We are going to apply the pointwise ergodic theorem to deduce a famous result of Borel that says almost every number is normal in base 10. We will do this by studying the map \[ T(x) = 10 x \bmod 1 \] on $[0,1)$ equipped with Lebesgue measure $\lambda$ on the Borel subsets of $[0,1)$.
First we check that Lebesgue measure is $T$ invariant. It suffices to check that \[ \lambda(T^{-1}([a,b))) = \lambda([a,b)) = b-a \] for all $0 \le a \l b \le 1$ because the collection of sub-intervals of $[0,1)$ forms, together with the empty set, a π-system. But \[ T^{-1}([a,b)) = \bigcup_{i=0}^9 \, [\tfrac{a + i}{10},\tfrac{b+i}{10}) \] is a disjoint union of intervals whose lengths do add up to $b-a$. Thus $\lambda$ is an invariant measure for $T$.
The map $T(x) = 10 x \bmod 1$ is ergodic with respect to Lebesgue measure.
Fix $B \subset [0,1)$ that is $T$ invariant. We wish to prove that $\mu(B)$ is either zero or one. Suppose it is positive and define a measure $\nu$ by \[ \nu(E) = \dfrac{\lambda(B \cap E)}{\lambda(B)} \] on the Borel subsets of $[0,1)$. We will check that $\nu(I) = \lambda(I)$ for every interval of the form \[ I = [ \tfrac{a}{10^r},\tfrac{a+1}{10^r} ) \] with $r \in \N$ and $0 \le a \l 10^r$. Intervals of the above kind are called decimal intervals. Since such intervals generate the Borel σ-algebra, this will then imply $\nu = \lambda$ and $\lambda(B) = 1$ by arguments we have seen before.
Fix a decimal interval $I$. To check the above we note that \[ \lambda(B \cap I) = \lambda(B \cap (T^n)^{-1} I) \] for all $n \in \N$ because $\lambda$ is $T$ invariant and $T^{-1}(B) = B$. Next, fixing $\epsilon \g 0$ we choose decimal intervals $J_1,J_2,\dots$ with \[ B \subset J_1 \cup J_2 \cup \cdots \] and \[ \lambda(B) + \epsilon \ge \lambda(J_1) + \lambda(J_2) + \cdots \ge \lambda(J_1 \cup J_2 \cup \cdots) \ge \lambda(B) \] using the definition of Lebesgue measure. We can then estimate \[ \left| \lambda(B \cap I) - \lambda \left( \bigcup_{t=1}^\infty J_t \cap (T^n)^{-1} I \right) \right| \l \epsilon \] for all $n \in \N$. One calculates for every $K \in \N$ that \[ \lim_{n \to \infty} \lambda \left( \bigcup_{t=1}^K J_t \cap (T^n)^{-1} I \right) = \lambda \left( \bigcup_{t=1}^K J_t \right) \lambda(I) \] and thus deduces \[ \left| \lambda(B \cap I) - \sum_{t=1}^\infty \lambda(J_t) \lambda(B) \right| \l \epsilon \] whence $\lambda(B \cap I)$ and $\lambda(B) \lambda(I)$ are within $2\epsilon$ of each other. Since $\epsilon \g 0$ was arbitrary we are done. ▮
Now that we know $T$ is ergodic we can apply the pointwise ergodic theorem. Let us first apply the theorem with $f = 1_{[0,0.1]}$ to get \[ \lim_{N \to \infty} \dfrac{1}{N} \sum_{n=0}^{N-1} 1_{[0,0.1]}(T^n(x)) = \int 1_{[0,0.1]} \intd \lambda = \dfrac{1}{10} \] for almost every $x \in [0,1)$. This means that almost every point $x \in [0,1)$ has the property that the sequence \[ n \mapsto T^n(x) \] spends about 10% of its time in the interval $[0,0.1]$. Formally, the pointwise ergodic theorem gives us a set $\Omega_0 \subset [0,1)$ with $\lambda(\Omega_0) = 1$ such that every $x \in \Omega_0$ has the desired property.
