A function $f(x)$ that Riemann integrable on $[a,b]$. Define a function $f(x)$ that Riemann integrable on $[a,b]$.
Let
$$g(x)=\begin{cases}
 f(x)&\text{if}&x\in[a,b],  \\ 
 f(a)&\text{if}&x<a,   \\ 
 f(b)&\text{if}&x>b.   
 \end{cases}$$
Let $\delta >0$,define$$F_{\delta}(x)=\frac{1}{\delta}\int_{0}^{\delta} (g(x+t)-g(x))dt,x\in[a,b].$$
Proof：
$$\lim _{ \delta\rightarrow {0}^{+}}\int_{a}^{b} F_{\delta}(x)dx=0.$$
I want to use some propostions from Intergration of an Intergral Dependig on Parameter to get something like $$\int_{0}^{\delta}\left(\int_{a}^{b}( g(x+t)-g(x))dx\right)dt=\int_{a}^{b}\left(\int_{0}^{\delta} (g(x+t)-g(x))dt\right)dx$$.but it seemingly not easy ! Maybe someone   have a best answear  to this quesion,any help would be appreciated!

I suppose that $0<\delta<b-a$.
Your last equality
$$\int_{0}^{\delta}\left(\int_{a}^{b}( g(x+t)-g(x))dx\right)dt=\int_{a}^{b}\left(\int_{0}^{\delta} (g(x+t)-g(x))dt\right)dx$$
is true by Fubini's theorem.
Now we have:
$$\int_{a}^{b}( g(x+t)-g(x))dx=\int_{a+t}^{b+t}g(u)du-\int_a^b g(x)dx=\int_b^{b+t}g(u)du-\int_a^{a+t}g(u)du$$
Or:
$$\int_{a}^{b}( g(x+t)-g(x))dx=tf(b)-\int_a^{a+t}f(u)du$$
Put $\displaystyle G(t)=\int_a^{a+t}f(u)du$. Then, as $f$ is Riemann-integrable, the function $G$, defined on $I=[0,b-a]$, is continuous on $I$.
Your expression is equal to
$$M(\delta)=\frac{\delta}{2}f(b)-\frac{1}{\delta}\int_0^{\delta}G(t)dt$$
Now as $G$ is continuous, $\displaystyle H(\delta)=\int_0^{\delta}G(t)dt$ is a differentiable function of $\delta$ with $H^{\prime}(\delta)=G(\delta)$. Thus $\displaystyle \frac{H(\delta)}{\delta} \to G(0)=0$ if $\delta\to 0$, and we are done. 
