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The questions was asked by me on Math StackExchange, but no answer appears, so I ask for help again.

Let $(X, d)$ be a complete (Hausdorff, separable, local compact and other nice properties you want) metric space and $\mathcal{M}$ be the space of local finite full supported Borel probability measures on $X$ (with Borel sigma-algebra $\mathcal{B}$). Then the total variation metric can be defined by $d_{TV}:= \sup_\limits{A}|\mu(A)- \nu(A)|$ where $A$ runs over $\mathcal{B}$ and $\mu \in\mathcal{M}, \nu\in \mathcal{M}$.

Questions:

  1. Is $d_{TV}$ a complete metric on $\mathcal{M}$? Is $(\mathcal{M}, d_{TV})$ a Hausdorff/separable space?
  2. When the subset $\mathcal{N}$ ($\subset \mathcal{M}$) is pre-compact?

  3. Let $(X,d)=([0, 1], | |)$, where $| |$ is the Euclidean metric, then if it has sequence $\{\mu_i\}$ such that $\{\mu_i\}$ weak converge to $\mu$, but not $d_{TV}$-converge to $\mu$? Here $\mu_i, \mu \in \mathcal{M} $.

How about the questions when $(X, d)$ is a Riemannian manifold or require the meassures are Radon measures?

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Answer to Question 3: Yes, there is such a sequence. E.g., for all $n=0,1,\dots$ and all Borel $B\subseteq[0,1]$, let
$$\mu_n(B):=\frac12\int_B(2+\sin2\pi nx)\,dx.$$ Then $\mu_n$ converges to $\mu_0$ weakly but not in total variation. (Also, $\mu_n$ is full supported for each $n$.)

Concerning Question 1: In general, $d_{TV}$ is not a complete metric on $\mathcal M$. Indeed, for all $n=0,1,\dots$ and all Borel $B\subseteq X:=[0,1]$, let
$$\mu_n(B):=(2-1/n)|B\cap[0,1/2]|+(1/n)|B\cap[1/2,1]|,$$ where $|\cdot|$ is the Lebesgue measure, and $$\mu(B):=2|B\cap[0,1/2]|.$$ Then $\mu_n\in\mathcal M$ for all $n$, $d_{TV}(\mu_n,\mu_k)\to0$ as $n,k\to\infty$, but $\mu_n\to\mu$ in total variation as $n\to\infty$ and $\mu\notin\mathcal M$.

Also concerning Question 1: The metric space $(\mathcal{M}, d_{TV})$ is Hausdorff, as any metric space.

The metric space $(\mathcal{M}, d_{TV})$ is not separable in general. Indeed, for all $x\in X:=[0,1]$ and all Borel $B\subseteq[0,1]$, let
$$\mu_x(B):=\tfrac12\,|B|+\tfrac12\,1_B(x).$$ Then the subset $\{\mu_x\colon x\in[0,1]\}$ of $\mathcal{M}$ is uncountable and $d_{TV}(\mu_x,\mu_y)=1/2$ for any distinct $x$ and $y$ in $[0,1]$. So, $(\mathcal{M}, d_{TV})$ is not separable (cf. e.g. the first paragraph in this answer).

Concerning Question 2: $\mathcal N$ will be pre-compact if and only if $\mathcal N$ is totally bounded and closed in $d_{TV}$ as a subset of the set of all probability measures on $X$.

NB: Try to avoid asking multiple questions in one post.

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