According to Lebesgue's differentiation theorem, $$\lim_{r\to 0}\frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)-f(x)|dx=0$$ holds for a.e. $x\in \Omega$
Then for a fixed point $x$, \begin{align}|f^{\epsilon}-f|&=|\int_{B(x,\epsilon)}\eta_{\epsilon}(x-y)[f(y)-f(x)]dy|\\ &\leq \frac{1}{\epsilon^n}\int_{B(x,\epsilon)}\eta(\frac{x-y}{\epsilon}|f(y)-f(x)|)dy\\ &\leq C\frac{1}{|B(x,\epsilon)|}\int_{B(x,\epsilon)}|f(y)-f(x)|dy\to0 \;as \; \epsilon\to 0 \end{align}
i.e. $f^{\epsilon}$converges to $f$ a.e. in $\Omega$.
$\eta$ is the standard mollifier
$$ \eta(x)=\begin{cases} C \exp(\frac{1}{|x|^2-1})\;\; if |x|<1\\ 0\;\;if |x|\geq 1 \end{cases} $$ the constant C is selected so that $\int_{R^n}\eta(x)dx=1$. and for each $\epsilon>0$. $$\eta_{\epsilon}(x)=\frac{1}{\epsilon ^n}\eta(\frac{x}{\epsilon})$$
f is locally integrable, $$f^{\epsilon}=\int_{\Omega}\eta_{\epsilon}(x-y)f(y)dy$$ for $x\in \Omega_{\epsilon}$.
Now for some questions I'm working on, I want to use the Lebesgue dominated convergence theorem, and for that, I need to show that$|f^{\epsilon}-f|$is a.e. uniformly bounded for small enough $\epsilon$ so that I can find a dominating function.
