Limit of Lebesgue integral: $\lim_{x \to \infty} \int_0^\infty |f(x+t)+f(t)|\mathrm dt=\int_0^\infty |f(t)|\mathrm dt$ I'm trying to show the following: Given $f$ integrable over $[0, \infty)$,  then $$\lim_{x \to \infty} \int_0^\infty |f(x+t)+f(t)|\mathrm dt=\int_0^\infty |f(t)|\mathrm dt\text .$$
I've tried to use the dominating convergence theorem, but it didn't work as far as I could tell. I'm not really sure about how to proceed from here.
 A: Hint:
$\int^\infty_0|f(x+t)|\,dt =\int^\infty_x |f(t)|\,dt$
which converges to 0 by dominated convergence as $x$ goes to infinity
A: This is a standard way of proving this.
Let $f_n = f \cdot 1_{[0,n]}$ and $\epsilon>0$.
Choose $n$ such that $\int |f-f_n| < \epsilon$. In particular, $| \int|f| -\int|f_n| | < \epsilon$.
For $x > n$ (and $t \in [0,\infty)$, of course) we have $f_n(t+x) = 0$ and so $\int_0^\infty |f_n(x)+f_n(x+t)| dt =  \int |f_n| $.
We have
\begin{eqnarray}
||f(t)+f(x+t)| - |f_n(t)+f_n(x+t)| |
&\le& |f(t)-f_n(t) + f(x+t)-f_n(x+t)| \\
&\le&  |f(t)-f_n(t)|+|f(x+t)-f_n(x+t)|
\end{eqnarray}
and so we have
$| \int_0^\infty |f(t)+f(x+t)|dt - \int_0^\infty |f_n(t)+f_n(x+t)|dt| \le 2 \epsilon $.
For $x>n$ we have $\int_0^\infty |f_n(t)+f_n(x+t)|dt = \int |f_n|$ and so
$| \int_0^\infty |f(t)+f(x+t)|dt - \int |f| | \le | \int_0^\infty |f(t)+f(x+t)|dt - \int |f_n|| + \int |f-f_n| < 3 \epsilon$.
Hence $\limsup_{x \to \infty} | \int_0^\infty |f(t)+f(x+t)|dt - \int |f| | \le 3 \epsilon$ and
since $\epsilon>0$ is arbitrary we have the desired result.
A: This is to address a question posted by the OP in a comment, and which is a little cumbersome to respond in the same way (as a comment).
This is simple exercise:
Lemma: Suppose $f:I\rightarrow\mathbb{\mathbb{R}}$. Let $a\in\overline{I}$ (or $a=\infty$ if $I=(\alpha,\infty)$).Then, $\lim_{x\rightarrow a}f(x)=L$ if and only if $\lim_{n\rightarrow \infty}f(x_n)=L$ for all $\{x_n\}\subset I$ such that $x_n\rightarrow a$.
Proof:
($\Longrightarrow$:) For any $\varepsilon>0$, there is neighborhood $U$ of $a$. (in the case $a=\infty$, one can take $U=(x_\varepsilon,\infty)$) such that
$$
|f(x)-L|<\varepsilon \quad\text{whenever}\quad x\in U$$
Hence, if $x_n$ is a sequence in $I$ that converges to $a$ as $n\rightarrow\infty$, then there is $N$ such that $n\geq N$ implies that $x_n\in U$ for all $n\geq N$. Consequently, $|f(x_n)-L|<\varepsilon$ for all $n\geq N$.
($\Longleftarrow$:) Suppose $\lim_{x\rightarrow a}f(x)\neq L$. Then, there exists $\varepsilon>0$ such that for any $n\in\mathbb{N}$, there is $x_n\in B(a;\frac{1}{n})$ (if $a=\infty$ we take $x_n>n$) such that $$|f(x_n)-L|\geq\varepsilon$$
This means that $f(x_n)$ does not converge to $L$.

How we use this in the setting of dominate convergence?
Suppose $I$ and $a$ are are in the Lemma above.
Lemma: Suppose $\{f_x:x\in I\}\subset L_1(\mu)$, and that $|f_x|\leq h$ $\mu$-a.s.  for some $h\in L_1$. If $f:=\lim_{x\rightarrow a}f_x$ exists $\mu$-a.s. and $f$ is $\mu$--measurable, then $f\in L_1(\mu)$ and
$$\lim_{x\rightarrow a}\int f_x\,d\mu=\int \lim_{x\rightarrow a} f_x\,d\mu =\int f\,d\mu$$
Proof:
Choose any sequence $\{x_n\}\subset I$ that converges to $a$. Then $g_n=f_{x_n}$ satisfies all the conditions of the (standard) dominated convergence theorem;  consequently
$$
\lim_n\int f_{x_n}  = \lim_n \int g_n=\int \lim_n g_n =\int f
$$
As this holds for any sequence $\{x_n\}\subset I$ converging to $a$, it follows that
$$
\lim_{x\rightarrow a}\int f_x\,d\mu =\int \lim_{x\rightarrow a}f_x=\int f$$

Proof: Recall that
