elementary number theory - Show that $lim_{n rightarrow infty} left(prod_{i=1}^{n} (a_i+1) right)^{1/n} $ using Birkhoff Ergodic Theorem

Show that for Lebesgue-almost every $x \in [0,1)$, the geometric mean

$$\lim_{n \rightarrow \infty} \left(\prod_{i=1}^{n} (a_i+1) \right)^{1/n} $$

exists and has common value. What is this? (no proof required)

I think this has something to do with the Birkhoff ergodic Theorem

I tried $$\begin{align} \log \left( \lim_{n \rightarrow \infty} \left(\prod_{i=1}^{n} (a_i+1) \right)^{1/n} \right) &= \lim_{n \rightarrow \infty} \log\left(\prod_{i=1}^{n} (a_i+1) \right)^{1/n} \\
&= \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \log (a_i+1) \\
&= ....???
\end{align}$$

It was shown in the part before that if $x = \sum_{i=1}^{\infty}\frac{a_i}{10^i}$ where $a_i \in \{0,1,\dots,9 \}$ that for Lebesgue-almost every $x \in [0,1)$ that

$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n}a_i=\frac{9}{2} $$

but I cannot see how this can be used.

Answer

The $n$th term is $\exp(S_n(x)/n)$ where $$S_n(x)=\sum\limits_{k=1}^nX_k(x),\qquad X_k(x)=\log(1+a_k(x)).$$ With respect to the Lebesgue measure on $[0,1)$, the sequence $(a_k)$ is i.i.d. hence $(X_k)$ is i.i.d. and $S_n\to E(X_1)$ almost surely, by the strong law of large numbers for i.i.d. integrable sequences. Furthermore, $a_1$ is uniform on $\{0,1,\ldots,9\}$ hence $E(X_1)=\frac1{10}\sum\limits_{i=0}^9\log(1+i)=\frac1{10}\log(10!)$.
Thus, $\exp(S_n(x)/n)\to\ell$ for almost every $x$, where $$\ell=\exp(E(X_1))=(10!)^{1/10}\approx4.5287,$$ and in particular, $\ell\ne9/2$.

Nota: One may replace "the strong law of large numbers for i.i.d. integrable sequences" above by "Birkhoff ergodic theorem".