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.

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    $\begingroup$ I'd say that in either variant the value is $\infty$, for $\sum_{k = 0}^\infty \mu(A_k)\nu(B_k)< \infty$ implies that $\cup_k A_k\times B_k$ is included in a set $(\mathbb{R}\times C ) \cup (N\times\mathbb{R}) $ with $\mu(N)=0$ and $C$ a countable set. $\endgroup$ – Pietro Majer Jul 22 '14 at 9:17
  • $\begingroup$ Can you give a (reference to a) proof of the fact that $(\mu\times\nu)(E)=\infty$ in the Borel case? It would be easier to see what changes when the $\sigma$-algebra changes than to start from scratch. $\endgroup$ – Joonas Ilmavirta Jul 22 '14 at 9:27
  • $\begingroup$ @PietroMajer Thank you for your comment. Your argument seems cogent, but I'm not completely convinced. I will think about it later. $\endgroup$ – triple_sec Jul 22 '14 at 9:29
  • $\begingroup$ @JoonasIlmavirta Please see here (attention is restricted to $[0,1]$, but I don't think this difference is essential). $\endgroup$ – triple_sec Jul 22 '14 at 9:30
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    $\begingroup$ @triple_sec. for a given sum $\sum_k \mu(A_k)\nu(B_k) <\infty$, define $N:=\cup_{k\in \mathbb{N}, \mu(A_k)=0}A_k$ and $C:=\cup_{k\in \mathbb{N}, \nu(B_k)<\infty}B_k$. $\endgroup$ – Pietro Majer Jul 22 '14 at 10:21

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 $$ \sum_{k\in\mathbb N}\mu(A_k)\nu(B_k)=\infty. $$

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  • $\begingroup$ Excellent, thank you very much! Does this mean that Bogachev is wrong? I'd be surprised because I thought of it as an authoritative resource in measure theory. Let me quote the text verbatim: “Let us consider the metric space $X=\Omega\times\mathbb R^1$, where $\Omega$ is the real line with the discrete metric and $\mathbb R^1$ is equipped with the standard metric. Then $X$ with the product topology is locally compact and $\mathscr B(X)=\mathscr B(\Omega)\otimes\mathscr B(\mathbb R^1)$. [cont'd below] $\endgroup$ – triple_sec Jul 22 '14 at 21:16
  • $\begingroup$ [cont'd from above] For every $B\in\mathscr B(X)$, we set $$\mu(B)\equiv\sum_{\omega\in\Omega}\lambda(B_{\omega}),$$ where $B_{\omega}=\{t\,|\,(\omega,t)\in B\}$ and $\lambda$ is Lebesgue measure, i.e., $\mu$ is the product of the counting measure on $\Omega$ and Lebesgue measure [italics added].” If this result were to be true, then, since $\mu(E)=0$ (where $E$ is the diagonal of $[0,1]\times[0,1]$ and $E_{\omega}$ is either empty or a singleton for every $\omega\in\Omega$) and the product measure should be 0, too. But you just proved that it must be infinite. What is going on here? $\endgroup$ – triple_sec Jul 22 '14 at 21:18
  • $\begingroup$ Upon reflection all I can think of (apart from Bogachev's being plain wrong) is that he and I are not on the same page on the definition of “product measure.” I interpret “the” product measure as the Carathéodory extension from the algebra $\mathscr A$ of finite disjoint unions of Cartesian products of measurable sets, as in $(*)$, and I'm tempted to think this is the standard definition. However, this is not the unique measure that extends the premeasure $(\mu\times\nu)|\mathscr A$ onto the product $\sigma$-algebra, since $\nu$ is not $\sigma$-finite. $\endgroup$ – triple_sec Jul 22 '14 at 21:35
  • $\begingroup$ Is it possible that what Bogachev means by “the product of the counting measure on $\Omega$ and Lebesgue measure” is just one particular measure that extends the premeasure $(\mu\times\nu)|\mathscr A$? $\endgroup$ – triple_sec Jul 22 '14 at 21:36
  • $\begingroup$ What Bogachev defines indeed seems to be a product measure, not the product measure. It corresponds to the iterated integral $\int_0^1\int_0^1f(x,y)d\mu(x)d\nu(y)$ of a function $f$, not the integral with respect to the standard $\mu\times\nu$. $\endgroup$ – Joonas Ilmavirta Jul 22 '14 at 22:15

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.

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  • $\begingroup$ By regular, do you mean that the measure of any set is given by the supremum of the measures of compact sets contained in it and the infimum of the measures of open sets containing it? BTW, I was aware that this Fubini-type result works whenever $f$ is continuous and compactly supported on $X\times Y$, but I didn't know it is also true for any $f\in L^1(\mu\times\nu)$. I will definitely check out the reference you provided. $\endgroup$ – triple_sec Jul 22 '14 at 21:30
  • $\begingroup$ No, in Fubini's theorem, no topology is involved : $\mu$ is regular on $X$ if set $A\subset X$, there exists a $\mu$ measurable set $B\subset X$ such that $A\subset B$ and $\mu(A)=\mu(B)$. $\endgroup$ – Paul-Benjamin Jul 23 '14 at 7:05

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