proof that translation of a function converges to function in $L^1$ Let $f \in L^1(\mathbb{R})$, for $a\in \mathbb{R}$ let $f_a(x)=f(x-a)$, prove that:
$$\lim_{a\rightarrow 0}\|f_a -f \|_1=0$$
I know that there exists $g\in C(\mathbb{R})$ s.t $\|f-g\|_1 \leq \epsilon$, this is also true for $f_a$ and $g_a$.
Now I have the next estimation:
\begin{align*}
\|f_a-f\|_1 &= \int_{\mathbb{R}} |f(x)-f(x-a)| dx \\
&= \int |f(x)-g(x)+g(x)-g(x-a)+g(x-a)-f(x-a)| \\
&\leq \|f-g\|_1 + \|f_a-g_a\|_1 + \int |g(x)-g(x-a)|
\end{align*}
my question is: I can argue that $\lim_{a\to0} \int |g(x)-g(x-a)| = \lim_{a\to0} \lim_{T\to\infty} |g(x_0)-g(x_0-a)| 2T$, now I think that I can change the order of the limits here, but I am not sure why?
P.s
$x_0$ is some point in $[-T,T]$, and the above is valid from the intermediate integral theorem for continuous functions, right?
Thanks.
 A: What you need to do is to be a little more picky in your choices. You can choose $g$ not only continuous, but also in $L^1$ and compactly supported. 
You have already reduced the problem to proving it for $g$. If you know that $g$ is compactly supported, then there exists a $T$ such that 
$$
\int_{\mathbb{R}}|g(x)-g(x-a)|=\int_{-T}^T|g(x)-g(x-a)|
$$
and then you don't need to take a limit on $T$. 
(By the way, I don't think you can exchange the limits at the end like you want)
A: You can't change the order of the limits since with the initial one it's $+\infty$, except particular case, and with the second one it's $0$. 
But in fact, you can approximate $f$ in $L^1$ by continuous function with compact support $g$ (i.e., the closure of the set of $x$ such that $g(x)\neq 0$ is compact). To see that, use $\psi_n$ a continuous function with compact support, with is $1$ on $[-n,n]$ and $0$ outside $[-n-1,n+1]$, for example using distance of a point from these sets) and $\psi_n g$ to get what you want. 
Once you have $g$ continuous with support contained in $[-R,R]$, use the fact that $g$ is uniformly continuous to get, for a fixed $\varepsilon>0$, $\delta$ such that if $|a|\leq \delta$ and $x\in\Bbb R$ then $|g(x+a)-g(x)|\leq \varepsilon$. Hence for $|a|\leq \min(\delta,1)$, 
$$\left|\int_{\Bbb R}(g(x+a)-g(x))dx\right|\leq \int_{[-R-1,R+1]}|g(x+a)-g(x)|dx\leq 2(R+1)\varepsilon.$$
