# Is a tight finite measure necessarily separately-valued and uniquely determined by its characteristic function?

Let $$E$$ be a Hausdorff space and $$\mu$$ be a tight$$^1$$ finite measure on $$E$$.

Is it possible to show that there is a closed separable $$E_0\subseteq E$$ such that $$\mu(E_0)=\mu(E)$$?

If not, I'm also interested in the same question under the additional assumption that $$E$$ is metrizable and/or $$\mu$$ is Radon$$^1$$.

Finally, what I would like to conclude is that if $$E$$ is a normed vector space, then $$\mu$$ is uniquely determined by its characteristic function $$\hat\mu:E'\to\mathbb C\;,\;\;\;\varphi\mapsto e^{{\rm i}\varphi}\:{\rm d}\mu.$$ In the case $$E=\mathbb R^d$$, this is a basic result and easy to verify. Once again, if it doesn't hold for an arbitrary normed vector space, I'm also interested in sufficient conditions for the claim to hold.

$$^1$$ Remember that $$\mu$$ is called

• tight if for all $$\varepsilon>0$$, there is a compact $$K\subseteq E$$ with $$\mu(K^c)<\varepsilon\tag1;$$
• Radon if for all $$B\in\mathcal B(E)$$ and $$\varepsilon>0$$, there is a compact $$K\subseteq E$$ with $$\mu(B\setminus K)<\varepsilon\tag2.$$

It is easy to verify that $$\mu$$ is tight iff $$\mu(E)=\sup_{K\subseteq E\text{ is compact}}\mu(K)\tag{1'}$$ and $$\mu$$ is Radon iff $$\forall B\in\mathcal B(E):\mu(B)=\sup_{\substack{K\subseteq E\text{ is compact}\\ K\subseteq B}}\mu(K)\tag{2'}.$$

• With regard to the first question, you could look at Haar measure on a non-separable compact group (e.g. an uncountable product of circle groups). For the second, note that in a metrisable space, compact subsets (and hence countable unions thereof) are separable. Commented Apr 29, 2021 at 13:35
• Another counterexample for question 1 is the Dieudonné measure on $E=\omega_1 + 1$. The tightness is trivial because $E$ is compact. Commented Apr 29, 2021 at 13:40
• For the third question, a sufficient condition is that $E'$ generates the Borel $\sigma$-algebra of $E$. This is true in particular when $E$ is separable. I don't have a reference with a proof off the top of my head, but it's a pretty routine application of the multiplicative system theorem. Commented Apr 29, 2021 at 13:54
• @bathalf15320 Small but important correction - "uncountable product" is not good enough, a continuum-sized product of separable spaces is separable. Commented Apr 29, 2021 at 13:59

I decided not to "comment-answer" this question, so other people have duplicated some of my answer in the comments.

Since there are several questions, I will separate them and answer them individually.

If $$E$$ is a Hausdorff space and $$\mu$$ is a finite Radon measure on $$E$$ is there a closed separable subspace $$E_0$$ with $$\mu(E_0) = \mu(E)$$?

The answer to this one is no, and is a good illustration of how separability does not behave as people expect it to outside of metrizable spaces. Take $$X = 2^{2^{\aleph_0}}$$, and let $$E = 2^X$$ with the product of the discrete topologies on $$2$$. This is a compact Hausdorff space, and the product measure $$\mu$$ of independent fair coin tosses on $$2$$ is a Radon probability measure on it (this can be deduced from various "big theorems", e.g. the Radonness of independent products of Radon measures, the Riesz representation theorem, the existence of Haar measures on compact Hausdorff groups, so I'll leave it to you).

The support of $$\mu$$ is $$2^X$$ itself, so for any closed set $$C$$ with $$\mu(C) = 1$$, $$C = 2^X$$. We can therefore finish the counterexample by observing that $$2^X$$ is not separable. The reason is that the cardinality of $$2^X$$ is $$2^{2^{2^{\aleph_0}}} > 2^{2^{\aleph_0}}$$, and any separable Hausdorff space has cardinality $$\leq 2^{2^{\aleph_0}}$$, because the set of ultrafilters on a countable set has cardinality $$2^{2^{\aleph_0}}$$, and every point in a separable space is the limit of an ultrafilter on a fixed countable dense set.

If $$E$$ is metrizable and $$\mu$$ is a finite tight measure on $$E$$, is there a closed separable $$E_0$$ such that $$\mu(E_0) = \mu(E)$$?

Here the answer is yes. For each $$N \in \mathbb{N}$$, Let $$K_n$$ be a compact set with $$\mu(E \setminus K_n) < 2^{-n}$$. Compact metrizable spaces are separable, so each $$K_n$$ is separable. Countable unions of separable spaces are separable, so $$D = \bigcup_{n=1}^\infty K_n$$ is separable. Finally, closures of separable subspaces are separable, so $$E_0 = \overline{D}$$ is separable. Then $$\mu(X \setminus E_0) \leq \mu(X \setminus D) = 0$$, by countable additivity, so $$\mu(E_0) = \mu(E)$$.

We note in passing at this point that since Borel measures on metrizable spaces are inner regular with respect to closed sets, tightness and Radonness are the same thing in this case.

If $$\mu_1$$ and $$\mu_2$$ are finite Radon measures on a normed space $$E$$ with the same characteristic function, is $$\mu_1 = \mu_2$$?

This is true (and in considerably more generality). See Corollary 1 on page 201 of Vakhania, Tarieladze and Chobanyan's Probability Distributions on Banach Spaces.