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What classes of functions are sufficient to determine whether two measures are equal? If $$\int_{R^d} f d\mu =\int_{R^d} f d\nu $$ for some functions $f$, when can we say that $\mu=\nu$? Obviously, if $f$ can be any indicator function this is easy but is there a theorem for continuous functions say? Also, what about if we integrate over more general spaces, e.g. a Banach space?

Thanks.

EDIT: Extra question - According to Stefan in the comments, in the special case that $(S,d)$ is a metric space and μ and ν are Borel probability measures, then the space of non-negative bounded and continuous functions works. Please can somebody suggest a good reference for this or a reason why this is true?

David
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    For what it's worth: If $(S,d)$ is a metric space, then the space of non-negative bounded and continuous functions has that property if you assume that $\mu$ and $\nu$ are Borel probability measures. If $S=\mathbb{R}^d$, then ${x\mapsto e^{-n|x-a|^2}\mid n\in\mathbb{N},,a\in\mathbb{R}^d}$ is enough. – Stefan Hansen Mar 23 '13 at 22:23
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    @Stefan: That is helpful to me thanks. Please could you recommend a reference? – David Mar 23 '13 at 23:21
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    Probability Measures on Metric Spaces by Parthasarathy is a good reference (in addition to Billingsley's book mentioned by Ju'x). – Quinn Culver Mar 24 '13 at 00:51

2 Answers2

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Let $(S,d)$ be a metric space and let $\mu$, $\nu$ be two Borel probability measures on $S$. Then $$ \int f\,\mathrm d\mu=\int f\,\mathrm d\nu,\;\forall\,f\in \mathrm{bC}(S)_+\iff\mu=\nu. $$

Proof: Assume that the integrals of every function in $\mathrm{bC}(S)_+$ are identical. Then to show $\mu=\nu$, it is enough to show that $\mu(U)=\nu(U)$ for all $U\subseteq S$ open. For each $n\geq 1$ we define $$ f_n(x):=n\cdot d(x,U^c)\wedge 1,\quad n\geq 1. $$ Then the $f_n$'s are bounded, non-negative and continuous (they are in fact Lipschitz continuous) and $f_n\uparrow 1_U$ pointwise. Thus, $$ \mu(U)=\int 1_U\,\mathrm d\mu=\lim_{n\to\infty}\int f_n\,\mathrm d\mu=\lim_{n\to\infty}\int f_n\,\mathrm d\nu=\nu(U). $$

We can relax the assumption that $\mu$ and $\nu$ need to be probability measures. The above also holds in the more general setting where $\mu$ and $\nu$ are just measures such that there exists a sequence $(A_n)_{n\geq 1}$ of open Borel sets such that $$ S=\bigcup_{n\geq 1}A_n,\quad \text{and}\quad \mu(A_n)=\nu(A_n)<\infty,\quad n\geq 1. $$

Stefan Hansen
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  • Thanks for this answer but where did you use the fact that $\mu$ and $\nu$ are probability measures? Also, I am a little confused about how you show that is Lipschitz (although it is probably trivial - it is in $R^d$ with open balls which is enough in that case). – David Mar 24 '13 at 11:52
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    @David: $d(x,F)$ for any subset $F$ of a metric space $X$ is Lipschitz. Look here: http://math.stackexchange.com/a/48879/12690 – Damian Sobota Mar 24 '13 at 13:16
  • @David: We used that $\mu$ and $\nu$ are probability measures when I said that it's enough to show $\mu(U)=\nu(U)$ for $U$ open. But this would also hold when just $\mu$ and $\nu$ are both finite measures. – Stefan Hansen Mar 24 '13 at 13:28
  • @StefanHansen: I thought that just used that it is a Borel measure. ..Is there a counter example if the measures are not finite? – David Mar 24 '13 at 13:36
  • @David: You actually don't need them to be finite (see my edit). This uses a classical uniqueness result which relies on Dynkin's lemma, and I'm pretty sure we can't relax any further on the assumptions. But I can't give you a concrete counter example (maybe that is a new question?). – Stefan Hansen Mar 24 '13 at 13:48
  • @StefanHansen: I don't completely understand but I will have a think about it. Please can you recommend a reference? – David Mar 24 '13 at 14:01
  • I don't have a reference for the first part, but Measures, Integrals and Martingales by Schilling (Chapter 5) treats the uniqueness of measures. – Stefan Hansen Mar 24 '13 at 14:07
  • @StefanHansen: Thanks. Billingsley's Convergence of Probability Measures as recommended by Ju'x has a reference for the first statement. – David Mar 24 '13 at 14:11
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If is enough to check the equality of integrals with $f(x) = e^{i\langle x,\theta\rangle}$ for every $\theta \in \Bbb R^d$. This is the Fourier transform.

The proofs of both this and the general result given by Stefan can be found in Billingsley's Convergence of Probability Measures.

Siméon
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