If a sequence of measures is weakly convergent outside each compact ball, the sequence itself is weakly convergent

Question

Let $E$ be a $\mathbb R$-Banach space and $\mathcal M_+(E)$ denote the space of finite nonnegative measures on $\mathcal B(E)$.

If $\lambda\in\mathcal M_+(E)$, let $$\left.\lambda\right|_\delta(B):=\lambda\left((B\cap\left\{x\in E:\left\|x\right\|_E\le\delta\right\}\right)$$ and $$\left.\lambda\right|_\delta^c(B):=\lambda\left((B\cap\left\{x\in E:\left\|x\right\|_E>\delta\right\}\right)$$ for $B\in\mathcal B(E)$.

How can we show that if $(\lambda_n)_{n\in\mathbb N}\subseteq\mathcal M_+(E)$ is bounded in total variation norm (i.e. $\sup_{n\in\mathbb N}\lambda_n(E)<\infty$) and $(\left.\lambda_n\right|_\delta^c)_{n\in\mathbb N}$ is relatively compact with respect to the topology of weak convergence of measures for all $\delta>0$, then $(\lambda_n)_{n\in\mathbb N}$ itself is relatively compact?

I honestly don't know how we need to argue, but I'm sure that at some point the boundedness in total variation norm must be important.

Remark: Please feel free to assume that each $\lambda_n$ is Radon, i.e. $$\forall B\in\mathcal B(E):\forall\varepsilon>0:\exists K\subseteq E\text{ compact}:\lambda_n(B\setminus K)<\varepsilon\tag3.$$

Answer 1

I suppose here that $E$ is separable.

Let $\epsilon > 0$ and fix a sequence $\epsilon_i > 0$ such that $\sum_{i=1}^\infty \epsilon_i < \epsilon$.

Let $B_r$ denote the closed ball of radius $r$. Set $A_1 = B_1^c = \{x : \|x\| > 1\}$ and $A_i = \{x : \frac{1}{i} < \|x\| \le \frac{1}{i-1}\}$ for $i \ge 2$, so that $\bigcup_{k=1}^m A_k = B_{1/m}^c$. By assumption, for each $i$, the measures $\lambda_n|_{B_{1/i}^c}$ are relatively compact and therefore, by Prokhorov's theorem, tight. So there is a compact set $K_i$ such that $\lambda_n|_{B_{1/i}^c}(K_i^c) = \lambda_n(B_{1/i}^c \cap K_i^c) < \epsilon_i$ for all $n$. In particular, since $A_i \subset B_{1/i}^c$, we have $\lambda_n(A_i \setminus K_i) < \epsilon_i$.

Set $K_i' = \overline{A_i} \cap K_i$, which is still compact, and let $K = \bigcup_{i=1}^\infty K_i' \cup \{0\}$, which is also compact. (Consider an infinite subset $S$ of $K$; I claim it has a limit point in $K$. If $S$ has 0 as a limit point, we are done. If not, then $S$ is contained in $B_{1/m}^c$ for some $m$, and then $S \subset \bigcup_{i=1}^m K_i'$ which is compact, so that $S$ has a limit point in $\bigcup_{i=1}^m K_i' \subset K$). Note that since $0 \in K$, we have $K^c = \bigcup_{i=1}^\infty (A_i \cap K^c) \subset \bigcup_{i=1}^\infty (A_i \setminus K_i)$. Thus for every $n$, we have $$\lambda_n(K^c) \le \sum_{i=1}^\infty \lambda_n(A_i \setminus K_i) < \sum_{i=1}^\infty \epsilon_i < \epsilon.$$ So the sequence $\lambda_n$ is tight. Since it is tight and bounded in total variation norm, by Prokhorov again it is relatively compact.