# Measure on topological spaces

So, given a topological space $S$ we can construct its Borel sigma-algebra $\mathcal{B}(S)$. Does it mean that we can construct a measure $\mu$ on this sigma-algebra as well? Say, discrete topology on the circle implies $\mathcal{B}(S) = 2^S$, and hence we know that we cannot construct a measure.

This leads to the idea that certainly we need some restrictions on the topological space. What are useful sufficient conditions?

I am especially interested in the conditions like - separable; - second countable; - metrizable;

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You can always construct a measure on a $\sigma$-algebra. For your example of the circle, you could use counting measure, or the zero measure. But I suppose you want your measure to satisfy some other property; what is it? – Nate Eldredge Jun 6 '11 at 13:14
But what are arguments like "there is no measure for a sigma-algebra induced by power set" about then? – Ilya Jun 6 '11 at 13:42
@Gortaur: Well, that statement isn't true as it stands. I'm not sure exactly what sort of argument you are thinking about, but if you have one in mind, I suggest you look closely at what it is actually saying. – Nate Eldredge Jun 6 '11 at 14:15
@Nate, thanks - I've tried to find it before answering you, but I couldn't find :( that should be a dream of mine. – Ilya Jun 6 '11 at 14:16
@Gortaur: An educated guess of mine: Maybe you were thinking about the fact that Lesbegue measure cannot be extended to the power set of the real numbers. – Tim van Beek Jun 6 '11 at 15:23

This question is an oldie, but I feel that it deserves a more elaborate answer, so here goes.

As you said, to every topological space $X$ one can associate the Borel $\sigma$-algebra $\mathcal{B}_X$, which is the $\sigma$-algebra generated by all open sets in $X$. Now $(X,\mathcal{B}_X)$ is a measurable space and it is desirable to find a natural Borel measure on it. By Borel measure I simply mean a measure defined on $\mathcal{B}_X$ and by "natural" I mean that it should be compatible with the topology of $X$ in some sense (otherwise, $X$ is just an abstract set). There are several compatibility conditions one can impose, which are motivated from the fact that the Euclidean spaces (with Lebesgue measure) satisfy all of them. The following are most often encountered:

• Strict positivity: this means that the measure of every nonempty open set is positive.
• Local finiteness: this means that every point has some open neighborhood $V$ with $\mu(V) < \infty$.
• Finiteness on compacta: this means that $\mu (K) < \infty$ for every compact set $K \subset X$.
• Outer regularity: This means that the measure of every Borel set $E \subset X$ is equal to the infimum of $\mu (G)$ over all open sets $G \subset X$ which contain $E$.
• (Weak) inner regularity: This means that the measure of every Borel set $E \subset X$ is equal to the supremum of $\mu (C)$ over all closed sets $C \subset X$ which are contained in $E$.
• (Strong) inner regularity, also known as tightness: The measure of every Borel set $E \subset X$ is equal to the supremum of $\mu (K)$ over all compact sets $K \subset X$ which are contained in $E$.
• Regularity: this is just outer regularity + (strong) inner regularity.
• Radon: $\mu$ is Radon if it is satisfies a certain combination of the above properties. Different texts often use different combinations, which often coincide if the topological space $X$ is nice enough (certainly if it is, say, $[0,1]$). One popular combination is "Radon = locally finite + inner regular" but some texts omit the inner regularity hypothesis, some replace it with outer regularity, some add outer regularity, etc. . The thing to keep in mind is that a Radon measure is a Borel measure which has some nice relation to the topology, but not necessarily every relation you want it to have (especially if the topology is somewhat ill-behaved).

In addition, there are some general measure-theoretical conditions one can impose on $\mu$ which ensure that it obeys to some general measure-theoretical theorems (e.g. Fubini). The most useful ones are:

• $\sigma$-finiteness: this means that $X$ can be written as a countable union of Borel sets such that each of these sets has finite measure.
• Finiteness: $\mu (X) < \infty$.

Also, you probably want $\mu$ to be nondegenerate ($\mu \ne 0$).

The above conditions are certainly not mutually independent and even for the most general topological spaces there are some obvious implications. For instance:

1. Strict positivity implies nondegeneracy.
2. If $X$ is Hausdorff then strong inner regularity implies weak inner regularity.
3. If $X$ is Hausdorff and locally compact then finiteness on compacta implies local finiteness.
4. If $X$ is $\sigma$-compact (i.e. equal to a countable union of compact subsets) and $\mu$ is finite on compacta, then $\mu$ is $\sigma$-finite.
5. If $\mu$ is finite then outer regularity is equivalent to (weak) inner regularity.
6. etc.

