Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In measure theory we frequently see the following definitions:

$$\limsup_{n\to\infty} A_n = \bigcap_{n=1}^{\infty}\left(\bigcup_{j=n}^{\infty} A_j\right)$$

$$\liminf_{n\to\infty} A_n = \bigcup_{n=1}^{\infty}\left(\bigcap_{j=n}^{\infty} A_j\right)$$

where $(A_n)_n$ is a sequence of measureable sets i.e. $\forall n: A_n\in\mathcal{M}$, where $\mathcal{M}$ is a $\sigma$-algebra on $X$, for example $\mathcal{M} = 2^X$. Therefore it makes sense to also define:

$$\lim_{n\to\infty}A_n = \limsup_{n\to\infty} A_n = \liminf_{n\to\infty} A_n$$

when the last two agree. If $\mu$ is a finite (positive, to keep things simple) measure, it is easy to see that under such definition we have $\mu(\lim_{n\to\infty}A_n) = \lim_{n\to\infty}\mu(A_n)$, whenever $\lim_{n\to\infty}A_n$ exists, which looks like some kind of continuity.

Does this kind of convergence of sequences of measurable sets arise from a (preferably Hausdorff, so that limits are unique) topology on $\mathcal{M}$? If such a topology exists, is $\mu:\mathcal{M}\to[0,\infty)$ in fact a continuous function?

(A related question that may be of interest would be: what happens if we allow arbitrary sets? Can we make the Von Neumann universe $V$ into a topological space in such a way?)

share|cite|improve this question
There is no natural way to put a topology on an arbitrary sigma algebra. – user18063 Jan 8 '12 at 18:52
@user18063: What do you mean by "natural"? – Nate Eldredge Jan 8 '12 at 19:31
I had asked this myself some time ago as well, but does it really have anything to do with measure theory? One could formulate the question more generally: given a set $X$, does there exist a topology $\tau$ on $P(X)$ such that the notion of limit with respect to $\tau$ is the one defined in the question? – Emilio Ferrucci Jan 8 '12 at 19:55
@Emilio: You're right, we can formulate most of this question without mentioning $\sigma$-algebras. I think that would be just as general, since $\mathcal{M} = P(X)$ is a $\sigma$-algebra. The tag (measure-theory) is there mostly because I'm also interested in the connection with measures. (I could also ask about arbitrary functions $\mathcal{M}\to[0,\infty)$, but measures seem particularly nice, since they preserve the limits.) – Dejan Govc Jan 8 '12 at 23:24
up vote 7 down vote accepted

There is such a topology. Simply give $\mathcal{M}$ the subspace topology induced by the product topology on $2^X$.

It may help to think of $2^X$ as the set of functions from $X$ to $\{0,1\}$, by identifying a set with its indicator function. Then we have $1_{\limsup A_n} = \limsup 1_{A_n}$ and so on. Since the product topology is just the topology of pointwise convergence, this behaves as desired.

However, the map $A \mapsto \mu(A)$ is not in general continuous with respect to this topology. For instance, the finite sets are dense in $\mathcal{M}$ with this topology, and so any nontrivial non-atomic measure gives a discontinuous map.

share|cite|improve this answer
Thanks, this indeed solves the main part of the question. Interestingly, for a countable $X$ it also solves the part about the measures in the positive, if I'm not mistaken, since it then gives a first-countable topology. I wonder if we can modify it somehow to work for uncountable $X$ also ... – Dejan Govc Jan 9 '12 at 0:42
@DejanGovc: Right, if $X$ is countable then $2^X$ is first countable in the product topology. In fact it is homeomorphic to the Cantor set and hence is a compact metrizable space. The map $A \mapsto \mu(A)$ will be continuous, because a measure on a countable set must be purely atomic. If $X$ is uncountable, however, then it will be hard to avoid problems, unless the $\sigma$-algebra $\mathcal{M}$ is very simple. – Nate Eldredge Jan 9 '12 at 3:56
I've decided to accept this answer, since I think it's pretty fascinating that such a simple topology already works. I initially suspected there to be no such topology, so this example was quite an eye-opener. Also the part about non-atomicity of measures is something I probably wouldn't have thought about, so thanks again for the helpful suggestions. – Dejan Govc Jan 11 '12 at 16:54

