Q: measure of the set consisting of those numbers in (0,1) whose decimal expansion contains hundred consecutive 4s It's a problem from Sheldon Axler;s Measure Theory book.
What I did so far is-
E= set of all numbers in (0,1) that have decimal expansion containing hundred consecutive 4s 
$A_{n}$ = set of numbers in (0,1) such that the hundred consecutive 4s start from nth position after decimal point and before the nth position there is no hundred consecutive 4s in that expansion.
Countable union of $A_{n}$ is the set E.
So E is borel set. But I am struggling to find the measure of this set.
I tried calculating the measure of the set of points which don't contain any hundred consecutive 4s in their decimal expansion. But there also I got stuck.
I have seen the similar question has been asked here before. But I could not totally understand the probability approach to the problem.
What is the Lebesgue Measure of the set of numbers in (0,1) that contain 100 consecutve 4's?

A: It might be easier to think about the complement of this set, i.e. the set of all numbers $x \in (0, 1)$ that don't have $100$ consecutive $4$'s lurking somewhere in their decimal expansion.
Lemma. For any base $b \geq 3$ and digit $d \in \{0, 1, ..., b-1 \}$, the Lebesgue measure of the set $$S_d := \{ x \in (0, 1): x \text{ contains no digit } d \text{ in base } b \}$$ is zero.
Proof of lemma. The set of all points that don't have $d$ as their first digit has measure $1 - 1/b$. The set of all points that don't have $d$ as their first or second digit has measure $(1 - 1/b)^2$. The set of all points that don't have $d$ as their first, second, third, ..., or $n$th digit has measure $(1 - 1/b)^n$. So $$\mu(S_d) \leq (1 - 1/b)^n \text{ for all } n,$$ which implies $\mu(S_d) = 0$.
Now think about some particular bases and digits that might apply to this problem, like, oh, I don't know, $b = 10^{100}$, $d = 4(10^{100}-1)/9$....

A: This is not an answer, just an elaboration.
Let $A$ be the set of sequences that do not contain a run of one hundred $4$s.
Let $A_k$ be the set of sequences such that at least one of $x_k,...,x_{k+99}$ is not $4$. Note that $P A_k = 1-{1 \over 10^{100}} <1$.
It is clear that $A \subset A_1 \cap A_{101} \cap A_{201} \cap ...$ and so  we have $PA \le P(A_1 \cap A_{101} \cap \cdots \cap A_{100n+1} )$ for any $n$.
By independence we have $P(A_1 \cap A_{101} \cap \cdots \cap A_{100n+1} ) =(P A_1)^n$ and so $PA = 0$.

A: Here is an argument for which you can fill in the details.
Look at the decimal expansion of $x\in (0,1)$ and break the expansion into contiguous blocks, each of length $d=100$. Let $B_k(x)$ be the function that maps $x\in(0,1)$ to the $k$th such block.
Let $S$ be the set of all sequences of $d$ decimal digits, so $S$ has $10^d$ elements. Argue that for any $s_1,\ldots,s_n \in S$ (with the $(s_k)$ not necessarily distinct):
$$
\lambda\left(\bigcap_{k=1}^n\left\{x\in(0,1):B_k(x)=s_k\right\}\right)=\frac1{10^{dn}}.\tag1
$$
In words, the set of numbers whose decimal expansion begins with $0.s_1s_2\cdots s_n$ has Lebesgue measure $1/10^{dn}$. It follows that for any subsets $A_1,\ldots,A_n$ of $S$, we have
$$
\lambda\left(\bigcap_{k=1}^n\left\{x:B_k(x)\in A_k\right\}\right)=\prod_{k=1}^n\lambda\left(\left\{x:B_k(x)\in A_k\right\}\right).\tag2
$$
(Proof: Express the set on the LHS of (2) as a disjoint union of sets of the form appearing in the LHS of (1).) In probabilistic terms, this establishes that the $B_k$ are independent random variables.
To finish the proof, argue that the complement of your set $E$ is a subset of the set
$$\bigcap_{k=1}^n\left\{x: B_k(x)\ne 44\cdots 4\right\}.\tag3$$
Compute the measure of (3) using (2), giving an upper bound for $\lambda(E^c)$. This upper bound holds for every $n$, so we are done.