In the highest generality (arbitrary topological spaces), there is almost nothing nontrivial one can say, and one may not be able to find any nice Borel measure on the space. Therefore one restricts to some nice class of topological spaces, in which an interesting theory can be developed. Among them we have:

1. $\sigma$-compact, locally compact Hausdorff spaces. On such spaces there are strictly positive, finite on compacta, regular measures (which are then also $\sigma$-finite, and finite if the space is compact). In fact, there are plenty of them. A theorem of Riesz asserts that in this case, Borel measures on $X$ are in 1-1 correspondence with positive linear functionals on $C_c (X)$(the normed space of all continuous real-valued functions on $X$ with compact support) and Borel measures satisfying the above regularity conditions. Every such functional acts as integration against a measure on $X$ satisfying the above.
2. Polish spaces, which are topological spaces homeomorphic to separable, complete metric spaces, e.g. infinite dimensional Banach spaces and countably infinite products of finite discrete spaces. In such spaces it is often difficult to find a geometrically appealing Borel measure, especially if you want the measure to be invariant under many isometries of a given metric. But as Tim mentioned, they frequently appear in functional analysis and probability theory, and so find their uses.

Almost any space which shows up in applications belongs to one of these two classes. For instance, real (and p-adic?) Lie groups and their homogeneous spaces always belong to the first class.

Here are some references: for the theory of measures on locally compact topological spaces, there's a book called "Measure and Integration" by König. It is somewhat technical but very general. For Polish spaces, there's a very nice book by Parthasarathy called "Probability Measures on Metric Spaces". "Handbook of measure theory" probably discusses these subjects in length too.

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Minor quibble: at the end the norm on $C_c(X)$ is completely immaterial for Riesz-Markov (it is only of use if you want to have finite measures). What counts is the order structure. Moreover, I think that two excellent references with outstanding generality are Bogachev and Fremlin. Also, the locally compact setting receives an extremely thorough treatment in Hewitt-Ross. – t.b. Jul 13 '11 at 8:55
Thank you very much for such an extended answer. – Ilya Jul 13 '11 at 9:49
@Theo: I agree with your remark. Another nice reference I just remembered of is Folland's "Real Analysis", which has a chapter discussing Radon measures on locally compact Hausdorff spaces. – Mark Jul 13 '11 at 11:02

As Nate already pointed out, one can always construct a measure on a $\sigma$-algebra, but maybe you had some special applications in mind, and asked what kind of spaces people usually use for those?

One possible answer would be: Polish spaces, which are the most general spaces usually used in functional analysis and probability theory (a Polish space is isomorphic to a separable complete metric space).

Also see this blog post: ncafe.

Also see Radon measure (Wikipedia) for some reasons why general topological spaces are usually not used, but at least locally compact Hausdorff spaces.

See for example the commutative Gelfand-Naimark theorem, which says that the spectrum of a commutative $C^*$- algebra is a locally compact Hausdorff space.

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I also read that a Polish space is isomorphic to either $\mathbb{R}$ or $\mathbb{Z}$ or a finite set. If there is an isomorphism $I:\mathbb{R}\to C([0,1])$ where the last one is endowed with uniform convergence metric? – Ilya Jun 6 '11 at 13:53
@Goraur: In this context one has to be precise about what structure the term "isomorphism" refers to: In this case we are talking about "isomorphic as measure spaces", not "isomorphic as metric spaces". Up to measurable bijection, there’s one Polish space for each cardinality that’s countable, and one whose cardinality is that of the continuum, that's true. – Tim van Beek Jun 6 '11 at 15:27
"the most general spaces usually used in functional analysis" ... ??? Some non-separable Banach spaces are used in functional analysis. Also weak topologies of various kinds. – GEdgar Jun 6 '11 at 17:30
@GEdgar: There is an implicit "on an introductory level", of course, or an "in a first approximation", and also "with regard to applications of measure spaces" not with regard to topological vector spaces used in functional analysis, since the question was somewhat inprecise with regard to the applications of measure theory. – Tim van Beek Jun 7 '11 at 7:19