show that this function is continuously differentiable ( application to Lebesgue theorem ?) Consider $R>0$ and $u \in C_{0}^{\infty}(B(0,R))$ (this is the set of smooth functions with compact support contained in $B(0,R)$).
Let $|\cdot|$ be the Lebesgue measure in $\mathbb R^n$.
Fix $  y \in \mathbb R^n$ and define the function :
$$ f(t) = \frac{\displaystyle\int_{B(y,t)} u(x) \ dx}{\left|B(y,t)\right|}$$
for $t>0$ . 
Show that $f$ is continuously diferentiable in the interval $(0,\infty)$.
My idea :
With some calculus i obtain this :
$$ \frac{f(t) - f(t_0)}{t - t_0} = \frac{1}{t-t_0}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} -  \frac{{t}^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)  $$
where $t_0$ is fixed in the interval above. I am trying to aply the Lebesgue diferentiation 
theorem in the quotient above ... 
Someone can help me? Thanks 
 A: The function $t\mapsto \vert B(y,t)\vert$ is positive and $\mathcal C^1$ (in fact, it is $c\, t^n$ for some constant $c$); so you just need to show that $\varphi(t)=\int_{B(y,t)} u(x)\, dx$ is $\mathcal C^1$. If you put $x=y+t\xi$, then the change of variable formula gives 
$\varphi (t)=t^n\int_{B(0,1)} u(y+t\xi )\, d\xi $, so you have to show that $\psi(t)=\int_{B(0,1)} u(y+t\xi)\, d\xi$ is $\mathcal C^1$. But this follows from the usual differentiation theorems under the integral sign.
A: Hint:
Let's begin with the expression
$$
\frac{f(t) - f(t_0)}{t - t_0} = \frac{1}{t-t_0}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} -  \frac{{t}^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)
$$
And rewrite that, saying $t=t_0+\Delta t$, as
$$
\frac{f(t_0+\Delta t) - f(t_0)}{\Delta t} = \frac{1}{\Delta t}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} -  \frac{(t_0+\Delta t)^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)
$$
Expand the second term and you can cancel the ${t_0}^n$ term, then use Lebesgue's theorem to evaluate the limit.
