Let $X$ be the set of all Borel probability measures on [0,1] (include those Dirac Delta distributions), is $X$ locally compact Hausdorff?

Question

Let $X$ be the set of all Borel probability measures on [0,1], endowed with the topology of weak convergence. Is $X$ locally compact Hausdorff?

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I learned that a normed space is locally compact iff it is finite dimensional. $X$ is a normed space I guess. So $X$ is not locally compact?

Possible answer: Theorem 4.3 of this note https://math.nyu.edu/~varadhan/Spring11/topics4.pdf suggests that the compactness of $[0,1] $ implies the compactness of space $X$.

Answer 1

It's not just locally compact. It's compact.

Let $K$ be compact, Hausdorff, metrizable ($K=[0,1]$ in this case). Let $C(K)$ be the Banach space of continuous, real-valued functions on $K$ and let $\mathcal{M}(K)$ be the space of all regular Borel measures on $K$. By (one version of) the Riesz representation theorem, $\mathcal{M}(K)$ is isometrically isomorphic to $C(K)^*$, the space of all continuous, linear functionals on $C(K)$ with the action $\mu(f)=\int_K fd\mu$. The dual norm is given by $\|\mu\|=\sup_{\|f\|\leqslant 1}|\mu(f)|$, where $\|f\|=\max_{x\in K}|f(x)|$. It is part of the Riesz representation that this norm is equal to the total variation norm $$|\mu|(K)=\sup\Bigl\{\sum_{n=1}^\infty |\mu(B_n)|:(B_n)_{n=1}^\infty\text{ a Borel partition of }K\Bigr\}.$$ The dual ball $B=\{\mu:\|\mu\|\leqslant 1\}$ is compact metrizable in the weak$^*$-topology by the Banach-Alaoglu theorem. But this is just the topology of pointwise convergence on every member of the predual, ie $(\mu_\lambda)$ is weak$^*$-convergent to $\mu$ if and only if $\lim_\lambda \int_K fd\mu_\lambda=\int_K fd\mu$ for all $f\in C(K)$. This is the topology of weak convergence of probability measures (which is distinct from the weak topology on a Banach space).

In order to provide an affirmative answer to your question, we need to know that $X$ is a weak$^*$-closed subset of $B$, since then it is a weak$^*$-closed subset of a compact metric space, and is also therefore a compact metric (Hausdorff) space.

Let $P$ denote the set of all positive measures in $\mathcal{M}(K)$. A measure $\mu$ is positive if and only if $\mu(B)\geqslant 0$ for all Borel $B\subset K$, which is equivalent to $\int fd\mu\geqslant 0$ for all non-negative $f$. One direction of this equivalence is clear. The other uses Jordan decomposition into positive negative parts, regularity, and Urysohn's lemma to show that if $\mu$ is not positive, there exists a non-negative $f\in C(K)$ with $\int fd\mu<0$.

Let $T$ denote the set of all $\mu\in \mathcal{M}(K)$ such that $\mu(K)=1$. This is the same as saying $\int_K 1_Kd\mu=1$, since $\mu(K)=\int_K 1_Kd\mu$.

We will argue that $P,T$ are weak$^*$-closed and $X=B\cap P\cap T$. Since $B$ is weak$^*$-compact, and therefore weak$^*$-closed, $X$ will be realized as a weak$^*$-closed subset of a weak$^*$-compact (metrizable, Hausdorff) space $B$, and $X$ is also compact Hausdorff metrizable.

1. Assume $(\mu_\lambda)_\lambda\subset P$ converges weak$^*$ to $\mu$. Fix a non-negative $f\in C(K)$. Since $\mu_\lambda\in P$, $\int_K fd\mu_\lambda\geqslant 0$. Therefore $\int_K fd\mu=\lim_\lambda fd\mu_\lambda\geqslant 0$. Since this holds for any non-negative $f\in C(K)$, $f\in P$ and $P$ is weak$^*$-closed.

2. Assume $(\mu_\lambda)\subset T$ converges weak$^*$ to $\mu$. Then $\int_K 1_Kd\mu =\lim_\lambda \int_K 1_Kd\mu_\lambda=1$. So $\mu\in T$ and $T$ is weak$^*$-closed.

On one hand, every Borel probability measure $\mu$ on a compact metrizable space $K$ is regular, positive, and has total variation measure $|\mu|(K)$ equal to the measure $\mu(K)$ equal to $1$. Regular with total variation equal to $1$ means $\mu\in B$. It's positive, so it's in $P$. It has $\mu(K)=1$, so it's in $T$. Thus $X\subset B\cap P\cap T$. On the other hand, any member $\mu$ of $B\cap P\cap T$ is a positive Borel measure with $\mu(K)=1$, so it is a Borel probability on $K$, and $B\cap P\cap T\subset X$.