How to prove $\lim_{n\to \infty}\int_{f^n(B)}g(x)dx=0$ I'm struggling to solve this question:

Let $f:\mathbb R^m\to \mathbb R^m$ be a diffeomorphism and $B$ be the closed unitary ball of $\mathbb R^m$ such that $f(B)\subset B$. Suppose $|\det f'(x)|\lt 1$ for every $x\in B$. Prove for every continuous function $g:B\to \mathbb R$ we have $$\lim_{n\to \infty}\int_{f^n(B)}g(x)dx=0$$

My attempt
Changing the variables we have for the case $n=2$:
\begin{align}
&  \int_{f^2(B)}g(y)dy\\[10pt]
= {} & \int_{f(B)}g(f(x))|\det f'(x)|dx \\[10pt]
= {} & \int_B g(f^2(x))|\det f'(f(x))||\det f'(x)|dx \\[10pt]
\end{align}
By induction we have:
$$\int_{f^n}g(y)dy=\int_B g(f^n(x))|\det f'(f^{n-1}(x))||\det f'(x)|^ndx$$
Using the fact $|\det f'(x)|\lt 1$ we have $\displaystyle\int_Bg^n(x)dx\lt \int_B1\cdot dx=\operatorname{vol} B$
So I don't know if my reasoning is right and how to do with this information I got:
$$\int_Bg^n(x)dx\lt \operatorname{vol} B$$
I need help
 A: By induction, the change of variables is
$$
\begin{align}
\int_{f^n(B)}g(x)\,\mathrm{d}x
&=\int_{f^{n-1}(B)}g(f(x))\left|\det\!\left(f'(x)\right)\right|\,\mathrm{d}x\\
&=\int_{f^{n-2}(B)}g\!\left(f^2(x)\right)\left|\det\!\left(f'(f(x))\right)\right|\,\left|\det\!\left(f'(x)\right)\right|\,\mathrm{d}x\\
&=\int_{f^{n-3}(B)}g\!\left(f^3(x)\right)\left|\det\!\left(f'\!\left(f^2(x)\right)\right)\right|\,\left|\det\!\left(f'(f(x))\right)\right|\,\left|\det\!\left(f'(x)\right)\right|\,\mathrm{d}x\\
&=\int_Bg\!\left(f^n(x)\right)\prod_{k=0}^{n-1}\left|\det\!\left(f'\!\left(f^k(x)\right)\right)\right|\,\mathrm{d}x\tag{1}
\end{align}
$$
Since $f$ is a diffeomorphism on $\mathbb{R}^m$ and $B$ is compact, $\left|\det\!\left(f'(x)\right)\right|$ attains is maximum, $\lambda\lt1$, on $B$. $|g(x)|$ also attains its maximum, $\gamma$. Therefore,
$$
\left|\int_{f^n(B)}g(x)\,\mathrm{d}x\right|
\le\gamma\,\lambda^n\int_B1\,\mathrm{d}x\tag{2}
$$
and since $\lambda\lt1$, $(2)$ tends to $0$ as $n\to\infty$.
A: This is to answer to comments above. The function $f:\Bbb R^m\rightarrow \Bbb R^m$ is a global diffeomorphism, so we can use it to change the variables on integrals.
Now:
$$\int_{f^n(B)}g(y)dy=\int_{f^{n-1}(B)}g(f(x))|\det f'(x)|dx=\dots =\int_Bg(f^n(x))\Big (\prod_{k=0}^{n-1}|\det f'|(f^k(x))\Big)dx$$
Since $|\det f'(x)|<1$ then $|\det f'(x)|\rightarrow 0$ for $n\rightarrow \infty$.
Essentially applying the change of variable $n$ times, you obtain $|\det f'(x)|$ inside the integral.
So $$\lim_{n\to \infty}\int_{f^n(B)}g(x)dx=0$$
A: Assuming that you have access to the dominated convergence theorem, the following solution is also available:
Proof. By the assumption, there is $\lambda \in [0, 1)$ such that
$$|f(x) - f(y)| \leq \lambda|x - y|, \qquad \forall x, y \in B.$$
(Indeed, you may take $\lambda = \sup_{x \in B} |\det f'(x)|$. By the continuity of $f'$ and the compactness of $B$, we have $\lambda < 1$. Then the claim follows from the mean value inequality.) Then by the contraction mapping theorem, there is a unique fixed point $x_0$. Now
$$ \max_{x \in B} |f^n(x) - x_0| \leq \lambda^n \max_{x \in B} |x - x_0| $$
and thus $f^n (B) \to \{x_0\}$ in the sense that $\mathbf{1}_{f^n(B)}(x) \to \mathbf{1}_{\{x_0\}}(x)$ for each $x \in B$. Since $g$ is integrable on $B$, by the dominated convergence theorem we have
$$ \int_{f^n(B)} g = \int_{B} g \mathbf{1}_{f^n(B)} \xrightarrow[n\to\infty]{} \int_{B} g \mathbf{1}_{\{x_0\}} = 0. $$

Remark. This argument allows us to weaken the assumption on $f$. For instance, it suffices to assume that $f : B \to B$ is continuous and satisfies $|f(x) - f(y)| < |x - y|$ for all distinct $x, y \in B$.
