Random Numbers are Normal (Random Numbers are Normal). Let $X$ be a uniformly distributed random
variable on (0, 1). Let $X_1$ be the first digit in the decimal expansion of $X$. Let $X_2$ be the
second digit in the decimal expansion of$X$. And so on.
• Show that the random variables $X_1$, $X_2$, . . . are uniform on $\{0, 1, 2, . . . , 9\}$ and independent.
• Fix $m \in \{0, 1, 2, . . . , 9\}.$ Using the Strong Law of Large Numbers, show that with
probability one, the fraction of appearances of the number m in the first n digits of
$X$ converges to $1/10$ as $n\to\infty$.
I have no idea how to do a formal proof of this. I know that a decimal can be $0-9$ and that any $X_i$ can be one of these 10 numbers with equal probability. Any help would be appreciated. 
 A: Given $0\leq m\leq 9$ and $k\geq 1$, what is the probability that $X_k=m$?
If $x\in (0,1)$, which is not of the form $m10^{-n}$ for any $n$ or $0\leq m\leq 9$, the $k$'th digit of the decimal expansion of $x\in (0,1)$ is equal to $m$ if and only if $x$ belongs to an interval of the form
$$(R_{k-1}(x)+\frac{m}{10^k},R_{k-1}(x)+\frac{m+1}{10^k})\enspace\enspace (1)$$
where $R_{k-1}(x)=\sum_{j=1}^{k-1}\frac{m_j(x)}{10^{j}}$, is the number obtained from $x$ by taking only its first $k-1$ digits,
 $0\leq m_j(x)\leq 9$. There are $10^{k-1}$ choices for the first $k-1$ digits, and so there are $10^{k-1}$ possibilities to choose an interval of length $10^{-k}$ of the above form where $x$ will lie, and all these intervals are mutually disjoint, so for a uniformly distributed $x$, the probability will be $10^{k-1}\times 10^{-k}=\frac{1}{10}$. Since $0\leq m\leq 9$ was arbitrary, we see that $X_k$ are equally distributed, and take every value with probability $\frac{1}{10}$. To see that they are mutually independent, consider the probability:
$$P(X_k=m_1\,\text{ and } X_l=m_2)$$
for some $k<l$ and $0\leq m_1, m_2\leq 9$. We are looking at all the numbers $x\in (0,1)$ for which the $k$'th digit is $m_1$ and the $l$'th digit is $m_2$. This means that $x$ belongs to an interval of the form $(1)$ with $m=m_1$ (because its $k$ digit is $m_1$). Furthermore, since its $l$'th digit is $m_2$, it must belong to some $m_2$'th interval of length $10^{-l}$ of the form 
$$(R_{l-1}(x)+\frac{m_2}{10^l},R_{l-1}(x)+\frac{m_2+1}{10^l})$$
 So let's count: we have $10^{l-1-k}$ intervals of length $10^{-l}$ each that can occur in each one of $10^{k-1}$ intervals. All these intervals are disjoint, so the probability is 
$$10^{l-1-k}\times 10^{k-1}\times 10^{-l}=10^{-2}$$
which is also the product of the probabilities $P(X_k=m_1)\times P(X_l=m_2)$. Thus the $X_k$'s are mutually independent.
For integers $k\geq 1$, and $0\leq m\leq 9$ define a random variable on $(0,1)$ by: $$U^m_k(x) = \left\{\begin{array}{lr}
        1, & \text{the $k$'th digit is $m$, }\\
        0, & \text{otherwise. }
        \end{array}\right.
$$
Then we have $X_k=m$ if and only if $U_k^m=1$, so for fixed $m$ the random variables $U_k^m$ are mutually independent for different $k$'s. Moreover, they are equally distributed -- they all take the value $1$ with probability $1/10$, and so by the law of large numbers :
$$\lim_{n\to\infty}\frac{U^m_1+\cdots +U^m_n}{n}=\frac{1}{10}\, \text{with probability 1}$$
Note however that the average $\frac{1}{n}\sum_{j=1}^nU_j^m$ is precisely the frequency with which the digit $m$ appears among the $n$ first digits.

