$f\in L^1$ , $F_h(x) = \frac{1}{2h} \int_{-h}^h f(x-y)d\lambda(y)$ is absolutely continuous $f\in L^1(\mathbb{R})$ I wish to show that $F_h(x) = \frac{1}{2h} \int_{-h}^h f(x-y)d\lambda(y)$ is absolutely continuous over any close interval $[a,b]$
What I tried so far:
$|F_h(d)-F_h(c)| = |\frac{1}{2h}\int_{-h}^hf(d-y)-f(c-y)d\lambda(y)| 
 =$
$=|\frac{1}{2h}\int_{h}^{-h}f(d+y)-f(c+y)d\lambda(y)| $
I changed integration order just for convenience.
Due to translation invariance we can shift by $c$ and get:
$|\frac{1}{2h}\int_{h+c}^{-h+c}f((d-c)+y)-f(y)d\lambda(y)|$
$\le \frac{1}{2h}\int_{h+c}^{-h+c}|f((d-c)+y)-f(y)d\lambda(y)|$
Using the translation continuity for $f$ : $||  f(y+\delta)-f(y)||_{L^1}\rightarrow0$  , we get that for $\varepsilon >0$  exists $\delta >0$ such that for $d-c<\delta \frac{1}{2h}\int_{h+c}^{-h+c}|f((d-c)+y)-f(y)d\lambda(y)|<\varepsilon/2h$.
However when taking $\{(c_i,d_i)\}_{i=1} ^{i=n}$ interval is $[a,b]$ such that $\sum_{i=1}^{i=n}(d_i-c_i) <\delta$ we get $\sum_{i=1}^{i=n}|F(d_i)-F(c_i)| < \frac{1}{2h}\varepsilon\cdot n$. So it doesn't respond to absolutely continuous definition.
I am not so sure that this is the right approach for this problem, I even have a strong feeling that it should be more simple, but I don't find how to solve it. 
 A: The following characterization might simplify the problem:

A function $u:[a,b]\to \mathbb{R}$ is absolutely continuous if and only if there is a function $g\in L^1(a,b)$ and a real constant $c\in \mathbb{R}$ such that
  $$\int_a^x g(t)\,dt=u(x)+c\qquad x\in [a,b] $$

Here you have
\begin{align*}u(x)=\frac{1}{2h}\int_{-h}^hf(x-y)\,dy=-\frac{1}{2h}\int_{x-h}^{x+h}f(y)\,dy
\end{align*}
Heuristically, if $f$ were continuous then by the fundamental theorem of calculus we could say 
\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x}\left[-\frac{1}{2h}\int_{x-h}^{x+h}f(y)\,dy\right]=-\frac{1}{2h}\left(f(x+h)-f(x-h)\right)\qquad x\in [a,b]\end{align*}
This hints that a correct choice of $g$ is
$$g(x):= -\frac{1}{2h}\left(f(x+h)-f(x-h)\right)\qquad x\in [a,b]$$
Clearly, $g\in L^1(\mathbb{R})$. Moreover,
\begin{align*}\int_a^xg(t)\,dt&=-\frac{1}{2h}\int_a^x[f(t+h)-f(t-h)]\,dt=-\frac{1}{2h}\left[\int_{a+h}^{x+h}f(t)\,dt-\int_{a-h}^{x-h}f(t)\,dt\right]=\\ 
&=-\frac{1}{2h}\left[c+\int_{x-h}^{x+h}f(t)\,dt\right]=-\frac{c}{2h}+\frac{1}{2h}\int_{-h}^{h}f(x-t)\,dt
\end{align*}
Where $c:=\int_{a-h}^{a+h}f(t)\,dt$ is a real constant depending only on $h$ (and not on $x$). This proves that $u$ is absolutely continuous.
