prove 0.1234567891011... is a normal number In the book Probability by Geoffrey Grimmett and Dominic Welsh, One of the exercise in page 19 says: Any number $w\in[0,1]$ has decimal expansion $w=0.x_1x_2...$,
and we write $f_k(w,n)$ for the proportion of times that the integer $k$ appears in the first $n$ digits in this expansion. We call a number normal if $f_k(w,n) \rightarrow\frac{1}{10}$ as $n \rightarrow \infty$ for $k=1,2,3,4,5,6,7,8,9$. Prove that 0.123456789101112... is normal.
So the number of times $1$ appear in the first $9$ digits after the decimal point is 1, then in the next $90$ digits is $19=10+1+8*1$, then the number of times $1$ appears in the next 900 digits is $100+19+19*8$...
Here is what I tried:So if we look at the first $10^i$ digits after the decimal expansion, and we know the number of 1 in the first $i-1$ digits is $f_{i-1}$. we get the number of ones that appear is $\frac{10^{i-1}+f_{i-1}+f_{i-1}*8}{10^i}$.
Here $f_i=10^{i-1}+f_{i-1}+f_{i-1}*8$ with $f_1=1$. So it boils down to solve the recurrence. And relate the recurrence equation to the case of analyzing the first $n$ digits for arbitrary $n$ and not just powers of 10.
 A: I think this is a great problem. Although the Champernowne constant $C$ is well-known as one of the few real numbers proven to be normal, I could not find a concise proof online. It is somewhat "obvious" that $C$ should be normal, but I think it is a challenge to prove it.
My approach was not to consider all indices $n \in \mathbb N$, but only those indices $(n_i)_{i\in\mathbb N}$ for which the Champernowne number truncated at $n_i$-th position reads off the first $i$ integers. For example, we set $n_{99}=189$ since the first $189$ digits of the Champoernowne number read off the first $99$ integers. Hopefully this diagram will make the sequence $(n_i)$ clear:
$$ \begin{array}{c|cccccc}
C:& 1 & 2 & 3 &  & 9 & 1 & 0 & 1 & 1 & & 9 & 9 & 1 & 0 & 0 & \ \\
\text{Label} & n_1 & n_2 & n_3 & \cdots & n_9 & & n_{10} & & n_{11} & \cdots & & n_{99} & & & n_{100} & \cdots\\
\text{Value} & 1 & 2 & 3 & & 9 & & 11 & & 13 & & & 189 & & & 192
\end{array} $$
A formula for $n_i$ is given by
$$ \begin{split}
 n(i) & = (i + 1 - B^{L(i)-1}) \cdot L(i) + (B-1)\sum_{d=1}^{L(i)-1} d B^{d-1} \\
& = (i+1) \cdot L(i) - \frac{B^{L(i)} - 1}{B-1}
\end{split} $$
where $B$ is the base, and $L(i) = 1 + \lfloor \log_B(i) \rfloor$ gives the length of an integer (here, $\lfloor \cdot \rfloor$ is the floor function).
The change of perspective from the first $n$ indices to the first $i$ integers is helpful in deriving a counting function $c_k(n_i)$ which will count the number of times the digit $k$ appears in the first $i$ integers.
First, I derived a preliminary function $c_k^{(d)}(n_i)$ which counts the number of times the digit $k$ appears specifically in the $d$-th place of some integer, where $d=1,2,3,\dots$ (in base 10, the ones place, tens place, etc.) This one is not too difficult to calculate:
$$ c_k^{(d)}(n_i) = B^{d-1} \left \lfloor \frac{i+1}{B^d} \right \rfloor + \begin{cases}
0 & \text{if } 0 \leq m < kB^{d-1} \\
m - k B^{d-1} & \text{if } k B^{d-1} \leq m < (k+1) B^{d-1} \\
B^{d-1} & \text{otherwise}
\end{cases} $$
where $0 \leq m < B^i$ satisfies $m \equiv i+1 \quad (\bmod B^d)$. This function is notationally dense, but seeing its plot, perhaps in Desmos, makes it clear. Now, the function $c_k(n_i)$ is the sum over each $d$:
$$ c_k(n_i) = \sum_{d=1}^{\infty} c_k^{(d)}(n_i) $$
(For practical purpose, you need only to compute the terms $d \leq 2 + \log_B(i)$.) Below I have plotted $c_k(n_i)$ in blue and $\frac{1}{10} n(i)$ in red (the expected limit) with $k=1$ and $i$ in  $[0,10^{11}]$.

The convergence is quite erratic, at least by absolute difference, but I think the relative error is indeed vanishing. Now, we may write
$$ f_k(n_i) = \frac{c_k(n_i)}{n_i} $$
The solution is not complete, however, for taking the limit as $i \rightarrow \infty$ only addresses a narrow subsequence of all $n \in \mathbb N$. And although $f_k(n_i)$ is precisely equal to $1/B$ for all digits $k$ whenever $i$ is a power of $B$, this likewise does not imply convergence to $1/B$.
So, how to proceed? I imagine the counting function $c_k(n_i)$ could be extended to a more general function which takes as input any index $n$, not just the $n_i$ (which are relatively sparse). Perhaps, for arbitrary $n$, an improved counting function could take the form
$$ c_k(n_i) + r_k(n) $$
where $n_i$ is the most recent integer index, satisfying $n_i \leq n < n_{i+1}$, and $r_k(n)$ acts like a "remainder" term, counting up any digits between $n_i$ and $n$ which are not yet part of a fully formed integer. Maybe you can make some progress on this problem. 
