Suppose $f:[0,1)\times[0,1)\to [0,\infty]$ is the function where $f(x,y)$ is the number of digits (base 2) where $x$ and $y$ disagree. Ignore cases where we have infinitely many $1$'s to ensure uniqueness of binary expansion. I want to show that $f$ is measurable, and indeed, find the Lebesgue measure of $f^{-1}([0,\infty))$.
My idea to show $f$ measurable was to write $$f(x,y) = \sum_{n=1}^{\infty}1_{R_n(x)\ne R_n(y)}$$
Where $R_n(x)$ is the $n^{th}$ digit of $x$. I have previosuly shown that $R_n$ measurable for all $n$. So I want to show that each $1_{R_n(x)\ne R_n(y)}$ measurable, i.e. that $$\{(x,y)|R_n(x)\ne R_n(y)\}$$ measurable, and then presumably use some lemma to the effect that the infinite sum of non-negative measurable functions is measurable. But I'm not sure how to show the above set is measurable, or even to describe it really.
Let $A(n)$ be $f^{-1}(\ \{n\}\ )$.
A description of $A(n)$ is as follows. Consider all possible ordered $n$-tuples $k$ in $\Bbb N$, i.e. of indices for the decimals of a binary number in $I:=[0,1]$, $$k=(k_1<k_2<\dots<k_n)\ .$$ They build a countable set $K(n)$, a subset of $\Bbb N^{\times n}$. For each such $k\in K(n)$ consider $$A(n,k)=\Big\{\ (x,y)\in[0,1]^2\ :\ x,y\text{ differ exactly at the "places" }k_1,k_2,\dots,k_n\ \}\ . $$ It is clear that $A(n,k)$ is measurable, and $$ f^{-1}(n)=A(n)=\bigsqcup _{k\in K(n)}A(n,k)\ . $$ For this, one may consider the "$k$-switch" map $S(n,k):I\to I$, defined almost everywhere, which associates to an $x\in I$ the (almost always) unique element $y=S(n,k)(x)\in I$, which differs in its binary representation from $x$ at the places in $k$.
Then $A(n,k)$ is the graph of $S(n,k)$, so it has measure zero.
The second part of the problem asks for the measure of $$ \{f<\infty\}\ =\ \bigsqcup _{n\ge 0} \bigsqcup _{k\in K(n)} A(n,k) \ .$$ A countable union of zero sets.