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Let me make my question clear. I want to define a measure $\mu$ on a space $X$. But instead of telling you what value I assign for some subset of $X$ (measurable sets that form a $\sigma$-algebra), I tell you that for each $f$ continuous, what $\int_X f(x)d\mu (x)$ is.

Then, is this measure uniquely determined? I know if I tell you how to integrate all measurable functions, then this measure is of course uniquely determined. Because integrate characteristic functions will give you measure of that respective set. But is it also true if I only define integration with continuous functions?

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What is the topology on $X$? – Davide Giraudo Jun 8 '12 at 20:31
This is the Daniell Integral approach. – ncmathsadist Jun 8 '12 at 20:58
up vote 8 down vote accepted

In general this is false. Here are some examples to think about:

  1. If the $\sigma$-algebra on $X$ is not the Borel $\sigma$-algebra, there is generally no hope. (What if $X$ has the trivial topology but the $\sigma$-algebra is not trivial?) Hence you should restrict your attention to Borel measures.

  2. Take $X = \{a,b\}$ with the topology $\tau = \{\emptyset, \{a\}, \{a,b\}\}$. The Borel $\sigma$-algebra is $2^X$ but the only continuous functions $f : X \to \mathbb{R}$ are constant, so $\mu_1 = \delta_a$ and $\mu_2 = 2 \delta_a - \delta_b$ agree on all continuous functions. Thus you probably want a Hausdorff space.

  3. Take $X = \mathbb{R}$. Let $\mu$ be counting measure and $\nu = 2\mu$. So you probably want to look at $\sigma$-finite measures.

  4. As I mentioned in the above comment, on $X = \omega_1 + 1$ (which is compact Hausdorff), one can find two distinct finite measures which agree on all continuous functions.

However, here is a positive result.

Proposition. Let $\mu, \nu$ be finite Borel measures on a metric space $(X,d)$. If $\int f d\mu = \int f d\nu$ for all bounded continuous $f$, then $\mu = \nu$.

Proof. Let $E$ be a closed set, and let $f_n(x) = \max\{1 - n d(x,E), 0\}$. You can check that $f_n$ is continuous and $f_n \downarrow 1_E$ as $n \to \infty$. So by dominated convergence, $\mu(E) = \nu(E)$, and $\mu, \nu$ agree on all closed sets.

Now we apply Dynkin's $\pi$-$\lambda$ theorem. Let $\mathcal{P}$ be the collection of all closed sets; $\mathcal{P}$ is closed under finite intersections, and $\sigma(\mathcal{P})$ is the Borel $\sigma$-algebra $\mathcal{B}$. Let $\mathcal{L} = \{ A \in \mathcal{B} \colon \mu(A) = \nu(A)\}$. Using countable additivity, it is easy to check that $\mathcal{L}$ is a $\lambda$-system, and we just showed $\mathcal{P} \subset \mathcal{L}$. So by Dynkin's theorem, $\mathcal{B} = \sigma(\mathcal{P}) \subset \mathcal{L}$, which is to say that $\mu,\nu$ agree on all Borel sets, and hence are the same measure.

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This is a really nice answer, Nate! Thank you. – henryforever14 Jun 8 '12 at 21:59
An alternative way to conclude the proposition's proof is by regularity of finite Borel measures on metric spaces: the measure of Borel sets can be approximated by closed sets from below, so if $\mu=\nu$ on closed sets then they agree on all Borel sets. But Dynkin is ofcourse also very neat. – T. Eskin Jun 9 '12 at 22:53
@ThomasE.: Well, I would probably use Dynkin's theorem to prove the regularity also :) – Nate Eldredge Jun 9 '12 at 22:55
Yeah; there's no doubt that Dynkin is a powerful tool :-) You could, however, also prove it manually by showing that the collection of sets for which such approximation holds true is a $\sigma$-algebra that contains Borel sets. But as you said, Dynkin probably does the job easier. – T. Eskin Jun 9 '12 at 23:01
@PPR: These are finite measures, so you may just use 1 as your dominating function. – Nate Eldredge Jul 18 at 21:48

Check out the Riesz Representation Theorem, for example in $\textit{Real and Complex Analysis}$ by Rudin page 40. At least in the form presented in Rudin, if $X$ is a locally compact Hausdorff space and $\Lambda$ is a positive linear functional on $C_c(X)$, the continuous functions with compact support, then there exists a unique sigma algebra and a unique measure on the algebra such that $\int_X f \ d\mu = \Lambda(f)$ for all $f \in C_c(X)$.

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A unique Radon measure. It is possible that there are also non-Radon measures $\nu$ for which $\int f d\mu = \int f d\nu$ for all continuous $f$. The classic example is on $X = \omega_1 + 1$. This is Exercise 7.15 in Folland's Real Analysis. – Nate Eldredge Jun 8 '12 at 21:08
The part of the proof which shows uniqueness of $\mu$ is nice and fits the OP's question very well: it shows that if $\int f\,d\mu=\int f\,d\nu$ for all $f\in C_{c}(X)$ then $\mu=\nu$. It's done basicly with Urysohn's lemma, and you also need regularity of $\mu,\nu$ or Dynkin's theorem in the proof, which Rudin does not specify explicitly. He just says it is enough to deal closed sets which, however, isn't a trivial remark. – T. Eskin Jun 9 '12 at 23:18

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