A problem on a bounded and integrable function Let $f$ be a bounded and integrable function. Define $g(h) = \int_{\Bbb R} |f(x+h) - f(x)|dx$. Prove that $\lim_{h \to 0}g(h) = 0$
I tried to approximate $f$ using bounded simple functions, but I don't have any common set of finite measure here to use BCT. How to proceed from here ? Hint Enough.
I have proved the result for step functions. And then I have approximated $f$ using sequence of step functions. I have divided $\Bbb R$ into $E$ and $E^c$ where $m(E)$ is finite and in the remaining part the integral is less than $\epsilon$. Now for the set $E$ I ca use BCT and change limit $n \to \infty$ and integral. But, I need to change $\lim_{n->\infty}$ and $\lim_{h -> 0}$. How can someone do this ?
 A: Use the fact that $C_c(\mathbb{R})$ (the space of continuous functions with compact support) is dense in $L^1(\mathbb{R})$.
If $f \in L^1(\mathbb{R})$, then there is a sequence of $f_n \in C_c(\mathbb{R})$ such that $\|f-f_n\|_1 \to 0$ (that is, $f_n \to f$ in the $L^1$ norm).
Let $\epsilon>0$ and choose $n$ such that $\|f-f_n\| < { 1 \over 3} \epsilon$.
Then
\begin{eqnarray}
g(h) &\le& \int |f(t+h)-f_n(t+h)|dt+ \int |f_n(t+h)-f_n(t)|dt+ \int |f(t)-f_n(t)|dt \\
&<& {2 \over 3} \epsilon + \int |f_n(t+h)-f_n(t)|dt
\end{eqnarray}
Suppose the support of $f_n$ is contained in $[-B,B]$, and $f_n$ is bounded by $M$.
Then if $|h| <1$, we see that $|f_n(t+h)| \le M 1_{[-B-1,B+1]}(t)$, and the 
right hand side is integrable. Since $f_n$ is continuous, we have $f_n(t+h) 
\to f_n(t)$ and hence the dominated convergence theorem shows that $\lim_{h 
\to 0} \int|f_n(t+h)-f(t)|dt = 0$.
Now choose $\delta>0$ such that if $|h| < \delta$, then $\int|f_n(t+h)-f(t)|dt < {1 \over 3} \epsilon$, then the above shows that $g(h) < \epsilon$.
