\nu}= 0,&\text{on }\partial B_{1-{\varepsilon}}.
\end{cases}$$

where $\nu$ denotes the inward-pointing normal on the sphere $\partial B_{1-{\varepsilon}}$. One can show that ${\lambda}_{\varepsilon}$ is simple and strictly positive and that $u_{\varepsilon}$ is radially symmetric and has a constant sign, say positive. In particular, the restriction of $u_{\varepsilon}$ to $\partial B_{1-{\varepsilon}}$ equals a positive constant, $c_{\varepsilon}> 0$. We continuously extend $u_{\varepsilon}$ inside $B_{1-{\varepsilon}}$ by $c_{\varepsilon}$ and denote the resulting function, defined on ${\overline}B_1$, by $v_{\varepsilon}$.

Since the normal derivatives of $v_{\varepsilon}$ on both sides of $\partial B_{1-{\varepsilon}}$ vanish, $\Delta  v_{\varepsilon}$ is an $L^2$-function. More precisely, one has $$-\Delta v_{\varepsilon}= {\lambda}_{\varepsilon}u_{\varepsilon}1_{A_{\varepsilon}}, \quad \mbox  {in } {\mathcal D}'(B_1),$$ where ${\mathcal D}'(B_1)$ denotes the space of distributions on $B_1$. By adding on both sides the $L^2$ function $ {\lambda}_{\varepsilon}v_{\varepsilon}1_{{\overline}B_{1-{\varepsilon}}}$ we get\
$$-\Delta v_{\varepsilon}+  {\lambda}_{\varepsilon}v_{\varepsilon}1_{{\overline}B_{1-{\varepsilon}}}= {\lambda}_{\varepsilon}v_{\varepsilon},\quad \mbox  {in } {\mathcal D}'(B_1).$$ We view $\mu :=  {\lambda}_{\varepsilon}1_{{\overline}B_{1-{\varepsilon}}}$ as a capacitary measure on $B_1$. Then formally, we get $$-\Delta v_{\varepsilon}+ \mu  v_{\varepsilon}= {\lambda}_{\varepsilon}v_{\varepsilon},\quad \mbox  {in } {\mathcal D}'(B_1).$$ Combined with the fact that $v_{\varepsilon}>0$, this means that ${\lambda}_{\varepsilon}$ is the first Dirichlet eigenvalue of $- \Delta + \mu$. We assume from now on that $v_{\varepsilon}$ is normalized by $\|v_{\varepsilon}\|_{L^2(B_1)}=1$. Since the measure $\mu$ is finite with support in $B_1$, 