Calculate integral for every Lebesgue point of a function $f \in L^1(-\infty, \infty)$ I'm currently studying for a real analysis qual and got stuck on the following question:

Let $k(y) = c \exp\{-y^2\}$, $-\infty < y < \infty$, where $c$ is chosen so that $\int_{-\infty}^\infty k(y)dy = 1$. If $f \in L^1(-\infty, \infty)$, prove that for every point $x$ of the Lebesgue set of $f$, there holds
  $$\lim_{t \to 0+} \int_{-\infty}^\infty f(x-y) \frac{1}{t} k\left(\frac{y}{t}\right) dy = f(x)$$

I know the definition of a Lebesgue point, but I'm not sure how to use that information (or really any of the information) in solving the problem. Any help would be appreciated!
 A: Here is a proof, adapted from Folland's more general result: Set $k_t(y)=\frac{1}{t} k(\frac{y}{t})$ and note that $(1+|y|)^{3/2}k(y)$ is bounded by some positive constant $C$ and that $$\int_{-\infty}^\infty k(y)dy = \int_{-\infty}^\infty k_t(y)dy = 1.$$
If $x$ is a Lebesgue point of $f$, then for all $\epsilon>0$ there is a $\delta>0$ such that $$\int _{|y|<r}|f(x-y)-f(x)|dy<r\epsilon \ \text{whenever}\ r<\delta. $$
Consider $$I_1=\int _{|y|<\delta}|f(x-y)-f(x)|\cdot k_t(y)dy $$ and $$I_2=\int _{|y|\ge \delta}|f(x-y)-f(x)|\cdot k_t(y)dy .$$ Then, $$\int_{\mathbb R} |f(x-y) -f(x)|\cdot k_t(y) dy\le I_1+I_2.$$
So it suffices to show that $I_1,I_2\to 0$ as $t\to 0.$
Fix $t>0$ and let $K$ be the integer such that $2^{K}\le\frac{\delta }{t}<2^{K+1}$ if $\frac{\delta }{t}>1$ and set $K=0$ if not.
Then, on $2^{-k}\delta \le |y|<2^{1-k}\delta $ we have $$k_t(y)\le C\left ( \frac{1}{t} \right )\cdot \left ( \frac{1}{1+\left | \frac{y}{t} \right |} \right )^{3/2}\le C\left ( \frac{1}{t} \right )\cdot \left ( \frac{|y|}{t} \right )^{-3/2}\le C\left ( \frac{1}{t} \right )\cdot \left ( \frac{2^{-k}\delta }{t} \right )^{-3/2}$$
whereas on $|y|<2^{-K}\delta $ we have  $$k_t(y)\le C\left ( \frac{1}{t} \right )\cdot \left ( \frac{1}{1+\left | \frac{y}{t} \right |} \right )^{3/2}\le Ct^{-1}\ \text{(this is true in any event).} $$ 
But $|y|<\delta$ is the disjoint union $A\cup B=\left \{ 2^{-k}\delta \le |y|<2^{1-k}\delta  \right \}^{K}_{k=1}\cup \left \{ |y|<2^{-K}\delta \right \}$ 
so 
\begin{align}
(I_1)_A &\le \sum ^{K}_{k=1}C\left ( \frac{1}{t} \right )\cdot \left ( \frac{2^{-k}\delta }{t} \right )^{-3/2}\int _{2^{-k}\delta \le |y|<2^{1-k}\delta }|f(x-y)-f(x)|dy\\
&\le \sum ^{K}_{k=1}C\left ( \frac{1}{t} \right )\cdot \left ( \frac{2^{-k}\delta }{t} \right )^{-3/2}\cdot \epsilon 2^{(1-k)}\cdot \delta
=C\epsilon\sum ^{K}_{k=1}\left ( \frac{\delta}{t} \right )^{-1/2}\cdot 2^{-3k/2}\cdot 2^{(1-k)}\\
&\le 2^{K/2}C\left ( \sum ^{K}_{k=1}2^{-3k/2}\cdot 2^{(1-k)} \right )\epsilon
\end{align}
and
\begin{align}
(I_1)_B&\le Ct^{-1}\int _{|y|<2^{-K}\delta }|f(x-y)-f(x)|dy\\
&\le Ct^{-1}\cdot 2^{-K}\delta\cdot \epsilon=2^{-K}C\epsilon\cdot \left ( \frac{\delta }{t} \right )\le 2^{-K}C\epsilon 2^{K+1}= 2C\epsilon.
\end{align} 
We conclude that $I_1\le (I_1)_A+(I_1)_B=0.$
Now set $A=\left \{ |y|>\delta \right \}.$ Then, applying Holder's inequality, we have  $$I_2\le \|f\|_1\cdot \|\chi_A\cdot k_t\|_{\infty }+|f(x)|\cdot \|\chi_A\cdot k_t\|_{\infty }.$$
But $$\|\chi_A\cdot k_t\|_{\infty }\le Ct^{-1}\cdot \frac{1}{(1+\frac{\delta}{t})^{3/2}}\le \frac{Ct^{1/2}}{\delta ^{3/2}}\to 0$$ as $t\to 0.$
Thus, $I\le I_1+I_2\to 0$ as $t\to 0$ and we are done. 
