Show $f_n \rightarrow f$ in $L^2(\mathbb R)$ if $f_n(x) = f(x + x_n)$ where $x_n \rightarrow 0$ This is an old qualifying exam question. My attempt was to say 
\begin{align}\|f_n - f\|_{L^2(\mathbb R)}^2 &= \int_{\mathbb R} (f(x+x_n) - f(x))^2\\ &= \|f(x+x_n)\| + \|f(x)\| - 2 \int_{\mathbb R} f(x+x_n)f(x) \mathsf dx\end{align}
Now I want to say that $$f(x+x_n)f(x) \stackrel{n\to\infty}\longrightarrow f(x)^2$$ but I don't think that is valid unless $f$ was continuous in order to bring the limit inside to the $x_n$. I also tried thinking about $$f(x+x_n) - f(x) = x_n\frac{f(x+x_n) - f(x)}{x_n}$$ which would approach $f'$ but again, this need not exist. Anyway, I am stumped. Suggestions? Thanks!
 A: By Lusin's Theorem, for any $\delta>0$, there exists an open set $E$ such that $\mu(\mathbb R\setminus E)<\delta$ and $f$ is continuous on $E$. Then, $f(x+x_n)f(x)\to f^2(x)$ on $E$. So by dominated convergence,
$$\int_E f(x+x_n)f(x)\mathsf dx\to\int_E f^2(x)\mathsf dx.$$
On the other hand by Cauchy-Schwarz,
\begin{align}
\left|\int_{\mathbb R\setminus E} f(x+x_n)f(x)dx\right|&\le\left(\int_{\mathbb R\setminus E} f^2(x+x_n)\mathsf dx\right)^{1/2}\left(\int_{\mathbb  R\setminus E} f^2(x)\mathsf dx\right)^{1/2}
\\
&\le\|f(x)\|_{L^2(R)}\left(\int_{\mathbb R\setminus E} f^2(x)dx\right)^{1/2}\\&<\epsilon\|f(x)\|_{L^2(\mathbb R)},
\end{align}
for some $\epsilon>0$. Since $\mu(\mathbb R\setminus E)$ can be arbitrarily small, $\epsilon$ can be arbitrarily small. Then we have
\begin{align}
&\limsup_{n\to\infty}\|f(x+x_n)-f(x)\|_{L^2(\mathbb R)}^2
\\
&\le 2\|f(x)\|_{L^2(\mathbb R)}^2-2\liminf_{n\to\infty}\int_{\mathbb R} f(x+x_n)f(x)\mathsf dx
\\
&\le 2\|f(x)\|_{L^2(\mathbb R)}^2-2\liminf_{n\to\infty}\int_E f(x+x_n)f(x)\mathsf dx-2\liminf_{n\to\infty}\int_{\mathbb R\setminus E} f(x+x_n)f(x)\mathsf dx
\\
&=2\int_{\mathbb R\setminus E} f^2(x)\mathsf dx-2\liminf_{n\to\infty}\int_{\mathbb R\setminus E} f(x+x_n)f(x)\mathsf dx\\&\le 2\epsilon^2+2\epsilon\|f(x)\|_{L^2(\mathbb R)}.
\end{align}
Since $\epsilon$ can be arbitrarily small, $$\limsup_{n\to\infty}\|f(x+x_n)-f(x)\|_{L^2(\mathbb R)}=0.$$
A: One can also use an approximation by simple function: fix $\varepsilon>0$, and $g=\sum_{i=1}^nc_i\chi_{A_i}$ a simple function. $g_n(x):=g(x+x_n)$. 
So it's enough to prove the result when $f$ is simple. By regularity of Lebesgue measure, it's enough to do it when $f=\chi_O$, where $O$ is an open set of finite measure.
In this can, a dominated convergence argument will work.
