Prove set of points in $[0,1]$ with prime digits is measurable Let $A$ be the subset of $[0,1]$ consisting of all numbers such that every digit of $x$ after the decimal point is a prime number i.e. $x=0.p_1p_2p_3\ldots$, where $p_i\in\{2,3,5,7\}$. For example, $0.2$, $0.32$, $0.557$, $0.7532$, et cetera.
I want to prove that $A$ is a Lebesgue measurable set and evaluate the Lebesgue measure $m(A)$.
I notice that $A$ is an infinite set: $A = \{0.2, 0.22, 0.222,\ldots , 0.7777\ldots\}$. $A$ is bounded above by $0.777\ldots$, and bounded below by $0.2$ . Is that $A$ is compact and therefore measurable?
I know that $m(X) = \text{length of }X=\lambda(X)$ when $X$ is an interval
Also for $X \subseteq (0,1)$ and \lambda(X) = \lambda((0,1)) - \lambda((0,1)\X)
I have not idea about how to evaluate $\lambda((0,1)\setminus A)$. 
Any ideas?

The number of sets with first digit $2,3,5,7$ is $4/10$ of the interval ($10$ possible digits, four of them are primes), so the measure of the set $A_1$ with first digit $d_1\in \{2,3,5,7\}$ is $2/5$. Note that $A_1$ is a union of intervals, hence is a Borel set. Similarly for the second digit, we have $2/5$ of the remainder in a set $A_2$ which breaks our intervals from $A_1$ into four intervals each hence $A_2$ is a Borel subset of $A_1$. We continue in this fashion, inductively defining $A_n$ from $A_{n-1}$ each one being a finite union of intervals, hence Borel sets.
If we also define $A_0=[0,1]$ just for simplicity we have $m(A_n)={2\over 5}m(B_{n-1})$. Then the set we care about is just

$$A=\bigcap_{n=0}^\infty A_i$$

which is a countable intersection of Borel sets, hence is itself a Borel set and has measure

$$0=\lim_{n\to\infty}m(A_n)=\lim_{n\to\infty}\left({2\over 5}\right)^n.$$

If you cannot see the limiting directly, you can also note $\varnothing\subseteq A\subseteq A_n$ for all $n$ hence
$$0\le A\le m(A_n)=\left({2\over 5}\right)^n$$
which also gives the conclusion.
