Let $\phi\in\ L^1([0,1])$ and $f:\mathbb R\rightarrow\mathbb R$ defined by integral
$f(t)=\int_{0}^{1}|\phi(x)-t|\ dx$, show that
$m(\{\phi=t\})=0\ ,\forall t\ ,in\ an\ interval\Longleftrightarrow f\ continuously\ differentiable\ on\ that\ interval $
1.Condition $\ x\rightarrow |\phi(x)-t|$ is integrable, since f is Lipschitz continuous by
$|f(t)-f(s)|=\int_{0}^{1}||\phi(x)-t|-|\phi(x)-s||\le\int_{0}^{1}|t-s|\le|t-s|$
if $\phi(x)$ is outside of a $\delta-neighbourhood$ it seems clear that its differentiable but how to show if $\{t-\delta\le\phi(x)\le t+\delta\}$ than the differential converges to zero.