Here is the problem:
Let $\phi=\phi(x)$ be a bounded measurable function in $\mathbb{R}^d$ such that $\phi(x)=0$ for $|x|\geq 1$ and $\int_{\mathbb{R}^d}\phi=1$. Let $\phi_\varepsilon(x)=\varepsilon^{-d}\phi(x/\varepsilon)$, where $\varepsilon>0$. Define $$ f*\phi_\varepsilon(x)=\int_{\mathbb{R}^d}f(x-y)\phi_\varepsilon(y)dy. $$ Show that if $x$ is a Lebesgue point of $f$, then $$ \lim_{\varepsilon\to0}f*\phi_\varepsilon(x)=f(x). $$ NOTE: We say $x$ is a Lebesgue point of $f$ if $$ \lim_{\substack{m(B)\to0\\B\ni x}}\frac{1}{m(B)}\int_B|f(y)-f(x)|dy=0. $$
I am very stuck on this problem, and I am not sure where to begin. I am pretty sure the way to go about this is to show $$ \lim_{\varepsilon\to0}\;(f*\phi_\varepsilon(x)-f(x))=0, $$ but I am not quite sure how to implement this. Any suggestions?