Prove that $\lim_{n\to \infty}\|f+g_n\|_1 = \|f\|_1+\|g\|_1$ 
Let $f,g,g_n\in L^1(\mathbb{R})$, where
  \begin{align}
g_n(x) = g(x-n)\text{ for all }x \in \mathbb{R}.
\end{align}
  Prove that
  $$\lim_{n\to \infty}\left|f+g_n\right|_1 = \left|f\right|_1 + \left|g\right|_1 .$$

It is probably simple but not clicking me right now, appreciate a hint.
 A: Recall that the space of continuous compactly supported functions $C_{00}$ is $L^{1}$ dense. Now pick $\varphi,\psi\in C_{00}$ such that $\|f-\varphi\|_{L^{1}}$ and $\|g-\psi\|_{L^{1}}$ are small.
We see that
\begin{align*}
&\big|\|f+g_{n}\|_{L^{1}}-(\|f\|_{L^{1}}+\|g\|_{L^{1}})\big|\\
&\leq\big|\|f+g_{n}\|_{L^{1}}-\|\varphi+\psi_{n}\|_{L^{1}}\big|+\big|\|\varphi+\psi_{n}\|_{L^{1}}-(\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}})\big|\\
&~~~~~~~~+\big|\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}}-(\|f\|_{L^{1}}+\|g\|_{L^{1}})\big|,
\end{align*}
and 
\begin{align*}
\big|\|f+g_{n}\|_{L^{1}}-\|\varphi+\psi_{n}\|_{L^{1}}\big|\leq\|f-\varphi\|_{L^{1}}+\|g_{n}-\psi_{n}\|_{L^{1}},
\end{align*}
and
\begin{align*}
\|g_{n}-\psi_{n}\|_{L^{1}}=\int_{\mathbb{R}}|g(x-n)-\psi(x-n)|dx=\int_{\mathbb{R}}|g(x)-\psi(x)|dx=\|g-\psi\|_{L^{1}},
\end{align*}
so the term $\big|\|f+g_{n}\|_{L^{1}}-\|\varphi+\psi_{n}\|_{L^{1}}\big|$ is small, whereas 
\begin{align*}
\big|\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}}-(\|f\|_{L^{1}}+\|g\|_{L^{1}})\big|\leq\|f-\varphi\|_{L^{1}}+\|g-\psi\|_{L^{1}},
\end{align*}
which is also small.
So we need only to concentrate on $\big|\|\varphi+\psi_{n}\|_{L^{1}}-(\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}})\big|$. In other words, 
\begin{align*}
\lim_{n\rightarrow\infty}\|\varphi+\psi_{n}\|_{L^{1}}=\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}}
\end{align*}
is what we are to show for.
Now let $M>0$ such that $\text{supp}(\varphi),\text{supp}(\psi)\subseteq\{|x|\leq M\}$, then for $n\geq 3M$, we have
\begin{align*}
\|\varphi+\psi_{n}\|_{L^{1}}&=\int_{\mathbb{R}}|\varphi(x)+\psi(x-n)|dx\\
&=\int_{|x|\leq M}|\varphi(x)+\psi(x-n)|dx+\int_{|x|>M}|\varphi(x)+\psi(x-n)|dx,
\end{align*}
while for $|x|\leq M$, we have $|x-n|\geq n-|x|\geq 3M-M=2M$, so $\psi(x-n)$ vanishes for all such $x$, this yields that
\begin{align*}
\int_{|x|\leq M}|\varphi(x)+\psi(x-n)|dx=\int_{|x|\leq M}|\varphi(x)|dx=\|\varphi\|_{L^{1}}.
\end{align*} 
On the other hand, 
\begin{align*}
\int_{|x|>M}|\varphi(x)+\psi(x-n)|dx&=\int_{|x|>M}|\psi(x-n)|dx\\
&=\int_{\mathbb{R}}|\psi(x-n)|dx-\int_{|x|\leq M}|\psi(x-n)|dx\\
&=\int_{\mathbb{R}}|\psi(x)|dx-\int_{|x|\leq M}|\psi(x-n)|dx\\
&=\|\psi\|_{L^{1}}-\int_{|x|\leq M}|\psi(x-n)|dx,
\end{align*}
once again we have $|x-n|\geq 2M$, so 
\begin{align*}
\int_{|x|>M}|\varphi(x)+\psi(x-n)|dx=\|\psi\|_{L^{1}},
\end{align*}
this shows that $\|\varphi+\psi_{n}\|_{L^{1}}$ is eventually $\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}}$, we are done.
A: By the triangle inequality and translation invariance, we have $$\|f+g_n\| \leq \|f\| + \|g_n\| = \|f\|+ \|g\|,$$ which proves one inequality.  For the other, suppose $f, g \geq 0$.  Note $$f+g_{2n} \geq f \cdot 1_{(-\infty,n)} + g_{2n} \cdot 1_{(n, \infty)}= f \cdot 1_{(-\infty,n)} + g \cdot 1_{(-n, \infty)}.$$  Taking norms and sending $n \to \infty$ (with the dominated convergence theorem) gives the other inequality.  Try to do the general case yourself.
Edit:  The intuition for why the statement in the problem is correct as follows.  Integrable functions behave like the "bell curve" $e^{-x^2}$ in the sense that most of their mass is "in the middle of $\mathbb{R}$," i.e. for any integrable function $f$, we have $$\lim_{R \to \infty} \int_{|x| \geq R} f = 0.$$  Thus, if $f$ and $g$ are integrable and we translate $g$ by a large number $n$, then the "area" under $f+g$ should roughly be that under $f$ plus that under $g$.