We can reinterpret this by expressing \[ x = \sum_{j=1}^\infty \dfrac{d_j(x)}{10^j} \] for natural numbers $d_1(x), d_2(x),\dots$ between zero and nine i.e., by writing the decimal expansion of $x$. In those terms we have $x \in [0,0.1)$ if and only if $d_1(x) = 0$. Moreover \[ \begin{align} T(x) & {} = 10 \left( \dfrac{d_1(x)}{10} + \dfrac{d_2(x)}{10^2} + \dfrac{d_3(x)}{10^3} + \cdots \right) \bmod 1 \\ & {} = d_1(x) + \dfrac{d_2(x)}{10} + \dfrac{d_3(x)}{10^2} + \cdots \bmod 1 \\ & {} = \dfrac{d_2(x)}{10} + \dfrac{d_3(x)}{10^2} + \cdots \end{align} \] and $T(x)$ belongs to $[0,0.1]$ if and only if the second digit in the decimal expansion of $x$ is a zero. Repeatedly applying $T$ as above, we see that \[ \sum_{n=0}^{N-1} 1_{[0,0.1]}(T^n x) \] counts how many of the first $N$ digits in the decimal expansion of $x$ are equal to 0 and the pointwise ergodic theorem is telling us that almost every point $x \in [0,1]$ has, asymptotically, the property that about 10% of its decimal digits are equal to 0.
We can, for example, repeat the above argument with $f = 1_{[0.47,0.48]}$. The sum \[ \sum_{n=0}^{N-1} 1_{[0.47,0.48]}(T^n x) \] counts the occurrences of 47 amongst the decimal digits of $x$ and the pointwise ergodic theorem provides us with a set $\Omega_{47}$ of full measure such that every point in $x$ has 47 appear in its decimal expansion 1% of the time.
In fact, for every finite string $\sigma$ of digits between zero and nine we get a set $\Omega_\sigma$ with $\mu(\Omega_\sigma) = 1$ and the property that every $x \in \Omega_\sigma$ has the expected frequency - $1/10^\ell$ where $\ell$ is the number of digits in $\sigma$ - of appearences of $\sigma$ in its decimal expansion. A number with this property is sometimes called normal in base 10 and the above argument proves the following theorem.
There is a set $\mathcal{N} \subset [0,1)$ with $\lambda(\mathcal{N}) = 1$ and the property that every $x \in \mathcal{N}$ is normal in base 10.
Let $\Sigma$ be the set of all finite strings of digits between zero and nine. The intersection \[ \bigcap_{\sigma \in \Sigma} \Omega_\sigma \] is a countable intersection of sets with full measure and therefore has full measure itself. Every number in the intersection is normal. ▮
Despite the theorem guaranteeing an abundance of normal numbers - one would, for example, have to be unlucky not to come across a normal number by randomly choosing digits to define a number's decimal expansion - the theorem does not provide us with a specific example of a normal number! The somewhat artificial number \[ \mathtt{0.1234567891011121314151617\cdots} \] known as Champernowne's constant is normal by direct verification, and the artificial but interesting number \[ \mathtt{0.23579111317192329313741\cdots} \] was proved normal by Copeland and Erdős. It is an infamous problem to determine whether $\sqrt{2}$ or $\pi$ or $e$ are normal.
One can show that the map \[ T(x) = \dfrac{1}{x} \bmod 1 \] is ergodic with respect to the measure \[ \nu(E) = \dfrac{1}{\log 2} \int 1_E \cdot \phi \intd \lambda \] where $\phi(x) = (1+x)^{-1}$. Applying the pointwise ergodic theorem to functions such as $[\tfrac{1}{k+1},\tfrac{1}{k}]$ one can show that almost every number $x \in [0,1)$ has a normal continued fraction expansion, if one appropriately weights each digit according to the measure $\nu$.