Space of probability measures total bounded?

Question

I want to consider a space of probability measures on some set $\Omega$. It's complete (am I right?). But I don't know whether it's total bounded. Actually, I want to prove that the space of probability measures over $(\Omega,\Sigma)$ is compact w.r.t. total variation metric, where $\Omega$ is a compact Hausdorff space and $\Sigma$ is $\sigma$-algebra. I'm new in studying probability measures, so I can't tell -- How can I prove it?

Thanks!

Answer 1

It is complete but not compact.

Here is an interesting example. In the compact space $[0,1]$, let measure $\mu_n$ have mass $1/n$ at each of the $n$ equally-spaced points $1/n, 2/n, 3/n, \cdots, n/n$. The variation distance from $\mu_n$ to $\mu_{n+1}$ is approximately $2$, so our space of probability measures is not totally bounded. In the narrow topology the sequence $\mu_n$ converges to Lebesgue measure $\lambda$ on $[0,1]$. That is, $$ \lim_{n\to\infty} \int_{[0,1]} f\;d\mu_n = \int_{[0,1]} f\;d\lambda $$ for all continuous $f \colon [0,1] \to \mathbb R$.

The space $P[0,1]$ of all Borel probability measures on $[0,1]$ is compact in the narrow topology.

Answer 2

Endow the unit interval $[0,1]$ with its Borel-$\sigma$-algebra. For each positive natural number $n$, let $\mu_n$ be the probability measure with density $$d(x)=\max\{2n-2n^2x,0\}.$$

For all $n$, $\mu_n\big(\{0\}\big)=0$ and for all $\epsilon>0$, $\mu_n((\epsilon,1])=0$ for $n$ large enough. So any convergent subsequence would have to converge to the zero measure which is not a probability measure.