If $f \in \mathcal{L}^1(\mathbb{R})$ then $\lim_{h\rightarrow 0}\int_{-\infty}^\infty |f(x+h)-f(x)|dx\rightarrow 0$ I am working on some past prelim problems and one of them is the following: If $f \in \mathcal{L}^1(\mathbb{R})$ then $\lim_{h\rightarrow 0}\int_{-\infty}^\infty |f(x+h)-f(x)|dx\rightarrow 0$ 
Update: Following Jackie Chong's advice, I am attempting this problem with the dominated convergence theorem. Since $f\in L^1(\mathbb{R})$, $f$ can be approximated from below by a sequence of continuous functions $f_n$. Then, I believe I can say that, for all $n$ sufficiently large, $|f_n(x+\frac{1}{n})-f_n(x)|\le 2|f(x)|$. (I'm not sure of this part thought). Then $2|f(x)|$ is integrable and hence a dominating function for $|f_n(x+\frac{1}{n})-f_n(x)|$. So by the dominated convergence theorem we can take the limit inside $\lim_{n\rightarrow \infty}\int|f_n(x+\frac{1}{n})-f(x)|dx=\int \lim_{n\rightarrow \infty}|f_n(x+\frac{1}{n})-f(x)|dx=0$ Is this on the right track?
Another approach that I thought may work, but which I'm less certain of is the following: If $f$ can be approximated by $C^1$ functions (I have no real justification for this, so please correct me if its wrong), then for each $n$ and each $x$ there is some $y(x,n)\in B_{1/n}(x)$ such that $|f(x+\frac{1}{n}-f(x)|=\frac{1}{n}|f'(y(x,n)|$ but from here I don't see how to proceed, because just because $|f|$ is integrable, I can't justify $|f'|$ being integrable, but if there is some way to fix this approach, hints would be appreciated. 
Another thing I was thinking was to first approximate $f$ from below by continuous functions $f_n$ then to define $g_{m,n}=|f_n(x+\frac{1}{m})-f_n(x)| \chi_{[-m,m]}$. Then, for each $n$, the function $g_{m,n}$ is continuous on its support, and since it support is compact, it is uniformly continuous there. Can this be used to justify bringing the limit $\lim_{n\rightarrow \infty}$ inside the integral? Then apply the monotone convergence theorem? I know this idea is probably even more sketchy than my previous one, but if it can be fixed please let me know :)
Thanks! 
 A: Call a function $g:\mathbb{R}\to\mathbb{R}$ ``very simple'' if
there is a positive integer $n$ and there are some disjoint bounded intervals $I_1,\ldots,I_n$ and real constants $y_1,\ldots,y_n$ such that
$$
g(x)=\begin{cases} y_i & \text{if $x\in I_i$} \\ 0 &\text{if $x\notin\bigcup I_i$.}\end{cases} 
$$
A useful lemma: The set of very simple functions is dense in $L_1(\mathbb{R})$.
Proof: For every function $f\in L_1$ and $\varepsilon>0$ we need a very simple function $g$ such that $\|f-g\|<\varepsilon$.


*

*Replace $f$ by a simple function
$$
h(x)=\begin{cases} y_i & \text{if $x\in A_i$} \\ 0 &\text{if $x\notin\bigcup A_i$}\end{cases} 
$$
where $A_1,\ldots,A_n$ are measuable sets and $y_1,\ldots,y_n$ are reals;

*Replace the sets $A_1,\ldots,A_n$ by countable unions of bounded intervals;

*Omit the small intervals and keep only finitely many intervals;

*Replace each multiply covered part by a single interval (take the sum of the corresponding function values).
For every $\varepsilon>0$, these steps can be performed such that the accumulated error is less can $\varepsilon$.

The statement is trivial for very simple sets. 
By the lemma, for every $\varepsilon>0$ there is a very simple function $g$ such that $\|f(x)-g(x)\|<\varepsilon/3$. Then there is some $\delta>0$ such that for every $d$ with $|d|<\delta$, $\|g(x)-g(x+d)|<\varepsilon/3$. Hence,
$$
\| f(x+d)-f(x) \| \le 
\| f(x+d)-g(x+d) \| + \| g(x+d)-g(x) \| + \| g(x)-f(x) \| <
\varepsilon.
$$
