If a measure only assumes values 0 or 1, is it a Dirac's delta?

Solution 1:

As delfonics says, the answer is: yes, if and only if there does not exist a measurable cardinal, and an outline of the proof was given by hot_queen in an answer to a question of mine from 2014 (Consistency strength of 0-1 valued Borel measures), which I somehow completely forgot about. In the meantime, I had written out the argument below, which essentially fills in the details in hot_queen's argument.

One direction is easy. Suppose $\kappa$ is a measurable cardinal, so that there exists a $\kappa$-additive (in particular, countably additive) $\mu : 2^{\kappa} \to \{0,1\}$ which is not a Dirac mass. Let $M$ be $\kappa$ equipped with the discrete metric (or for that matter, any other metric that $\kappa$ might happen to admit). Then $\mu$ (or its restriction to the Borel sets, if we are using a non-discrete metric) is the desired counterexample.

The other direction I found in the paper [1]. The argument, for the current case, goes as follows.

Working in ZFC, suppose there is a metric space $M$ and a 0-1 valued Borel measure $\mu$ which is not a point mass. Using the axiom of choice, $M$ is in one-to-one correspondence with some cardinal $\kappa$, so we can write $M = \{x_\alpha : \alpha \in \kappa\}$. (In other words, we have well-ordered $M$.) Note each $\alpha \in \kappa$ is itself an ordinal.

For every $\alpha$, we have $\mu(\{x_\alpha\})=0$, and $\{x\}$ is a decreasing intersection of the open balls $B(x_\alpha,1/n)$. So by continuity from above, there is an open ball $B_\alpha$ centered at $x_\alpha$ with $\mu(B_\alpha) = 0$. Set $H_\alpha = B_\alpha \cap \left( \bigcup_{\beta \in \alpha} B_\beta\right)^c$. (Some of the $H_\alpha$ might be empty but that is okay.) Note that $H_\alpha$ is the intersection of an open (hence $F_\sigma$) set and a closed set, so $H_\alpha$ is Borel (indeed $F_\sigma$). And since $H_\alpha \subset B_\alpha$ we have $\mu(H_\alpha) = 0$. By construction, the $H_\alpha$ are pairwise disjoint, and $\bigcup_{\alpha \in \kappa} H_\alpha = M$. Now they cite a result of D. Montgomery [2] which asserts that any arbitrary union of the $H_\alpha$ is in fact an $F_\sigma$; in particular it is Borel.

Now define a measure $\nu : 2^{\kappa} \to \{0,1\}$ by $\nu(Y) = \mu\left(\bigcup_{\alpha \in Y} H_\alpha\right)$. Since $\mu$ is countably additive and the $H_\alpha$ are disjoint, we have that $\nu$ is countably additive. Moreover, for any $\alpha$ we have $\nu(\{\alpha\}) = \mu(H_\alpha) = 0$, and $\nu(\kappa) = \mu(M) = 1$. Thus $\kappa$ admits a nontrivial countably additive 0-1 valued measure on all its subsets.

A measurable cardinal $\lambda$ has to have a measure which is not only countably additive but actually $\lambda$-additive, so to finish, we use an argument due to Ulam, mentioned on the Wikipedia page above. Since we have shown there is a cardinal (namely $\kappa$) with a nontrivial countably additive 0-1 valued measure on all its subsets, there is a minimal cardinal with this property; call it $\kappa_1$ and let $\nu_1 : 2^{\kappa_1} \to \{0,1\}$ be the corresponding countably additive measure.

Suppose $\nu_1$ were not $\kappa_1$-additive; that means there is a collection $\mathcal{C} \subset 2^{\kappa_1}$, having $\kappa_0 := |\mathcal{C}| < \kappa_1$, such that $\mathcal{C}$ consists of pairwise disjoint sets of $\nu_1$-measure zero, and yet $\nu_1\left(\bigcup \mathcal{F}\right) = 1$. Fix a bijection $\phi : \kappa_0 \to \mathcal{C}$ and define a measure $\nu_0 : 2^{\kappa_0} \to \{0,1\}$ by $\nu_0(B) = \nu_1\left(\bigcup_{\beta \in B} \phi(\beta)\right)$. Then $\nu_0$ is countably additive; for any $\beta \in \kappa_0$ we have $\nu_0(\{\beta\}) = \nu_1(\phi(\beta)) = 0$ since every set in $\mathcal{C}$ had measure zero; and $\nu_0(\kappa_0) = \nu_1(\bigcup \mathcal{F}) = 1$. So $\nu_0$ is nontrivial, and this contradicts the minimality of $\kappa_1$.

We have thus shown that $\kappa_1$ is a measurable cardinal.

[1] Marczewski, E.; Sikorski, R. Measures in non-separable metric spaces. Colloquium Math. 1, (1948). 133–139. MR 25548

[2] Montgomery, D. Non-separable metric spaces. Fundamenta Mathematicae 25, (1935). 527–533.

Note: I wasn't able to find a copy of Montgomery's paper online, and it doesn't seem to be indexed in MathSciNet. If someone has a copy of this paper, or knows where to find another proof of the result, I would be interested to hear it. I found a number of other references mentioning this result, so it seems to be fairly well established.

Solution 2:

The answer is yes iff there is no measurable cardinal. For the less trivial direction, see this.