Let $d\in\mathbb N$, $\eta\in C_c^\infty(\mathbb R^d)$ with $\operatorname{supp}\eta\subseteq B_1(0)$, $\eta\ge0$, $\int\eta(x)\:{\rm d}x=1$ and $$\eta_\varepsilon(x):=\frac1{\varepsilon^d}\eta\left(\frac x\varepsilon\right)\;\;\;\text{for }x\in\mathbb R^d.$$
Now let $x\in\mathbb R^d$, $r>0$, $f:B_r(x)\to\mathbb R$ be Lebesgue integrable, $$\tilde f(y):=\begin{cases}f(y)&\text{, if }y\in B_r(x)\\0&\text{, otherwise}\end{cases}$$ for $y\in\mathbb R^d$ and $f_\varepsilon:=f\ast\eta_\varepsilon$ for $\varepsilon>0$. Note that $$\left\|f_\varepsilon-f\right\|_{L^1(B_r(x)}\le\left\|f_\varepsilon-\tilde f\right\|_{L^1(\mathbb R^d)}\xrightarrow{\varepsilon\to0+}\tag10.$$
In particular, there is a sequence $(\varepsilon_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $\varepsilon_n\xrightarrow{n\to\infty}0$ and $$f_{\varepsilon_n}\xrightarrow{n\to\infty}\tilde f\;\;\;\text{almost everywhere}\tag2.$$
Question: Can we use the dominated convergence theorem to conclude that, for all $t\in(0,r)$, $$\int f_{\varepsilon_n}\:{\rm d}\sigma_{\partial B_t(x)}\xrightarrow{n\to\infty}\int f\:{\rm d}\sigma_{\partial B_t(x)}\tag3,$$ where $\sigma_{\partial B_t(x)}$ denotes the surface measure on $\mathcal B(\partial B_t(x))$?
Most probably we need to assume that $f$ is bounded to apply Lebesgue's dominated convergence theorem.
If $f$ is continuous, this convergence is true and follows from the uniform convergence of the mollified versions. Without continuity, this fails in general: You can change the values of $f$ on the null set $\partial B_t(x)$ without changing the values of the mollified functions $f_\epsilon$.