Let $(\delta\_k)\_{k\in\mathbb N}\subseteq(0,\infty)$ be nonincreasing with $\delta\_k\xrightarrow{k\to\infty}0$ and $(\varepsilon\_k)\_{k\in\mathbb N}\subseteq(0,\infty)$ with $\sum\_{k\in\mathbb N}\varepsilon\_k<1/2$ and $\sum\_{k\in\mathbb N}\varepsilon\_k/\delta\_k\le1/2$.


Let $E$ be a Banach space, $(\mu\_n)\_{n\in\mathbb N},(\nu\_n)\_{n\in\mathbb N}$ be sequences of probability measures on $\mathcal B(E)$ and $K\_k\subseteq E$ be compact with $$\sup\_{n\in\mathbb N}(\mu\_n\ast\nu\_n)(K^c\_k)<\varepsilon\_k\tag1,$$ where $\mu\_n\ast\nu\_n$ denotes the convolution, for $k\in\mathbb N$.


Now let $$B\_n:=\bigcap\_{k\in\mathbb N}\left\{x\in E:\mu\_n(K\_k-x)>1-\delta\_k\right\}$$ for $n\in\mathbb N$.



> 
> Why does it follow that $\nu\_n(B\_n)\ge1/2$ for all $n\in\mathbb N$?
> 
> 
> 


We may clearly write $$(\mu\_n\ast\nu\_n)(K\_k)=\int\nu\_n({\rm d}y)\mu\_n(K\_k-y)\ge\int\_{B\_n}\nu\_n({\rm d}y)\mu\_n(K\_k-y)\ge(1-\delta\_k)\nu\_n(B\_n)\tag2,$$ but that doesn't help, since $(1)$ only yields an inequality in the same direction: $$(\mu\_n\ast\nu\_n)(K\_k)\ge1-\varepsilon\_k\tag3.$$


**EDIT**:
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I still wasn't able to figure it out, but it somehow must follow from $$\mu\_n(K\_k-y)<\delta\_k\Leftrightarrow\frac1{\delta\_k}<\frac1{\mu\_n(K\_k-x)}\Leftrightarrow\frac{\varepsilon\_k}{\delta\_k}<\frac{\varepsilon\_k}{\mu\_n(K\_k-x)}<\frac{(\mu\_n\ast\nu\_n)(K\_k)}{\mu\_n(K\_k-x)}\tag4,$$ which holds for all $x\in B\_n$, and $$\nu\_n(B\_n)=\int\_{B\_n}\nu\_n({\rm d}x)\frac{\mu\_n(K\_k-x)}{\mu\_n(K\_k-x)}\ge\frac1{\delta\_k}\int\_{B\_n}\nu\_n({\rm d}x)\mu\_n(K\_k-x)\tag5.$$ The right-hand side is equal to $$\underbrace{(\mu\_n\ast\nu\_n)(K\_k)}\_{\ge\:1-\varepsilon\_k}-\underbrace{\int\_{B\_n}\nu\_n({\rm d}x)\mu\_n(K\_k-x)}\_{\le\:1}\ge-\varepsilon\_k\tag6,$$ but that's clearly useless, since it trivially holds $\nu\_n(B\_n)\ge0>-\varepsilon\_k/\delta\_k$ ...




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*Remark*: I've asked [this question on MSE](https://math.stackexchange.com/q/3962424/47771) before, but didn't receive an answer; even with a bounty. (The question on MSE will be deleted after the bounty grace period has ended.)



How can we show this estimate for the convolution of two probability measures?
$\newcommand\ep\varepsilon\newcommand\de\delta$Let $\mu:=\mu\_n$, $\nu:=\nu\_n$, $B:=B\_n$,
$$C\_k:=\{x\in E:\mu(K\_k-x)>1-\de\_k\},$$
so that
$$B=\bigcap\_k C\_k.$$
We have
$$1-\ep\_k<(\mu\*\nu)(K\_k)=\int\mu(K\_k-x)\nu(dx) \\
=\int\_{C\_k}\mu(K\_k-x)\nu(dx)+\int\_{C\_k^c}\mu(K\_k-x)\nu(dx) \\
\le\nu(C\_k)+(1-\de\_k)\nu(C\_k^c)=1-\de\_k+\de\_k\nu(C\_k),$$
whence
$$\nu(C\_k)\ge1-\ep\_k/\de\_k.$$
Thus,
$$\nu(B)\ge1-\sum\_k\ep\_k/\de\_k\ge1/2,$$
as desired.


