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 $B\in\mathcal B(E)$.



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> 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](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures) for all $\delta>0$, then $(\lambda\_n)\_{n\in\mathbb N}$ itself is relatively compact?
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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. $$\forall B\in\mathcal B(E):\forall\varepsilon>0:\exists K\subseteq E\text{ compact}:\lambda\_n(B\setminus K)<\varepsilon\tag3.$$



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


