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: $$(\mu\times\nu)(E)\equiv\inf\left\{\sum_{k=1}^{\infty}\mu(A_k)\cdot \nu(B_k)\,\Bigg|\,E\subseteq\bigcup_{k=1}^{\infty}A_k\times B_k,\,A_k\in\mathscr B_{\mathbb R},\,B_k\subseteq\widetilde{\mathbb R}\,\forall k\right\}\tag{*}$$
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 $(*)$ 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 $(*)$ (*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 (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 $(*)$. This, if true, readily implies that $(\mu\times\nu)(E)=0$, but no proof is provided.