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.
Suppose that $E\in\mathscr B_{\mathbb R\times\widetilde{\mathbb R}}$ is a Borel-measurable set on the product space. For each $y\in\widetilde{\mathbb R}$, define $$E^y\equiv\{x\in\mathbb R\,|\,(x,y)\in E\}.$$ Assume also that $$G\equiv\{y\in\widetilde{\mathbb R}\,|\,\mu(E^y)>0\}$$
is a countable set and also that $$\sum_{y\in G}\mu(E^y)<\infty.$$ I am trying to prove the following: $$\boxed{(\mu\times\nu)(E)=\sum_{y\in G}\mu(E^y),}\tag{*}$$
where $$(\mu\times\nu)(E)\equiv\inf\left\{\sum_{k=1}^{\infty}\mu(A_k)\cdot \nu(B_k)\,\Bigg|\,E\subseteq\bigcup_{k=1}^{\infty}A_k\times B_k,\,A_k\in\mathscr B_{\mathbb R},\,B_k\subseteq\widetilde{\mathbb R}\,\forall k\right\}\tag{**}$$
is the product measure on $\mathscr B_{\mathbb R}\otimes\mathscr B_{\widetilde{\mathbb R}}=\mathscr B_{\mathbb R\times\widetilde{\mathbb R}}$. (Remark: I have already proved that these latter two $\sigma$-algebras coincide and also that $E^y\in\mathscr B_{\mathbb R}$ for any $y\in\widetilde{\mathbb R}$).
Attempts:
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It is not difficult to show that $(\mu\times\nu)(E)\geq\sum_{y\in G}\mu(E^y)$, given that $$E=\bigcup_{y\in\widetilde{\mathbb R}}E^y\times\{y\}\tag{***}$$
and this union is disjoint. -
As for the other direction, note that $\{E^y\times\{y\}\}_{y\in\widetilde{\mathbb R}}$ cover $E$ according to $(***)$, so one could use the infimum property of the product measure $\mu\times\nu$ in $(**)$ to establish the desired inequality. The problem is that this union is uncountable (even though only countably many $y\in\widetilde{\mathbb R}$ have $m(E^y)>0$), so it may not be a suitable cover based on which $(**)$ could be made use of.
Any hints/thoughts are appreciated.
UPDATE #1: Possible counterexample: Let $E\equiv\{(x,x)\,|\,x\in[0,1]\}$. Clearly, $E^y=\{y\}$ if $y\in[0,1]$ and $\varnothing$ otherwise. It is clear also that $E\in\mathscr B_{\mathbb R}\otimes\mathscr B_{\widetilde{\mathbb R}}$. It follows that $G=\varnothing$ and the right-hand side of $(*)$ vanishes. However, it is well-known that $(\mu\times\nu)(E)=\infty$ whenever $\widetilde{\mathbb R}$ is endowed with the Borel $\sigma$-algebra and the counting measure. In the present setting, though, $\widetilde{\mathbb R}$ is endowed with the discrete topology and the corresponding $\sigma$-algebra $2^{\widetilde{\mathbb R}}$! Hence, there may still be some hope that the left-hand side of $(*)$ vanishes, too; to do this, one must cover $E$ with a countable union of rectangles of the form $(A_k,B_k)\in\mathscr B_{\mathbb R}\times 2^{\widetilde{\mathbb R}}$ the measures of whose products are small. To achieve the desired result, at least some of $B_k$'s must be non-Borel-measurable (say, cleverly chosen Vitali sets), in order to avoid the conclusion that $(\mu\times\nu)(E)=\infty$. Is such a choice of non-Borel sets possible, or is it the case that $(\mu\times\nu)(E)=\infty$ with the discrete topology, too?
UPDATE #2: The result should be true. According to Bogachev (2007) (Example 7.14.65, pp. 154–155), the measure $$E\mapsto\sum_{y\in\widetilde{\mathbb R}}\mu\left(E^y\right)$$ coincides with $(\mu\times\nu)$, as defined above in $(*)$, but this result is mentioned only; no proof is provided. If this is true, then the possible counterexample is obviated, too, and the diagonal of $[0,1]\times[0,1]$ should have a vanishing product measure.
