Finding $\lim\limits_{x→+∞}\int_{-∞}^{+∞}|f(t-x)-f(t)|\,\mathrm dt$ Let $f$ be integrable and continuous function on $\mathbb{R}$. Then I would like to find the value of:
$$\lim_{x \to +\infty} \int_{-\infty}^{+\infty} | f(t-x)-f(t) | \,\mathrm{d}t.$$
I have a lot of problem with these kind of "theorical" integration problems. Moreover, it is hard for me to even guess what the result is, maybe it is $0$, but I am not sure. So my little try, is the following:
$$\int_{-\infty}^{+\infty} | f(t-x)-f(t) | \,\mathrm{d}t \leq \int_{-\infty}^{+\infty}\sup_{\mathbb{R}} | f(t-x)-f(t) | \,\mathrm{d}t.$$
Here my goal is to prove that:
$$\sup_{\mathbb{R}} | f(t-x)-f(t) | \rightarrow 0,$$
yet the problem is that I do not know how to proceed, and it is possible that $0$ is not the right answer.
 A: Let us begin with the case where $f$ is an integrable step function. $f$ only takes a finite number of values $x_0=0$ and $x_1, \dots x_p \neq 0$. We note $A_i=f^{-1}({x_i})$. We note $\mu$ the Lebesgue measure on $\mathbb R$ and since $f$ is integrable the $A_i$ are measurable, and for $i \neq 0$, $\mu(A_i)$ is finite.
We note $B\doteq A_1 \cup \dots \cup A_p$ which is of finite measure. We note also $B_n = B \cap (-\infty, n]$ for $n \in \mathbb N$. $(B_n)$ is an increasing sequence of sets which converges to $B$. Therefore $\mu(B_n) \to \mu(B)$ and $\mu(B \cap [n, \infty)) \to_{n \to \infty} 0$. A similar argument can be done on the negative side and finally we have that for all $\varepsilon>0$ there exists $n >0$ such that $\mu(B \cap [-n,n]) \geq \mu(B)- \varepsilon$.
Let $\varepsilon>0$ and $n$ chosen as explained above. Let $x>2n$. We note $B_x=B+\{x\}$. We have $\mu(B \cap (B+\{x\}))\leq 3\varepsilon$.
For $t \notin C \doteq B \cap (B+\{x\})$, either $f(t)=0$ or $f(t-x)=0$. Therefore $|f(t)-f(t-x)|=|f(t)|+|f(t-x)|$ on $C$.
\begin{align*}
\int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t&=\int_{C}|f(t)-f(t-x)| \, \mathrm{d}t + \int_{C^c}|f(t)-f(t-x)| \, \mathrm{d}t\\
&=\int_{C}|f(t)-f(t-x)| \, \mathrm{d}t + \int_{C^c}|f(t)|+|f(t-x)| \, \mathrm{d}t
\end{align*}
We have $\int_{C^c}|f(t)|\, \mathrm{d}t=||f||_1-\int_{C}|f(t)|\, \mathrm{d}t \geq ||f||_1-3||f||_{\infty}\varepsilon$.
Consequently $\int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t \geq 2||f||_1-6||f||_{\infty}\varepsilon$.
And $\int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t \leq 6||f||_{\infty} \varepsilon+2 ||f||_1$.
This shows that $$\int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t \to_{x \to \infty} 2||f||_1.$$
Let us now consider the case where $f$ is an integrable function on $\mathbb R$. Let $\varepsilon >0$. Since the set of integrable step functions is dense in $L^1(\mathbb R)$ we now that there exists a step function $g$ integrable on $\mathbb R$ such that $||f-g||_1 \leq \varepsilon$.
Let $x \in \mathbb R$. 
\begin{align*}
\int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t&=\int_{\mathbb R}|f(t)-g(t)+g(t)-f(t-x)-g(t-x)+g(t-x)| \, \mathrm{d}t \\
&\leq \int_{\mathbb R}|g(t)-g(t-x)| \, \mathrm{d}t + \int_{\mathbb R}|f(t)-g(t)| \, \mathrm{d}t + \int_{\mathbb R}|g(t-x)-f(t-x)| \, \mathrm{d}t \\
&\leq \int_{\mathbb R}|g(t)-g(t-x)| \, \mathrm{d}t + 2\varepsilon
\end{align*}
Likewise,
\begin{align*}
\int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t&=\int_{\mathbb R}|f(t)-g(t)+g(t)-f(t-x)-g(t-x)+g(t-x)| \, \mathrm{d}t \\
&\geq \int_{\mathbb R}|g(t)-g(t-x)| \, \mathrm{d}t - \int_{\mathbb R}|f(t)-g(t)| \, \mathrm{d}t - \int_{\mathbb R}|g(t-x)-f(t-x)| \, \mathrm{d}t \\
&\geq \int_{\mathbb R}|g(t)-g(t-x)| \, \mathrm{d}t - 2\varepsilon
\end{align*}
Therefore $\lim_{x \to \infty} \int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t \in [2||g||_1-2\varepsilon, 2||g||_1+2\varepsilon] \subset [2||f||_1-4 \varepsilon, 2||f||_1+4\varepsilon]$. Since $\varepsilon$ is chosen as small as you want you obtain that
$$\int_{\mathbb R}|f(t)-f(t-x)| \, \mathrm{d}t \to_{x \to \infty} 2||f||_1.$$
