# Do limits of sequences of sets come from a topology?

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?)

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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

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.

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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.)

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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