Let $\varphi(x)$ be a smooth measurable function defined on $\mathbb{R}^n$ such that $\varphi(x)=0$ outside the ball centered at the origin with radius 1 and that $\int_{\mathbb{R}^n}\varphi(x)\,dx=1$. For $\varepsilon>0$, define $\varphi_{\varepsilon}(x):=\dfrac{\varphi(\frac{x}{\varepsilon})}{\varepsilon^n}$. Suppose $f\in L(\mathbb{R}^n)$. If $x$ is a Lebesgue point of $f$, then \begin{equation}\lim_{\varepsilon\to 0}f\ast\varphi_{\varepsilon}(x)=f(x),\end{equation} where \begin{equation}f\ast\varphi_{\varepsilon}(x):=\int_{\mathbb{R}^n}f(x-y)\varphi_{\varepsilon}(y)\,dy.\end{equation}
Does anyone know how to prove this?
\begin{align*} |f\ast\varphi_{\epsilon}(x)-f(x)|&=\left|\int f(y)\varphi_{\epsilon}(x-y)dy-\int\varphi_{\epsilon}(x-y)f(x)dy\right|\\ &=\left|\int(f(y)-f(x))\varphi_{\epsilon}(x-y)dy\right|\\ &=\dfrac{1}{\epsilon^{n}}\left|\int(f(y)-f(x))\varphi((x-y)/\epsilon)dy\right|\\ &=\dfrac{1}{\epsilon^{n}}\left|\int_{|x-y|<\epsilon}(f(y)-f(x))\varphi((x-y)/\epsilon)dy\right|\\ &\leq\dfrac{M}{\epsilon^{n}}\int_{|x-y|<\epsilon}|f(y)-f(x)|dy\\ &=\dfrac{Mv_{n}}{|B_{\epsilon}(x)|}\int_{B_{\epsilon}(x)}|f(y)-f(x)|dy\\ &\rightarrow 0. \end{align*}