Let $S$ be a metric space, $M\subset \bar{C}(S)$ ($\bar{C}(S)$ is the set of all bounded continuous functions on $S$) is called separating if whenever $P$, $Q$ are probability measures on $\mathcal B(S)$ and for every $f\in M$ $$\int f dP=\int f dQ$$ we have $P=Q$.

A claim in Ethier and Kurtz's Markov Processes Characterization and Convergence says that "Since $M$ separating, it is separating with respect to finite signed Borel measures as well".

I think at first we need to prove it in the condition that $P$ is a probability measure and $Q$ is a finite positive measure. In this condition, it is sufficient to prove $Q$ is a probability measure. But I am stuck with it. Any help please.


You are stuck because this is not true in general. Consider $S = \{0\}$. Then every two probability measures on $B(S)$ coincide, but not any two measures do.

Thus, the empty family (or the family consisting only of $x \mapsto 0$) is separating for probability measures, but not for general measures.

As a less trivial example, consider $S = \{0,1\}$ and $M = \{1_{\{0\}}\}$. Then $M$ is separating for probability measures, since if $$\mu(\{0\}) = \int 1_{\{0\}} \, d\mu = \int 1_{\{0\}} \, d\nu = \nu(\{0\}),$$ then also $$ \mu(\{1\}) = 1 - \mu(\{0\}) = 1 - \nu(\{0\}) = \nu(\{1\}), $$ but for general (finite) measures, this is obviously false.

If we require that the constant function $x \mapsto 1$ (or any constant, nonzero function) is contained in $M$, then $M$ is also separating for general finite (signed) measures, since (first case) if $\mu, \nu$ are positive measures, we get $$ \mu(S) = \nu(S). $$ Thus, if $\mu=0$, then $\nu=0$. Otherwise, we can normalize $\mu,\nu$ to be probability measures and get $\mu = \nu$, since $M$ is separating for probability measures.

Finally, if $P,Q$ are general finite signed measures, we have $P = P_1 - P_2$ and $Q = Q_1 - Q_2$ for positive measures $P_1, P_2, Q_1, Q_2$ and furthermore $$ \int f \, dP_1 - \int f \, dP_2=\int f \, dP = \int f \, dQ = \int f \, dQ_1 - \int f \, dQ_2 $$ and thus $$ \int f \, dP_1 + \int f \,dQ_2 = \int f \, dQ_1 + \int f \, dP_2 $$ for all $f \in M$. Hence, $P_q + Q_2 = Q_1 + P_2$ by what we just saw. Hence, $P=Q$.

  • $\begingroup$ You are right...But how about if $S$ is not trivial as in your example? $\endgroup$ – Danielsen Nov 30 '15 at 15:06
  • $\begingroup$ @Danielsen: I expanded my answer a bit. $\endgroup$ – PhoemueX Nov 30 '15 at 19:54
  • $\begingroup$ I get it. The claim in the book also requires $1\in M$ which I missed. Thank you very much! $\endgroup$ – Danielsen Dec 1 '15 at 2:20

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