Limit of sequence $\frac{1}{2^n} \sum\limits_\epsilon f(\epsilon_1\lambda+\dots+\epsilon_n\lambda^n)$

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.

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.