See, study the complement of the set. That is, look at integers which contain $5$ in their decimal expansion. Caveat : note that $0.6 = 0.5\overline{9}$ also counts as a decimal which is expressed with a $5$ as one of the digits, so belongs in the set. In particular, any terminating decimal which terminates with $6$ can be considered to belong to the set.
The first such category we can think of is: Those that have $5$ as the first digit following the decimal point. This is the set of numbers $[0.5,0.6]$. This has measure $0.1$, so the left over measure is $0.9$.
Now, from the remaining set, remove the set of all numbers with second digit $5$. This consists of $[0.05,0.06[, [0.15,0.16] \ldots [0.95,0.96]$ without $[0.55,0.56]$, since that was already removed earlier. Now, each of these has measure $0.01$, so we have removed $0.09$ more from the system. Hence, the left over is $0.81$.
By induction, prove the following : at the $n$th step, the set left over has measure $\frac{9^n}{10^n}$. Now, as $n \to \infty$, we see that the given set has measure zero (I leave you to rigorously show this, you can use the Borel-Cantelli lemma). This also incorporates the fact that the given set is measurable, since it's measure is computable(and is $0$).
I don't have a complete argument for it but I don't agree with your conjecture and believe the Lebesgue measure is 1. I might be mistaken but hope you could at least consider my position and maybe end up with a proof yourself.
Consider a much simpler system: an i.i.d sequence of integers, where each integer has a positive probability to appear. Then the probability of the Lusin event is obviously $1$. Indeed you just need one integer to be repeated infinitely often for the Lusin event to hold, but even if you insist on being a strict divisor, after you have seen an integer $k$, you are bound with probability $1$ to see $2k$ after some time, and then you could repeat the argument ensuring $4k$ will appear, etc.
Now what is the relevance of this toy model to the case at hand ? Surely the continued fraction expansion of a uniform random real number is not an i.i.d. sequence?
Well I just found these wikipedia pages:
Basically what they tell you is the following: you can easily define the continued fraction expansion of $x\in(0,1)$ with just two functions: $a_1(x) = \lfloor 1/x \rfloor$ and $h(x) = \{ 1/x\}$. You can then set $a_2 = a_1\circ h$, $a_3 = a_1 \circ h \circ h$, etc.
You can define a probability measure on $(0,1)$ by setting $\mu(dx) = \dfrac {dx}{\ln(2)(1+x)}$. Then $h : (0,1) \to (0,1)$ is a measure-preserving ergodic operator called the Gauss-Kuzmin-Wirsing operator.
Consider $X$ with distribution $\mu$. The "measure-preserving" part tells you that the sequence $X,h(X),h(h(X)), ...$ is identically distributed. The "ergodic" part tells you that the sequence, even though it's not independent, looks independent "in the long run". That is, $P(h^n(X) \in A \mid X \in B) \to P(X \in A)$ as $n$ goes to infinity.
If you now apply $a_1$ to this sequence, you end up with the continued fraction expansion of $X$, and we have seen that it is "approximately iid". Its distribution is the image of the measure $\mu$ by $a_1$, which is the above-mentioned Gauss-Kuzmin distribution, where all integers have positive probability. As a result, one could conjecture that the Lusin event has probability $1$ for $X$.
Finally, since the measure $\mu$ and the Lebesgue distribution on $(0,1)$ are mutually absolutely continuous, then what holds with probability $1$ for $\mu$ holds also with Lebesgue measure $1$.
Best Answer
Probability theory answer. The digits in the base $10$ expansion are i.i.d random variables with respect to Lebesgue measure.
Suppose I roll a 10-sided die repeatedly. What is the probability that I never get $100$ consecutive $4$'s? Of course that probability is zero by the Borel-Cantelli lemma. So (taking the complement) the set of decimals that contain $100$ consecutive $4$'s has measure $1$.
Requested, more complete explanation.
Divide into blocks of length $100$. Let $B_1$ be the first $100$ digits, let $B_2$ be the second $100$ digits and so on. Write $T_j$ for the event "Block $B_j$ consists of $100$ consecutive $4$'s". These events are independent since the blocks are disjoint. We claim
$$ \mathbb P\left(T_j\text{ occurs for infinitely many } j\right) = 1 . \tag 1$$
For any $j$ we have $\mathbb P(T_j) = 10^{-100}$. The exact value is not important, only that it is nonzero. It is the same for all $T_j$. The series $\sum_{j=1}^\infty \mathbb P(T_j)$ diverges to infinity. So by the Borel-Cantelli lemma, we get $(1)$.