim_{m\to\infty}\lim_{n\to\infty}a_{nm}$$ both exist and are equal to $L$ if and only if $\lim\limits_{n\to\infty}a_{nm}$ exists for almost all $m$ and $\lim\limits_{m\to\infty}a_{nm}$ exists for almost all $n$.

(Here, "almost all" means "all except perhaps for a finite number").

As I said, the condition above is pretty strong. You can have both iterated limits exist and be equal and yet for the double limit not to exist. To adapt the standard two-variable example, take $a_{nm}=\frac{nm}{n^2+m^2}$. The iterated limits both exist and are equal to $0$, but the double limit does not exist (for any $N\gt 0$ there exist $n,m\geq N$ such that $a_{nm}=\frac{1}{2}$ and there exist $n,m\geq N$ such that $a_{nm}=\frac{2}{5}$; just take $n=m=N$ for the first, and $n=2m=2N$ for the second).

You can get that the double limit exists and is equal to (one of) the iterated limits if you have some uniformness conditions.

I don't know of any necessary and sufficient conditions, and I suspect there won't generally be without some other overarching conditions. This is essentially the same problem as the problem of iterated limits in functions of two variables (just as the problem of finding the limit of a real function of real variable is closely connected to the problem of finding limits of sequences of real numbers). They are connected to the double limits, and you often have conditions based on uniform convergence that guarantee good things happen, but to state some general, simple condition for the iterated limits to exist and be equal seems difficult in general.

• I suspected that something like that might hold, though as you said, it is pretty strong. Very nice proof and examples, though. – asmeurer Dec 22 '10 at 23:35
• And now you've got me wondering what the set of numbers that $a_{nm}=\frac{nm}{n^2 + m^2}$ can converge to is. – asmeurer Dec 22 '10 at 23:37
• @asmeurer: Certainly, any number of the form $\frac{k}{k^2+1}$ with $k$ an integer; also $0$ (take $n=m^2$); if $n=\lfloor tm\rfloor$, you might be able to get any $\frac{t}{t^2+1}$. – Arturo Magidin Dec 22 '10 at 23:45
• I believe that the convergence of double sequences introduced in the beginning of this post is sometimes called Prigsheim convergence. – Martin Sleziak Jun 14 '11 at 15:59
• @Arturo: A question about Pringsheim convergence has been posted which, if I correctly understood OP's comment, is asking about a reference for the results mentioned in this answer. I thought it might be good to ping you about this. – Martin Sleziak Dec 24 '11 at 9:28

This result is similar to the result from Jesse Madnick's post.

Suppose that $a_{km}$ is non-decreasing in both variables, i.e., $a_{km}\le a_{k,m+1}$ and $a_{km}\le a_{k+1,m}$ for each $k$ and $m$.

Then $\lim\limits_{k\to\infty} \lim\limits_{m\to\infty} a_{km} = \lim\limits_{m\to\infty} \lim\limits_{k\to\infty} a_{km}$.

Proof: Denote $b_k=\lim\limits_{m\to\infty} a_{km}$ and $c_m=\lim\limits_{k\to\infty} a_{km}$. Monotonicity implies that these limits exist (they might be $+\infty$) and the sequences $(b_k)$, $(c_m)$ are non-decreasing.

Put $b:=\lim\limits_{k\to\infty} b_k$, $c:=\lim\limits_{m\to\infty} c_m$.

We have $a_{km}\le c_m \le c$ $\Rightarrow$ $b_k \le c$ $\Rightarrow$ $b\le c$.

The proof that $c\le b$ is analogous.

• @MartinSleziak, why is $b_k \leq c$? I don't understand how you deduce that step. – Kamil Jul 11 '16 at 8:16
• @Kamil Using the fact that limits preserve inequality: math.stackexchange.com/questions/1850471/… If $a_{km} \le c_m$ for each $m$, then also $\lim\limits_{m\to\infty} a_{km} \le \lim\limits_{m\to\infty} c_m$ (assuming the limits exist). – Martin Sleziak Jul 11 '16 at 8:24

The Moore-Osgood Theorem might also be relevant here.

