Limit of Lebesgue integrable function Let $f$ be a real valued, Lebesgue integrable function on $\mathbb{R}$. Prove that
$$\lim_{t \to 0} \int_{\mathbb R} |f(x+t)-f(x)|\, dx=0.$$
 A: If $f$ were continuous and compactly supported (by a compact set $K$) it would be easy using 
the dominated convergence theorem, with the domination
$$
|t| < 1 \implies
|f(x+t)|\le 1_K(x+t) \max |f| \le \sup_{t\in[-1,1]} 1_K(x+t)\max |f|
$$
Let $\epsilon>0$. Then you can find some  continuous, compactly supported function $f_\epsilon$ such as
$$
\int |f_\epsilon(x) - f(x)| dx <\frac \epsilon 3
$$
Then, you can use the properties of $f_\epsilon$  to prove that if $|t|$ is small enough:
$$
\int |f_\epsilon(x+t) - f_\epsilon(x)| dx < \frac \epsilon3
$$
Then
$$\begin{align}
\int |f(x+t) -& f(x)| dx
\le \int |f(x+t) - f_\epsilon(x+t)|dx +\\
 &\int |f_\epsilon(x+t) - f_\epsilon(x)| dx + \int |f_\epsilon(x) - f(x)|dx < \epsilon
\end{align}$$
A: We will first prove the result when $f$ is continuous and compactly supported.
Let the support of $f$ be $[-a,a]$ and as $f$ is assumed to be continuous (and compact supported), it has to be bounded $\vert f\vert \leq M$  and when $t$ is small ($\vert t \vert < 1)$) we have
$$\int_{\mathbb{R}} \vert f(x+t) - f(x) \vert dx = \int_{[-a-1, a+1]}\vert f(x+t) - f(x) \vert dx$$
Now, observe that $\vert f(x+t) - f(x) \vert \leq 2M$ and $2M$ is integrable on $[-a-1, a+1]$. Hence by Lebesgue Dominated Convergence Theorem,
\begin{align}
\lim_{t \to 0}  \int_{\mathbb{R}} \vert f(x+t) - f(x) \vert dx &= \lim_{t \to 0}\int_{[-a-1, a+1]}\vert f(x+t) - f(x) \vert dx \\&= \int_{[-a-1, a+1]}\lim_{t \to 0}\vert f(x+t) - f(x) \vert dx \\&= 0  
\end{align}
Now, when $f$ is not continuous or compactly supported but Lebesgue integrable, we can approximate it with $f_{\epsilon}$ which is continuous and compactly supported such that $\int \vert f - f_{\epsilon} \vert < \epsilon/2$.
And,
$$\vert f(x+t) - f(x) \vert  \leq \vert f(x+t) - f_{\epsilon}(x+t) \vert + \vert f_{\epsilon}(x+t) - f_{\epsilon}(x) \vert + \vert f(x) - f_{\epsilon}(x) \vert $$
The sum of integrals of 1st and 3rd terms are less than $\epsilon$. And as $f_{\epsilon}$ is continuous and compact supported, by previous arguments the integral of 2nd term goes to $0$ as $t \to 0$. Hence, 
$$\lim_{t \to 0} \int_{\mathbb{R}} \vert f(x+t) - f(x) \vert dx = 0 $$
