Let's call a measure $\mu$ on a measurable space $(X, \mathcal{A})$ multiplicative if $$\mu(A \cap B) = \mu(A) \mu(B) \tag{$*$}$$ for all $A,B \in \mathcal{A}$. Any dirac measure $\delta_x$ for some $x \in X$ is multiplicative. Do there exist nonzero multiplicative measures that are not dirac measures?

If so, then can we make additional assumptions on the measurable space and the measure such that nonzero multiplicative implies dirac? The implication is true if $X$ is a compact Hausdorff space and $\mu$ is a regular Borel measure on $X$, but does it hold in more general circumstances?

EDIT: As Jonas pointed out, $(*)$ becomes meaningless if $\mu(A) = 0$ and $\mu(B) = \infty$, so we should not require $(*)$ to hold for pairs of measurable sets where one has measure zero and the other has measure infinity.

EDIT: Under no extra assumptions, $\mathcal{A} = \{\emptyset, X\}$ with $\mu(X) = \infty$ is a counter-example, though a rather uninteresting one. If we assume that at least one measurable set $A$ has finite positive measure then if $\mu(B)=\infty$ we get $\infty > \mu(A) \geq \mu(A \cap B) = \mu(A)\mu(B) = \infty$, a contradiction, so no sets can have infinite measure in this case. Let's therefore assume that the measure space is finite, i.e. all measurable sets have finite measure.

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    $\begingroup$ $\mu(A\cap A) = \left(\mu(A)\right)^2$, which means $\mu(A)$ is either $0$ or $1$ for any measurable set $A$. $\endgroup$
    – Qiyu Wen
    Feb 17 '16 at 9:00
  • $\begingroup$ Yes, that is a very important consequence of being multiplicative. It also implies $\mu(X) = 1$, because otherwise $\mu$ would be zero. $\endgroup$
    – Ulrik
    Feb 17 '16 at 9:02
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    $\begingroup$ Humm... $\mu(A)=\mu(A)^2$ also allows that $\mu(A)=\infty$. $\endgroup$
    – John B
    Feb 17 '16 at 9:33
  • $\begingroup$ Oh, you're right! Thanks $\endgroup$
    – Ulrik
    Feb 17 '16 at 9:34

Let $X$ be a topological space and let $\mu$ be a nonzero Borel measure on $X$ that is outer regular on Borel sets and multiplicative.

As discussed in the opening post, if some Borel set has measure $\infty$ then all Borel sets have either measure sero or $\infty$. To prevent this situation, we assume that $\mu$ is finite.

As already commented, finiteness and multiplicativity implies $\mu(A) = \mu(A \cap A) = \mu(A)^2$ implies $\mu(A) = 0$ or $\mu(A) = 1$ for every measurable set $A$. Since $\mu \neq 0$ we must have $\mu(X) = 1$.

Now assume that there exists a compact set $K \subseteq X$ with positive measure, i.e. $\mu(K) = 1$. This would for instance happen if $X$ was locally compact or if $\mu$ was inner regular.

Now for the proof. Suppose that $\mu$ is not a dirac measure. Then, for any $x \in K$, there exists a measurable set $A_x$ with $x \in A_x$ and $\mu(A_x) = 0$. By outer regularity, for each $x \in X$ there exists an open set $U_x \supseteq A_x$ with $\mu(U_x \setminus A_x) < 1$, i.e. $\mu(U_x) = 0$. Now we have an open covering $$K = \bigcup_{x\in K} U_x$$ of $K$, which then admits a finite subcovering $$K = U_1 \cup \cdots \cup U_n$$ But this is impossible, as all the $U_i$'s have measure zero while $X$ has measure one. Hence, $\mu$ must be a Dirac measure.


I'll assume that the measure has the property that any point has a neighborhood with finite measure. Otherwise one would have the awkward condition $$ 0=\mu(A\cap\emptyset)=\mu(A)\mu(\emptyset)=\infty\times 0 $$ taking $A$ nonempty.

Now let me reply to your question in the particular case when there is an atom $x$ and a decreasing sequence of sets $B_n$ with intersection $x$. Then replacing $B$ by $B_n$ and letting $n\to\infty$ we obtain $$ \mu(A\cap\{x\})=\mu(A)\mu(\{x\}). $$ For $A$ containing $x$ we get $\mu(A)=1$, and otherwise $\mu(A)=0$. So $\mu=\delta_x$ (in this particular case).

  • $\begingroup$ If $x \in A$ then $\mu(A) = \mu(A \cap \{ x\}) = \mu(A) \mu(\{x\})$, but why does this imply $\mu(A) = 1$? Are you assuming that $\mu(\{x\}) = 1$? $\endgroup$
    – Ulrik
    Feb 17 '16 at 10:17
  • $\begingroup$ @Svinepels Just rephrased my answer, not what you want but the best I can do for now. $\endgroup$
    – John B
    Feb 17 '16 at 10:39
  • $\begingroup$ If there exists a measurable singleton $\{x\}$ with $\mu(\{x\}) = 1$, then it follows that $\mu = \delta_x$ (at least if the measure space is finite) so in this case, the argument with the sequence $B_n$ is not necessary. $\endgroup$
    – Ulrik
    Feb 17 '16 at 11:15

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