Here's a fact I use all the time:

Let $a_{mn}$ be any double sequence in the extended real numbers $\overline{\mathbb{R}}$.

Then $$\sup_m \sup_n \, a_{mn} = \sup_n \sup_m \, a_{mn} = \sup_{m,n} \, a_{mn}$$

Proof. Let $A=\sup_{m,n} \, a_{mn}, \, B = \sup_m \sup_n \, a_{mn}$.

Then for all $m$ and $n$, $A\ge a_{mn}$, so that $A \ge B$.

To show the other direction, first consider the case that $A<\infty$. Then $\forall \varepsilon > 0 \exists m^*,n^*$ s.t. $a_{m^*n^*} \ge A-\varepsilon$. It follows that $$B=\sup_m \sup_n \, a_{mn} \ge \sup_m \, a_{mn^*} \ge a_{m^*n^*} \ge A-\varepsilon.$$ Taking the limit as $\varepsilon\to 0$, one obtains $B\ge A$.

In case $A=\infty$, $\forall M>0\exists m^*,n^*$ s.t. $a_{m^*n^*} \ge M$. Then $$B=\sup_m \sup_n \, a_{mn} \ge \sup_m \, a_{mn^*} \ge a_{m^*n^*} \ge M.$$ Sending $M$ to $\infty$, one obtains $B=\infty$.

$A=\sup_n \sup_m \, a_{mn}$ follows by a symmetric argument. $\square$

As a corrolary one obtains:

Let $a_{mn}$ be any double sequence in the extended real numbers $\overline{\mathbb{R}}$ which is non-decreasing in $m$ and $n$. Then $$\lim_{n\to\infty}\lim_{m\to\infty} a_{mn} = \lim_{m\to\infty}\lim_{n\to\infty} a_{mn}$$

This can be used with sums of non-negative numbers, for example.

• To prove the first assertion, we can also use the "universal property" definition of the supremum. Expanding the definition of the supremum, we get that for an arbitrary real $A$, we have $A ≥ \sup_n \sup_m a_{nm}$ if and only if $A≥\sup_n a_{nm}$ for all $m$, if and only if $A≥a_{nm}$ for all $n$ and all $m$. But this is exactly the definition of $\sup_{n,m} a_{nm}$. – Idéophage Feb 21 at 2:09

The following is a minor variation on Arturo Magidin's answer, but according to my experience it is more widely used.

Proposition: Assume $$\lim_{m \to \infty} \lim_{n \to \infty} a_{mn} = \lim_{m \to \infty} a_m = a \qquad\text{and}\qquad \lim_{m \to \infty} a_{mn} = a_n$$ i.e. all limits exist and converge to their indicated value. We further assume that the single-variable limits converge uniformly in the other variable, i.e. for every $\varepsilon>0$ there exists $M > 0$ such that $|a_{mn} - a_{n}| < \varepsilon$ for all $m > M$ and all $n$, and likewise for $a_m$. Then $\displaystyle \lim_{n \to \infty} \lim_{m \to \infty} a_{mn}$ exists and it holds $$\lim_{n \to \infty} \lim_{m \to \infty} a_{mn} = \lim_{m \to \infty} \lim_{n \to \infty} a_{mn} .$$

Proof: For any $\varepsilon > 0$, we can find $N > 0$ such that $|a_{mn} - a_m| < \tfrac{\varepsilon}{3}$ for all $n > N$ and all $m$. Similarly, we can find $M>0$ such that both $|a_n - a_{mn}| < \tfrac{\varepsilon}{3}$ for all $n$ and $|a_m - a| < \tfrac{\varepsilon}{3}$ holds for all $m > M$. Thus, it holds $$|a_n - a| \leq |a_n - a_{mn}| + |a_{mn} - a_{m}| + |a_{m} - a| < \varepsilon$$ for all $n > N$ and $m > M$.