Show that $\|f_n-f\|_1 \rightarrow 0$, where $f_n = n\int_x^{x+1/n}f(t) dt$ 
Let $f(x)$ be an integrable function on $\mathbb R$ and let $f_n(x) = n\int_x^{x+1/n}f(t)dt, n\in \mathbb N$. Prove that $$\lim_{n\rightarrow \infty}\int_{\mathbb R} |f_n(x)-f(x)|dx =0.$$

First of all, by the Monotone Convergence Theorem applied to the increasing sequence of functions $h_k = |f_n-f|\chi_{[-k,k]}$, we know that there exists $k\in \mathbb N$ such that
$$\int_{\mathbb R\setminus [-k,k]} |f_n(x)-f(x)|dx < \epsilon/2.$$
Therefore, we only have to prove that $$\int_{[-k,k]}|f_n(x)-f(x)|dx<\epsilon/2 \mbox{ as } n\rightarrow \infty.$$
By Lusin's Theorem, we know that there exists $F\subseteq [-k,k]$ such that $f$ is continuous on $F$ and $m(F)$ can be made as small as we want. In particular, in this set we have $f_n(x) \rightarrow f(x)$ point-wise. Moreover, by Egorov's Theorem, we find $F'\subseteq F$ such that $f_n(x)\rightarrow f(x)$ uniformly there and $m(F\setminus F')$ can be made as small as we want. In this way, we find $n\in \mathbb N$ such that $n\geq N$ implies $|f_n(x)-f(x)|<\epsilon/(6 m([-k,k]))$ on $F'$. Thus,
$$\int_{[-k,k]}|f_n(x)-f(x)|dx = \int_{F'}|f_n-f|+\int_{F\,\setminus F'}|f_n-f|+\int_{[-k,k]\setminus F}|f_n-f|.$$
The first integral is less than $\epsilon/6,$ because of uniform convergence. However, so far I can not control the other two remaning integrals. If we could show that $|f_n-f|$ is a uniformly integrable family of functions, since $m(F\setminus F')$ and $m([-k,k]\setminus F)$ can be made arbitrarly small, we could bound these two integrals.
Is it true that the family $|f_n-f|, n \in \mathbb N,$ is uniformly integrable? If so, how to show that? I've been trying to show that $|f_n-f|\leq 2 |f|$ because that would imply uniform integrability, but so far I cant see why this should be the case.
Any help would be much appreciated.
 A: We know that the set $C^0_c(\mathbb R)$ ( continuous function with compact support) it's dense in $L^1(\mathbb R)$ ( this comes from the fact that  $L^1$ it's separable), so for every $\varepsilon>0$ there exists $g\in C^0_c(\mathbb R)$ such that: $||g-f||_1<\varepsilon/2$. now we define the sequence: $$g_n(x):={n}\int_x^{x+1/n}\!\!\!\!\!\!g(t)\text dt,$$
$$||f-f_n ||_1=||f-f_n\pm g\pm g_n ||_1\le||f-g||_1+||f_n-g_n||_1+||g-g_n||_1.$$
for the second term we have:$$||f_n-g_n||_1=\int|f_n-g_n|\text dx\le\int n\int_x^{x+1/n}|f(t)-g(t)| \text dt \text dx=$$ $$\stackrel{t-x=s}{=} \int n\int_0^{1/n}|f(s+x)-g(s+x)| \text ds \text dx  \stackrel{\text{Fubini Th.}}{=}$$ $$ = n\int_0^{1/n}\int|f(s+x)-g(s+x)|  \text dx \text ds\le n\int_0^{1/n}\frac{\varepsilon}{2}\text ds=\frac{\varepsilon}{2}.$$
for the third term we use the Dominated convergence theorem (https://en.wikipedia.org/wiki/Dominated_convergence_theorem) in this way: we know that $|g_n|<\max g=:M$ and that the support of $g$ is a compact $K$ so $\text{ supp }(g_n)\subset K\pm\varepsilon=:K'$, then by using the D.C.Th. with $M\chi_{K'}\in L^1$ we get $||g-g_n||_1\to 0$.
By unifying all estimates we get the thesis.
