$F(h)=\int_0^1{h\left\lvert f(x+h)-f(x) \right\rvert}dx$ has derivative at 0 
Let $f$ be a Riemann integrable function defined on $[-2,2]$. Define a function
  $F:(-1,1)\to \Bbb{R}$ by $$F(h)=\int_0^1{h\left\lvert f(x+h)-f(x) \right\rvert}dx$$
  Show that the derivative $F'(0)$ exists.

I have tried to use the fundamental theorem of calculus after some change of variables, but it doesn't seem to work since there is no relation between $h$ and $x$.
Then I tried to prove by definition: proving the following limit exists
$$\lim_{h\to 0}\frac{F(h)-F(0)}h=\lim_{h\to 0}\int_0^1{\left\lvert f(x+h)-f(x) \right\rvert}dx$$
I guess the limit is $0$. But I don't know how to prove it. Could you please give me some hints? Thank you.
Edit: Well, after some trying, I think I can do it like this (a rough idea):
$$\left\lvert f(x+h)-f(x) \right\rvert\le \underset{P_h}{\sup f}-\underset{P_h}{\inf f}$$ where $P_h$ is a partition of $[-2,2]$ s.t. it has some relation to restrict the $\left\lvert f(x+h)-f(x) \right\rvert$ to be small enough. I think this is the right approach, how can I change it to a rigorous argument?
 A: Assuming there is no absolute value inside the integral, i.e., that
$$
F(h) = \int_{0}^{1} h\bigl(f(x + h) - f(x)\bigr)\, dx,
$$
here's a short, elementary proof.
For $|h| < 1$, you have
$$
\int_{0}^{1} f(x + h)\, dx = \int_{h}^{1 + h} f(x)\, dx,
$$
i.e., "the integral of the same function over a shightly shifted interval". Since $f$ is Riemann integrable, its absolute value is bounded on $[-2, 2]$ by some real number $M$, so
\begin{align*}
\left\lvert \frac{F(h) - F(0)}{h}\right\rvert
  &= \left\lvert\int_{0}^{1} \bigl(f(x + h) - f(x)\bigr)\, dx\right\rvert \\
  &= \left\lvert\int_{h}^{1 + h} f(x)\, dx - \int_{0}^{1} f(x)\, dx\right\rvert \\
  &= \left\lvert\int_{1}^{1 + h} f(x)\, dx - \int_{0}^{h} f(x)\, dx\right\rvert
  \leq 2M|h|.
\end{align*}

With the absolute value, it looks to me that Paramanand Singh is right: You need that $f$ is "almost uniformly continuous". 
Here's a workable strategy: Approximate $f$ by a continuous function $g$ with $\int_{-2}^{2} |f(x) - g(x)|\, dx$ small (this is a little delicate, but elementary enough to be an exercise in Spivak's Calculus, if memory serves). Then use uniform continuity of $g$ to show
$$
\int_{0}^{1} |g(x + h) - g(x)|\, dx
$$
can be made arbitrarily small by taking $|h|$ sufficiently small. Finally, triangle inequality and the standard algebraic trick
$$
f(x + h) - f(x)
  = \bigl[f(x + h) - g(x + h)\bigr]
  + \bigl[g(x + h) - g(x)\bigr]
  + \bigl[g(x) - f(x)\bigr]
$$
can be used to control
$$
\int_{0}^{1} |f(x + h) - f(x)|\, dx.
$$
