Convergence of images of a sequence of converging continuous functions Suppose that $\phi_n$ are continuous functions from a compact set $\Omega\subset \mathbb R^n$ to $\mathbb R^n$, and that they converge uniformly to a continuous function $\phi:\Omega \to \mathbb R^n$. 
Call $D_n=\phi_n(\Omega)$ and $D=\phi(\Omega)$. It is quite easy to check that they are compact sets and that $D_n$ converge to $D$ in Haussdorf distance. 
My question is: is it true that the symmetric difference $D_n\triangle D$ has measure that converge to zero?
I think I proved that $D_n\setminus D$ converges to zero in measure, using a property of Lebesgue measure, but to prove that $D\setminus D_n$ is also converging to zero, I had to add more hypothesis to $\phi$. 
Is it true with the current formulation? All my attempts failed and now I start to believe that there's some counterexample linked to (fat) Cantor sets.
 A: There is indeed a counterexample with fat Cantor sets.  Pick a fat Cantor set $\Omega\subset[0,1]$.  Let $\Omega_n$ be the $n$th stage of the construction of $\Omega$, so $\Omega_n$ is a disjoint union of $2^n$ closed intervals and $\Omega_{n+1}$ is obtained from $\Omega_n$ by removing an open interval from the middle of each of its intervals, and $\Omega=\bigcap_n \Omega_n$.  Let $A_n\subset \Omega_n$ be a subset consisting of one point from each of its intervals.
We can then consider the functions $\phi_n:\Omega_n\to A_n$ which just send each interval of $\Omega_n$ to the chosen point in that interval.  Considering each $\phi_n$ as a map $\Omega\to \mathbb{R}$, they converge uniformly to the inclusion map $\phi:\Omega\to\mathbb{R}$ (the convergence is uniform because the intervals in $\Omega_n$ each have length at most $1/2^n$ so $|\phi_n(x)-x|\leq 1/2^n$ for all $x\in\Omega$).  But the image of $\phi_n$ is just $A_n$ which has measure $0$ for all $n$, while the image of $\phi$ is $\Omega$ which has positive measure.
