Limit of Lebesgue integrable function

Question

Let $f$ be a real valued, Lebesgue integrable function on $\mathbb{R}$. Prove that

$$\lim_{t \to 0} \int_{\mathbb R} |f(x+t)-f(x)|\, dx=0.$$

Answer 1

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 $$\begin{align} \int |f(x+t) -& f(x)| dx \le \int |f(x+t) - f_\epsilon(x+t)|dx +\\ &\int |f_\epsilon(x+t) - f_\epsilon(x)| dx + \int |f_\epsilon(x) - f(x)|dx < \epsilon \end{align}$$

Answer 2

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,

$$\lim_{t \to 0} \int_{\mathbb{R}} \vert f(x+t) - f(x) \vert dx = 0 $$