Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be integrable with$\int_{\mathbb{R}}g(x)dx=1$ and $|g(x)| \leq \frac{C}{(1+|x|)^{1+h}}$ for $x \in \mathbb{R} $, where $C, h>0$ are constants.
Let $g_t(x)=\frac{1}{t} g(\frac{x}{t})$ for $x \in \mathbb{R}$, $t>0$.
I want to show that:
If $f\in L^p$, where $1\leq p\leq \infty$, then $f*g_t(x) \rightarrow f(x)$ a.e.
I have tried in this way:
Let $x\in \mathbb{R}$ be the Lebesgue point of $f$, that is $lim_{r\rightarrow 0} \frac{1}{r} \int_{B(x,r)} |f(y)-f(x)|dx=0$, then
$ |f*g_t(x)-f(x)|\leq \int_{\mathbb{R}} g_t(x-y)|f(y)-f(x)|dy =I_1+I_2, $
where
$I_1=\int_{B(x,t)} g_t(x-y)|f(y)-f(x)|dy \leq\frac{1}{t} \int_{B(x,t)} \frac{C}{(1+\|\frac{x-y}{t}\|)^{1+h}} |f(y)-f(x)|dy $
$ \leq C\frac{1}{t}\int_{B(x,t)} |f(y)-f(x)|dy \rightarrow 0 \ as \ t \rightarrow 0;$
$I_2=\int_{\mathbb{R}\setminus B(x,t)} g_t(x-y)|f(y)-f(x)|dy .$
I don't know how to estimate the integral $I_2$.