Prove that the set $A$ is measurable and find its Lebesgue measure.

Question

Let $A ⊂ [0, 1] × [0, 1]$ be the set of points $(x, y)$ with decimal representations $x = 0.x_1x_2 ..., y = 0.y_1y_2 ...$ such that $x_ny_n = 5$ for all $n ∈ \mathbb{N}.$ Prove that the set $A$ is measurable and find its Lebesgue measure.

Okay so I know the straightforward way do this by taking open sets that go to measure zero. However I have a slicker proof that seems to almost work. Let $f(x, y) = 0$ if $(x,y) \notin A,$ and $= 1$ if it is. Then $m(E) = \int_{I \times I}f(x,y)d(x \times y)$. It is clear that $\int_I\int_I f(x,y)dxdy = \int_I\int_If(x,y)dydx = 0$. This is because $x$ determines $y$ and vise versa so there is at most one $x$ for every $y$ so $\int_I\int_I f(x,y)dxdy = \int_I 0 dy$ and the same for the other one. But, we don't know that $f(x,y)$ is measurable and in fact it is measurable iff $A$ is, so can we actually say anything about $\int_{I \times I}f(x,y)d(x \times y)$?

Thanks

Answer 1

The map $x\mapsto x_n$ is measurable for each $n$ (it is ambiguous only on a set of measure zero, so the ambiguity doesn't matter). It follows that the set of $(x,y)$ with $x_ny_n=5$ is measurable for each $n$, and $A$ is a countable intersection of such.

Answer 2

Let $B=\{(x,y)\in [0,1]:\forall n (\{x_n,y_n\}\subset \{1,5\}) \}.$ $$\text { For each } n \text { let } B(n)=\{ \sum_{j=1}^{j=n}b_j10^{-j}:b_j\in \{1,5\} \}. $$ Note that $B_n$ has $2^n$ members.For each $x\in B(n)$ let $I(x,n) = (x-10^{-n},x+10^{-n}).$ $$\text { Let } J(n)=\cup_{x\in B(n)}I(x,n).$$ $$\text { For every } n \text { we have }A\subset B\times B\subset J(n)\times J(n).$$ Let $\mu$ denote 2-dimensional Lebesgue measure.$$\text { We have } \mu (J(n)\times J(n))\leq \sum \{ \mu (I(x,n)\times I(y,n) : (x,y)\in B(n)\}=$$ $$=(2\cdot 10^{-n})^2(2^n)^2=2(2/10)^{2n}.$$ $$\text { Letting } n\to \infty \text { we obtain } \mu (A)\leq \mu (B\times B)=0.$$