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Let $X$ be Banach and let $B(x,\varepsilon)$ be the closed ball of radius $\varepsilon>0$ around $x\in X$ and consider the sequence $$f_{n;x}(y)= \begin{cases} 1-n\cdot d(yB(x,\varepsilon)), & (1-n\cdot d(y,B(x,\varepsilon)))\ge0\\ 0, &\text{elsewhere}. \end{cases} $$ for $y\in X$ and $n=1,2,\ldots$ and where $d(y,B(x,\varepsilon))=\inf\{\parallel y-z\parallel\colon z\in B(x,\varepsilon)\}$. The pointwise limit of this sequence is the characteristic function on $B(x,\varepsilon)$.

Let $BC(X)$ denote the space of closed and bounded subset of $X$, then the map $$f\colon X\to BC(X): x\mapsto B(x,\varepsilon)$$ is continuous for the Hausdorff metric $d_H$ . And for $A\in BC(X)$ fixed, the map $$g_A\colon BC(X)\to\mathbb{R}: B\mapsto\delta(A,B),$$ is 1-Lipschitz, where $\delta(A,B)=\sup_{a\in A}\inf_{b\in B}\parallel a-b \parallel$.

Is there somebody who knows how to prove the following:

Let $\{x_n: n=1,2,\ldots\}$ be a convergent sequence in $X$ with limit $x_0$, then $f_{n,x_j}\to f_{n,x_0}$ pointwise as $j\to\infty$.

My main goal is to show that the map $p\colon X\to\mathbb{R}:x\mapsto\mu(B(x,\varepsilon))$ is Borel measurable for any (not necessarily finite or positive) Borel measure $\mu$ on $X$. If I would be able to prove $f_{n,x_j}\to f_{n,x_0}$, then I'm done.

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Assume first that $y\in B(x_0,\varepsilon)$. Then $f_{n,x_0}(y)=1$. If $y\in B(x_j,\varepsilon)$, then $f_{n,x_j}(y)=1$. Otherwise, let $t=\varepsilon/d(y,x_j)<1$. Then $y_j=x_j+t(y-x_j)\in B(x_j,\varepsilon)$, since \begin{eqnarray} d(y_j,x_j)=d(x_j+t(y-x_j), x_j)&=&\|t(y-x_j)\|=t\|y-x_j\|=\varepsilon. \end{eqnarray} Then (note that below we use in a critical way that the distance is given by a norm) \begin{eqnarray} d(y,B(x_j,\varepsilon))+\varepsilon&\leq& d(y,y_j)+\varepsilon=d(y,y_j)+d(y_j,x_j)=\|(1-t)(y-x_j)\|+\|t(y-x_j)\|\\ &=&(1-t)\|y-x_j\|+t\|y-x_j\|=\|y-x_j\|\leq\|y-x_0\|+\|x_0-x_j\|\\ &\leq&\varepsilon+\|x_0-x_j\|, \end{eqnarray} so $$ d(y,B(x_j,\varepsilon))\leq\|x_0-x_j\|. $$ For $j$ big enough, $\|x_0-x_j\|\leq 1/n$, and so $$ f_{n,x_j}(y)=1-n\,d(y,B(x_j,\varepsilon))\geq1-n\|x_0-x_j\|\xrightarrow{j\to\infty}1. $$

The other case to consider is when $y\not\in B(x_0,\varepsilon)$. Then $f_{n,x_0}(y)=0$, and $d(y,x_0)>\varepsilon$. Then there exists $\delta>0$ such that $d(y,x_0)>\varepsilon+\delta$, and $$ d(y,x_j)\geq d(y,x_0)-d(x_0,x_j)>\varepsilon+\delta-d(x_0,x_j). $$ So, for $j$ big enough, we get $d(y,x_j)>\varepsilon$, i.e. $y\not\in B(x_j,\varepsilon)$ and $f_{n,x_j}(y)=0$.

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