Best Answer
Maybe I am missing something, but the following seems to suffice.
By the definition of the product measure (or Tonelli's Theorem if you prefer), $$ (\mu\times\nu)(E)=\int 1_Ed(\mu\times\nu)=\int \left(\int 1_E(x,y)d\mu(x) \right) d\nu(y). $$ But $1_E(x,y)=1_{E^y}(x)$, so $$ \int \left(\int 1_E(x,y)d\mu(x) \right) d\nu(y)=\int \mu(E^y)d\nu(y)=\sum_{y}\mu(E^y). $$ Clearly $$ \sum_{y}\mu(E^y)=\sum_{y\in G}\mu(E^y), $$ which was what you wanted.
EDIT: My measure theory, as you may have already noticed, is a bit rusty. Hopefully, there are not too many errors in the following. I follow the notation of 'Real Analysis and Probability' by Dudley. From page 134, the set $\mathcal R$ of rectangles $A\times B$ with $A,B$ measurable is a semi-ring, and $\rho(A\times B)=\mu(A)\nu(B)$ is countably additive on $\mathcal R$. Then $\rho$ can be extended uniquely to a countably additive function (still denoted $\rho$) on the algebra $\mathcal A$ generated by $\mathcal R$. The product measure, $\mu\times\nu$, you have defined, as far as I can tell, is then the outer measure (Caratheodory extension) generated by this $\rho$. Another extension would be $m(E)=\sum_{y}\mu(E^y)$.
Observation: Any extension $\alpha$ of $\rho$ satisfies $m\leq\alpha\leq (\mu\times\nu)$: Let $E\subset\bigcup_{k}A_k\times B_k$. Then $$ \alpha(E)\leq \sum_k \alpha(A_k\times B_k)=\sum_k \mu(A_k)\nu(B_k). $$ Taking the infimum yeilds $\alpha(E)\leq(\mu\times\nu)(E)$. As for the other inequality, observe that if $F$ is a finite set such that $E\supset\bigcup_{y\in F}E^y\times \{y\}$, then $$ \alpha(E)\geq \sum_{y\in F} \alpha(E^y\times\{y\})=\sum_{y\in F} \mu(E^y). $$ Taking the supremum over finite sets $F$ yeilds $\alpha(E)\geq m(E)$. As a consequence, $m=(\mu\times\nu)$ if and only if the product measure is unique (which may already indicate that the assertion may not hold in general).
If we use your example from update 1, $m(E)=0$ but $(\mu\times\nu)(E)=\infty$. In this relation, see the answer to this question. I will now argue that $(\mu\times\nu)(E)=\infty$ indeed holds. Given any $A_k,B_k$ such that $E\subset\bigcup_{k=1}^\infty A_k\times B_k$, consider the family $$ \{A_k\times B_k\;|\mu(A_k)>0\;\}. $$ This family must cover uncountably many of the points of $E$ (since $\mu(\bigcup A_k)\leq\sum\mu(A_k)=0$ when the union/sum ranges over those $k$ such that $\mu(A_k)=0$, and the complement of a Lebesgue-measure-zero set in [0,1] is uncountable). It follows that there is some $k$ such that $\mu(A_k)>0$ and $B_k$ is uncountable, which in particular implies $\nu(B_k)=\infty$. But then $$ \sum_k \mu(A_k)\nu(B_k)=\infty. $$
EDIT: Since counting measure is strictly localizable and Lebesgue measure is $\sigma$-finite, Theorem 252B of 'Measure Theory Vol 2' by Fremlin applies. Perhaps Bogachev is talking about the c.l.d. product measure (in the words of Fremlin) when he says 'the' product measure.
EDIT: Using Lemma 417B from Fremlin Vol IV, one sees also that we can define an extension $\alpha$ of $\rho$ by $$\alpha(E)=∫ν(E_x)dμ(x).$$ Here, $x\mapsto ν(E_x)$ is measurable by the lemma. As for $\sigma$-additivity, note that if $E_n$ are disjoint sets, then $(E_n)_x$ are disjoint sets as well. Using that $m\leq\alpha\leq (\mu\times\nu)$, we see again that $m\neq (\mu\times\nu)$, by considering your example from update 1.