Is it meaningful to say take a random measurable subset of [0, 1] with measure 0.5? If it is, is it then true that for any $r\in[0, 1]$, the probability that $r$ is in a random measurable subset of $[0, 1]$ with measure 0.5 is $0.5$?
Context: I was studying entropy encoding and thought that one way to think about why entropy encoding efficient is this: for each finite code word $s$ of length $d$, assign a subset $A = f(s)$ of $[0, 1)$ such that $\{f(s) : s \text{ is a code word of length } d\}$ partitions $[0, 1)$, and
$$Pr(s' = s | s' \text{ is of length }d) = |A|$$
For example, in arithmetic encoding, we have $f(0) = [0, 1/2), f(1) = [1/2, 1), f(00) = [0, 1/4), ...$.
Then, knowing the length $d$, $f(s)$ can be represented by any real number $\bar s\in f(s)$.
Now, the crucial part of entropy encoding is that in general, for each code $s$, there exists a real number $\bar s\in f(s)$ that is representable in under $$-\log_2 (Pr(s' = s | s' \text{ is of length }d)) = -\log_2 |f(s)|$$ bits.
For example, the binary string $b_1...b_n$ represents the fraction $(b_1...b_n)_2/2^n$, where $(b_1...b_n)_2$ denotes the string decoded as the binary code of an integer.
If we assume that it is nonzero, then $b_n$ can always be chosen as $1$, that is, $(b_1...b_n)_2$ is an odd number. There are $2^{n-1}$ such numbers, and indeed, since $b_n$ is always $1$, it can be dropped. So the new encoding would have $b_1...b_{n-1} \mapsto (b_1...b_{n-1}1 )_2/2^n$.
However, I thought that it is not necessary to use binary fractions. The crucial thing is that we have an injective "coding" function $\{0, 1\}^n \to [0, 1)$, and on average, each $f(s)$ contains one image of the coding function.
Suppose it can be proven that for any $r\in [0, 1)$, the probability that a random measurable subset $S\in[0, 1)$ contains $r$ is $|S|$, then it is immediate that arithmetic coding would work, because on average each $f(s)$ would contain one number with codelength $-\log_2|f(s)|$.