looking for a diffeomorphism (not C1) Let $f\colon\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ diffeomorphism with $f(B[0,1])\subset B[0,1]$ and $| \det f^{\prime}(x) |<1/2$ for all $x\in B[0,1]$ then for every continuous function $h\colon B[0,1] \rightarrow \mathbb{R}^{n}$ $$\lim\limits_{n \to \infty } \int_{f^n(B[0,1])}h(x)dx =0$$
The statement works when $ f $ is a diffeomorphism of class $C^1$. Wherefore seek some diffeomorphism (not C1) to serve as counterexample, but I can not think of any, exist? or question the way it is stated can be proved?
thanks for the help
 A: There is no counter-example, and we only need to assume that $h$ is integrable on $B[0,1]$.
In fact, there is a change-of-variables formula for injective differentiable maps as follows.

Proposition: Let $U\subset \Bbb R^n$ be a non-empty open set, and let $T:U\to \Bbb R^n$ be differentiable and injective. Then for every measurable function $g:\Bbb R^n\to [0,+\infty)$, 
  $$\int_{T(U)} g~dm=\int_U (g\circ T)\cdot |\det T'|~dm,\tag{1}$$ where $m$ denotes the
  Lebesgue measure on $\Bbb R^n$.

The proposition above can be essentially found, for example, in Theorem 7.26 of Walter Rudin's Real and Complex Analysis, Third Edition.  
As a direct corollary of the proposition above, if $K\subset U$ is some compact set and $g$ is the indicator function of $T(K)$, then applying $(1)$ to $g$ yields
$$m(T(K))=\int_K|\det T'|~dm.\tag{2}$$
In our situation, $U=\Bbb R^n$ and for every $k\ge 1$, $T=f^k$ is differentiable and injective on $U$. Moreover, for $K=B[0,1]$, we know that $|\det T'|\le \frac{1}{2^k}$ on $K$. Then from $(2)$ we know that 
$$m\big(f^k(B[0,1])\big)\le \frac{1}{2^k}m\big(B[0,1]\big),\  \forall k\ge 1\Longrightarrow \lim_{k\to \infty}m\big(f^k(B[0,1])\big)=0.\tag{3}$$ 
Due to $(3)$ and the fact $f^k(B[0,1])\subset B[0,1]$ for every $k\ge 1$, if $h:B[0,1]\to \Bbb R$ is integrable, 
$$\lim_{k\to \infty}\int_{f^k(B[0,1])} h~dm=0.$$
