Limit of sequence $\frac{1}{2^n} \sum\limits_\epsilon f(\epsilon_1\lambda+\dots+\epsilon_n\lambda^n)$ 
Let $0<\lambda<1$ and $f\in C(\mathbb{R},\mathbb{R})$. Consider
  $I_{n,\lambda}=\frac{1}{2^n}  \sum\limits_{(\epsilon_1,\dots,\epsilon_n) \in \{-1,1\}^n} f(\epsilon_1\lambda+\dots+\epsilon_n\lambda^n)$.
  Show that the sequence $(I_{n,\lambda}(f))_{n\in\mathbb{N}}$ has a limit.

I found this exercise in a book (it is an exercise from ENS Paris), if someone has an idea please share it.
 A: Let $S_n=\sum\limits_{k=1}^n\lambda^kX_k$ for $(X_k)$ an i.i.d. Bernoulli sequence, thus, $P[X_k=-1]=P[X_k=1]=\frac12$ for every $k$. Then $I_{n,\lambda}=E[f(S_n)]$. Since $|\lambda|\lt1$, $S_n\to S$ almost surely, where $S=\sum\limits_{k=1}^\infty\lambda^kX_k$ is almost surely finite. For every $n$, $|S_n|\leqslant(1-|\lambda|)^{-1}$ with full probability and $f$ is continuous and bounded on bounded intervals, hence $f(S_n)\to f(S)$ almost surely and $(f(S_n))$ is dominated by a constant. Thus, $E[f(S_n)]\to E[f(S)]$. QED.
A: This is the proof of V. Rossetto fixed. 
As $f$ is continuous
on the compact $[-R,R]$ with $R=1/(1-\lambda)$, it is uniformly continuous on the compact set $[-R,R]$ and therefore, for every $\varepsilon>0$, there is a $\delta=\delta(\varepsilon)>0$, such that 
$$
|x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<\varepsilon,
$$
for all $x,y\in [-R,R]$.
Then let $\varepsilon>0$ and set
$$
S_{n+p}=\frac1{2^{n+p}}\sum_{\epsilon\in\{-1,1\}^{n+p}}
f\left(\sum_{k=1}^{n+p}\epsilon_k\lambda^k\right)
=\frac1{2^{n+p}}\sum_{\epsilon\in\{-1,1\}^{n+p}}
f\left(x_n(\epsilon)+\mu_{n,p}(\epsilon)\right)$$
where 
$x_n(\epsilon)=\sum_{k=1}^n\epsilon_k\lambda^k$ 
and 
$\mu_{n,p}(\epsilon)=\sum_{k=n+1}^{n+p}\epsilon_k\lambda^k$.
We have 
$$\left|\mu_{n,p}(\epsilon)\right|
\leq
\sum_{k=n+1}^{n+p}\lambda^k<\sum_{k=n+1}^\infty \lambda^k
=\frac{\lambda^{n+1}}{1-\lambda}.$$
Let $n_0$, such that $\dfrac{\lambda^{n_0+1}}{1-\lambda}<\delta(\varepsilon)$, then 
$$
|\mu_{n,p}(\epsilon)|<\varepsilon, \quad \text{for all $p\in\mathbb N$ and $n\ge n_0$.}
$$
We rewrite 
$$S_n=\frac1{2^{n+p}}\sum_{\epsilon\in\{-1,1\}^{n+p}}
f\left(x_n(\epsilon)\right),$$
and compute  $S_{n+p}-S_n$, for $n\ge n_0$:
$$
|S_{n+p}-S_n|\le
\frac1{2^{n+p}}\sum_{\epsilon\in\{-1,1\}^{n+p}}
\left|f\left(x_n(\epsilon)+\mu_{n,p}(\epsilon)\right)
-f\left(x_n(\epsilon)\right)\right|<2^n\cdot\frac{1}{2^n}\varepsilon=\varepsilon.
$$
Therefore the sequence $(S_n)_n$ is a Cauchy sequence a therefore it converges.
