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.




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**UPDATE:** In fact, $(\mu\times\nu)(E)$ *should* be zero. [Bogachev - Measure theory (2007)](http://books.google.com/books?id=hJ_BngEACAAJ) (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.



Product of Lebesgue and counting measures
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
$$
\sum\_{k\in\mathbb N}\mu(A\_k)\nu(B\_k)=\infty.
$$


