Product of Lebesgue and counting measures

Question

Let $\mathbb R$ be endowed with the standard Euclidean topology and let $\widetilde {\mathbb R}$ denote the line endowed with the discrete topology. Let $\mu$ and $\nu$ denote the Lebesgue and counting measures on these two spaces, respectively. Define the product measure in the usual way: $$(\mu\times\nu)(E)\equiv\inf\left\{\sum_{k=1}^{\infty}\mu(A_k)\cdot \nu(B_k)\,\Bigg|\,E\subseteq\bigcup_{k=1}^{\infty}A_k\times B_k,\,A_k\in\mathscr B_{\mathbb R},\,B_k\subseteq\widetilde{\mathbb R}\,\forall k\right\}\tag{$*$}\label{star}$$ for any $E\in\mathscr B_{\mathbb R}\otimes 2^{\widetilde{\mathbb R}}=\mathscr B_{\mathbb R\times\widetilde{\mathbb R}}$ (the latter two $\sigma$-algebras can be shown to be equal).

Let $E\equiv\{(x,x)\,|\,x\in[0,1]\}$. It is known that if $\widetilde{\mathbb R}$ is endowed with the Borel $\sigma$-algebra (instead of the discrete $\sigma$-algebra) and the counting measure, then, once one replaces “$B_k\subseteq\widetilde{\mathbb R}$” with “$B_k\in\mathscr B_{\mathbb R}$” in \eqref{star} above, one has $(\mu\times \nu)(E)=\infty$ (with the redefined product measure).

My question is whether this conclusion still holds if $\widetilde{\mathbb R}$ is endowed with the discrete topology and the associated discrete $\sigma$-algebra instead. (Note that the $\sigma$-algebras on $\widetilde{\mathbb R}$ are associated with the counting measure in both cases.) One might conjecture that allowing for more general sets in \eqref{star} (i.e., arbitrary subsets of $\widetilde{\mathbb R}$ instead of just Borel subsets) may bring the infimum down—perhaps even to zero—with some clever choice of a countable collection of rectangles that cover $E$ and involve non-Borel sets (say, some carefully chosen Vitali sets). Yet, a rigorous proof has eluded me so far.

Any thoughts are greatly appreciated.

---

UPDATE: In fact, $(\mu\times\nu)(E)$ should be zero. Bogachev - Measure theory (2007) (in Example 7.14.65, pp. 154–155) claims that the measure $$E\mapsto\sum_{y\in\widetilde{\mathbb R}}\mu\left(\{x\in\mathbb R\,|\,(x,y)\in E\}\right)$$ coincides with $(\mu\times\nu)$, as defined above in \eqref{star}. This, if true, readily implies that $(\mu\times\nu)(E)=0$, but no proof is provided.

Answer 1

The conclusion is still true.

Suppose $E$ is covered by the sets $A_k\times B_k$, $k\in\mathbb N$. Let $M\subset\mathbb N$ be the set of such indices $k$ that $\mu(A_k)=0$. Let $S=\bigcup_{k\in M}A_k$; we obviously have $\mu(S)=0$ so that $[0,1]\setminus S$ is uncountable.

Now the set $F=\{(x,x);x\in [0,1]\setminus S\}\subset E$ is covered by the sets $A_k\times B_k$, $k\in\mathbb N\setminus M$. Then for each $x\in[0,1]\setminus S$ there is $k\in\mathbb N\setminus M$ such that $(x,x)\in A_k\times B_k$, in particular $x\in B_k$. Thus $[0,1]\setminus S\subset\bigcup_k B_k$. Since $[0,1]\setminus S$ is uncountable, at least one $B_{k_0}$ is uncountable. But $k_0\notin M$, so $\mu(A_{k_0})>0$ and thus $\mu(A_{k_0})\nu(B_{k_0})=\infty$. Thus $$ \sum_{k\in\mathbb N}\mu(A_k)\nu(B_k)=\infty. $$

Answer 2

My explanation relies on the concept of outer measure : http://en.wikipedia.org/wiki/Outer_measure, I don't know if you are aware of this construction. I'll use the word measure instead of outer measure in the following.

You still do have $\mu\times\nu(E)=\infty$, because every subset in $\mathbb{R}$ is $\nu$ measurable ; indeed, let $A,T\subset \mathbb{R}$, then if $\nu(T)=\infty$, we have $\nu(T\cap A)=\infty$ or $\nu(T\cap A^c)=\infty$, so $A$ is $\nu$ measurable. If $\nu(T)=n<\infty$, then $T$ is finite, and there exist a positive integer $n$, $t_1,\cdots,t_n\in\mathbb{R}$, and $0\leq k\leq n$ such that $T=\{t_1,\cdots, t_n\}$, and $T\cap A=\{t_1,\cdots,t_k\}$, $T\cap A^c=\{t_{k+1},\cdots,t_n\}$. It follows that $\nu(T)=n=k+(n-k)=\nu(T\cap A)+\nu(T\cap A^c)$. We deduce that every subset of $\mathbb{R}$ is $\nu$ measurable.

And you can define the product of two arbitrary measures $\mu,\nu$ on $X$ and $Y$ in the same way you did, if you take $\mu$ measurable sets for the $A_k$ and $\nu$ measurable sets for $B_k$. With this construction, you get a Fubini theorem, that says that your measure $\mu\times\nu$ is regular, and we have this corollary : If $f$ is $\mu\times\nu$ integrable ($\mu\times\nu$ measurable, with finite integral), and $\{f\neq 0\}$ is $\sigma$-finite for $\mu\times\nu$, then you have $$ \int_{X\times Y}f(x,y)d(\mu\times\nu)(x,y)=\int_{X}\int_{Y}f(x,y)d\nu(y)d\mu(x)=\int_{Y}\int_{X}f(x,y)d\mu(x)d\nu(y). $$

Going back to your example, the collection of $\mu$ measurable sets is $\mathscr{L}(\mathbb{R})$ (it contains $\mathscr{B}(\mathbb{R})$), and the $\nu$ measurable sets are $\mathscr{P}(\mathbb{R})$. $\mu\times\nu$ is regular so $E$ is $\mu\times\nu$ measurable, but $$ \int_{\mathbb{R}}\int_{\mathbb{R}}\mathbb{1}_E d\mu(x)d\nu(y)=0, \int_{\mathbb{R}}\int_{\mathbb{R}}\mathbb{1}_E d\nu(x)d\mu(y)=1 $$ so $\{\mathbb{1}_E\neq 0\}$ is not $\sigma$-finite for $\mu\times\nu$, so $\mu\times\nu(E)=\infty$. For a reference, I suggest you to consult Federer's Geometric Measure Theory, p.117, 2.6.3.