Normal numbers

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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 $$T(x)=10xmod1$$ 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 $0\le a<b\le 1$ because the collection of sub-intervals of $[0,1)$ forms, together with the empty set, a π-system. But $$T^{-1}([a,b))=\bigcup _{i=0}^9[\frac{a+i}{10},\frac{b+i}{10})$$ is a disjoint union of intervals whose lengths do add up to $b-a$. Thus $\lambda$ is an invariant measure for $T$.

Theorem

The map $T(x)=10xmod1$ is ergodic with respect to Lebesgue measure.

Proof:

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 $$\nu (E)={\displaystyle \frac{\lambda (B\cap E)}{\lambda (B)}}$$ on the Borel subsets of $[0,1)$. We will check that $\nu (I)=\lambda (I)$ for every interval of the form $$I=[\frac{a}{{10}^r},\frac{a+1}{{10}^r})$$ with $r\in \mathbb{N}$ and $0\le a<{10}^r$. 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 $n\in \mathbb{N}$ because $\lambda$ is $T$ invariant and $T^{-1}(B)=B$. Next, fixing $ϵ>0$ we choose decimal intervals $J_1,J_2,\dots$ with $$B\subset J_1\cup J_2\cup \cdots$$ and $$\lambda (B)+ϵ\ge \lambda (J_1)+\lambda (J_2)+\cdots \ge \lambda (J_1\cup J_2\cup \cdots )\ge \lambda (B)$$ using the definition of Lebesgue measure. We can then estimate $$|\lambda (B\cap I)-\lambda (\bigcup _{t=1}^{\mathrm{\infty }}{J}_t\cap (T^n{)}^{-1}I)|<ϵ$$ for all $n\in \mathbb{N}$. One calculates for every $K\in \mathbb{N}$ that $$\underset{n\to \mathrm{\infty }}{lim}\lambda (\bigcup _{t=1}^K{J}_t\cap (T^n{)}^{-1}I)=\lambda \left(\bigcup _{t=1}^K{J}_t\right)\lambda (I)$$ and thus deduces $$|\lambda (B\cap I)-\sum _{t=1}^{\mathrm{\infty }}\lambda (J_t)\lambda (B)|<ϵ$$ whence $\lambda (B\cap I)$ and $\lambda (B)\lambda (I)$ are within $2ϵ$ of each other. Since $ϵ>0$ 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 $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{1}_{[0,0.1]}(T^n(x))=\int 1_{[0,0.1]}\mathsf{d}\lambda ={\displaystyle \frac{1}{10}}$$ 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 ${\mathrm{\Omega }}_0\subset [0,1)$ with $\lambda ({\mathrm{\Omega }}_0)=1$ such that every $x\in {\mathrm{\Omega }}_0$ has the desired property.

We can reinterpret this by expressing $$x=\sum _{j=1}^{\mathrm{\infty }}{\displaystyle \frac{d_j(x)}{{10}^j}}$$ 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 $$\begin{array}{rl}T(x)& =10({\displaystyle \frac{d_1(x)}{10}}+{\displaystyle \frac{d_2(x)}{{10}^2}}+{\displaystyle \frac{d_3(x)}{{10}^3}}+\cdots )mod1\\ & =d_1(x)+{\displaystyle \frac{d_2(x)}{10}}+{\displaystyle \frac{d_3(x)}{{10}^2}}+\cdots mod1\\ & ={\displaystyle \frac{d_2(x)}{10}}+{\displaystyle \frac{d_3(x)}{{10}^2}}+\cdots \end{array}$$ 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 ${\mathrm{\Omega }}_{47}$ 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 ${\mathrm{\Omega }}_{\sigma }$ with $\mu ({\mathrm{\Omega }}_{\sigma })=1$ and the property that every $x\in {\mathrm{\Omega }}_{\sigma }$ 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.

Theorem (Borel's normal number 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.

Proof:

Let $\mathrm{\Sigma }$ be the set of all finite strings of digits between zero and nine. The intersection $$\bigcap _{\sigma \in \mathrm{\Sigma }}{\mathrm{\Omega }}_{\sigma }$$ 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 $$T(x)={\displaystyle \frac{1}{x}}mod1$$ is ergodic with respect to the measure $$\nu (E)={\displaystyle \frac{1}{\log 2}}\int 1_E\cdot \varphi \mathsf{d}\lambda$$ where $\varphi (x)=(1+x{)}^{-1}$. Applying the pointwise ergodic theorem to functions such as $[\frac{1}{k+1},\frac{1}{k}]$ 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 $\nu$.