I feel like I made a mistake, is this correct?
If $$E=\{\sum_{i=1}^\infty\frac{a_i}{10^i}\in[0,1]: a_i\in\{0,1,2,\dots, 9\} \text{ and } \forall n>0,\exists m>n, a_m=4\}$$ then define $$E_n=\{\sum_{i=1}^\infty\frac{a_i}{10^i}\in[0,1]: a_n=4\}.$$ Now, $x=\sum_{i=1}^\infty\frac{a_i}{10^i}$ is such that $a_n=4$ if and only if $$4/10^n\leq x\leq 5/10^n.$$ This gives us the measure of $E_n$.
We have that $E=\bigcap_{N=1}^\infty\bigcup_{k=N}^{\infty}E_k$ as the set on the right is equivalent to $$\{x:x\in E_k \text{ for infinitely many $k$}\}.$$
If this is true then, $$m(E)=\lim m\bigcup_{k=n}^\infty E_k\leq\lim \sum_{k=n}^\infty m E_k=0.$$
$\bf{Correction}$: $x\in E_n$ then the first $n-1$ digits can be any number $0-9$ and this gives $10^{n-1}$ possiblities. Then $y+\frac{4}{10^n}\leq x\leq y+\frac{5}{10^n}$ for some $y=\sum^{n-1}_{i=1}\beta_{i}/10^{i}$. Thus there are $10^{n-1}$ intervals of length $10^{-n}$.
Then $\sum mE_n=\infty$.
Fix $m,k$ and consider $E_m\cap E_{m+k}$. There are $10^{m+k-1}$ intervals that make up $E_{m+k}$. Each interval is determined by the proceeding $m+k-1$ digits of the left endpoint, by what we saw above. Since the $m$th digit is $4$, there are $10^{m+k-2}$ possible intervals in total, each of length $10^{-m-k}$. This argument leads to showing that $\{E_n\}$ is independent.
Now, Borel-Cantelli asserts that in a probability space if we are given a sequence of events, and if both the sum of these events diverges and the events are pairwise independent, then the lim sup has probability 1.
The mistake is in thinking that $x \in E_n$ iff $\frac 4 {10^{n}} \leq x \leq \frac 5 {10^{n}}$. In fact $E_n$ is a union of $10^{n-1}$ disjoint intervals of length $\frac 1 {10^{n}}$ each, so $m(E_n)=\frac 1 {10}$. For each $n$. Using Borel - Cantelli Lemmas we can show that $m(E)=1$.