I believe I have found a topology that answers both of my questions positively. Let us introduce some notation in order to avoid confusion. Let $\mathrm{LIM}_{n\to\infty}A_n$ denote the common value of $\limsup_{n\to\infty}A_n$ and $\liminf_{n\to\infty}A_n$ if it exists. (Defined using intersections and unions as explained above in the question.) This is to distinguish this notion from the possibly different notion of limit arising from a topology which we will denote $\lim$ in case we need it. Next define the following sets: $$\mathbf{M}=\lbrace\mu:\mathcal{M}\to[0,\infty)|\hbox{ }\mu\textrm{ is a finite positive measure}\rbrace$$ $$\mathbf{A}=\lbrace A:\mathbb{N}^+\to\mathcal{M}|\hbox{ }\mathrm{LIM}_{n\to\infty}A(n)=A(\infty)\rbrace$$ Here $\mathbb{N}^+$ denotes the one point compactification of $\mathbb{N}$ and $\infty$ is the added point. Now we take $\tau_0$ to be the initial topology on $\mathcal{M}$ with respect to $\mathbf{M}$ and $\tau_1$ to be the final topology on $\mathcal{M}$ with respect to $\mathbf{A}$. Initial topology is by definition the smallest topology with respect to which all $\mu\in\mathbf{M}$ are continuous. This means that the topologies $\tau$ for which all $\mu\in\mathbf{M}$ are continuous are precisely those for which $\tau_0\subseteq\tau$ holds. Dually, topologies $\tau$ for which every $A\in\mathbf{A}$ is continuous, are characterised by: $\tau\subseteq\tau_1$.

So to solve the problem, all we have to do is prove that $\tau_0\subseteq\tau_1$. The topology $\tau_0$ is generated by sets of the form $\mu^{-1}(U)$ where $U$ is an open set in $[0,\infty)$ and $\mu\in\mathbf{M}$. So it suffices to prove that every such set is also contained in $\tau_1$. To do this, we take $\tau = \lbrace\emptyset, \mu^{-1}(U), X\rbrace$. Clearly this is a topology, so in order for it to be contained in $\tau_1$ we just need to show that every $A\in\mathbf{A}$ is continuous with respect to $\tau$.

So, suppose $A\in\mathbf{A}$. All we need to see is that $A^{-1}(\mu^{-1}(U)) = (\mu\circ A)^{-1}(U)$ is open. If $\infty\notin (\mu\circ A)^{-1}(U)$, this is true. So let $\infty\in (\mu\circ A)^{-1}(U)$. For such a set to be open, we need to show that $(\exists N\in\mathbb{N})(\forall n\geq N): n\in(\mu\circ A)^{-1}(U)$. But, as we know $\mathrm{LIM}_{n\to\infty} A(n) = A(\infty)$ implies that $\lim_{n\to\infty} \mu(A(n)) = \mu(A(\infty))$ so such an $N$ indeed exists.

Conclusion: $\tau_0\subseteq\tau_1$.

So, indeed, if we take the topology on $\mathcal{M}$ to be $\tau_1$, the convergent sequences are precisely those for which $\mathrm{LIM}$ exists (using the fact that the topology Nate Eldredge suggests is contained in $\tau_1$) and every finite positive measure $\mu$ is a continuous map.

(If I have made a mistake somewhere, corrections are more than welcome.)

share|cite|improve this answer
Interesting. My first instinct is that it sounds too good to be true. I'll try to look at this more carefully when I have some time. – Nate Eldredge Jan 11 '12 at 18:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.