Normal in base 10 We consider a decimal expansion $x=\sum^{\infty}_{i=1} \frac{d_i}{10^i}$ for $x \in [0,1)$. This expansion is generated by map $Tx=10x ($mod $  1)$ defined on $([0,1), \cal B,\lambda)$ with $\lambda$ the Lebesgue measure. 
Now I want to show that $\lambda$ a.e. $x \in [0,1)$ is normal in base $10$. Thus that we have
\begin{equation}
\lim_{k \to \infty} \frac{\#\{1 \leq j \leq k : d_j=a_1, d_{j+1}=a_2, ..., d_{j+n-1}=a_n\}}{k} = \frac{1}{10^n}
\end{equation}
for any finite block of digits $a_1a_2a_3 ... a_n, a_i \in \{0,1,...,9\}$
How can I proof this? I think I need Birkhoff's Ergodic Theorem? Is that right? And how?
Thanks in advance 
 A: I think it deserves a clean answer. I'll assume that all decimal expansions are proper (there is no infinite string of 9s). I'll assume that it is known that the system $([0,1),\lambda,T)$ is ergodic.
Let $n \geq 1$, and let $\{ a_1, \ldots, a_n \} \in \{0, 1, \ldots, 9\}^n$. Let us define the corresponding cylinder :
$$[a_1, \ldots, a_n ] := \left\{x = \sum_{k=1}^{+ \infty} 10^{-k} d_k \ : \ d_k = a_k \ \forall \ 1 \leq k \leq n \right\} = \sum_{k=1}^n 10^{-k} a_k + [0,10^{-n}).$$
Let $[a_1, \ldots, a_n ]$ be a cylinder, and let $f := 1_{[a_1, \ldots, a_n ]}$ be its characteristic function. Note that $\int_0^1 f (x) dx = 10^{-n}$. Since the system is ergodic, by Birkhoff's ergodic theorem, for almost every $x$,
$$\lim_{k \to + \infty} \frac{1}{k} \sum_{j=1}^k f\circ T^j (x) = 10^{-n}.$$
But $f \circ T^j (x)$ is $1$ if and only if $d_j = a_1, \ldots, d_{j+n-1} = a_n$. Hence, 
$$\sum_{j=1}^k f\circ T^j (x) = \# \{1 \leq j \leq k \ : \ d_j = a_1, \ldots, d_{j+n-1} = a_n\}.$$
Let $A_{[a_1, \ldots, a_n ]}$ be the set of points such that:
$$\lim_{k \to + \infty} \frac{\# \{1 \leq j \leq k \ : \ d_j = a_1, \ldots, d_{j+n-1} = a_n\}}{k} = 10^{-n}.$$
Then we have just proved that $A_{[a_1, \ldots, a_n ]}$ has full Lebesgue measure. But this is true for all cylinders, no matter the length. In addition, there are only countably many cylinders (the set of finite strings of integers is countable), so $A := \bigcap_{\overline{a} \ \text{cylinder}} A_{\overline{a}}$ has also full Lebesgue measure. But $A$ is exactly the set of normal numbers, so almost every number in $[0,1)$ is normal.
