Uniqueness of $\sigma$-finite measures over a product space
Consider \((X, \mathcal{A}, \mu)\) and \((Y, \mathcal{B}, \nu)\) two sigma-finite measure spaces. It is well known that the product measure \(\mu \otimes \nu\) is the only measure \(\xi\) over the product space \((X \times Y, \mathcal{A} \ox \mathcal{B})\) satisfying
\[\forall E \in \mathcal{A}, F \in \mathcal{B}, \quad \xi(E \times F) = \mu(E) \nu(F).\]Thus the product of two \(\sigma\)-finite measures is characterized by its value on what we will call measurable rectangles. This however is no longer true if \(\mu\) and \(\nu\) are only assumed to be \(s\)-finite, i.e. to be a countable sum of finite measures. Indeed, consider \(\mu = \nu := \infty \cdot \mathrm{Leb}_{[0,1]}\), i.e. the measure which to a Lebesgue-measurable set associates \(\infty\) if it has positive Lebesgue measure, and zero otherwise. Denote by \(\lambda\) the Lebesgue measure on the diagonal of \([0,1]^2\). Then for any \(E, F \in \mathcal{L}([0,1])\) (the Lebesgue \(\sigma\)-algebra), if either \(E\) or \(F\) has Lebesgue measure \(0\), then
\[(\mu \ox \nu) (E \times F) = (\mu \ox \nu + \lambda) (E \times F) = 0.\]Otherwise, both measures give infinite measure to \(E \times F\), therefore the two measures \(\mu \ox \nu\) and \(\mu \ox \nu + \lambda\) coincide on measurable rectangles. However \(\mu \ox \nu\) gives measure zero to the diagonal, while \(\mu \ox \nu + \lambda\) gives it measure \(\sqrt{2}\), so the two measures are different.
Consider now the more general case of a \(\sigma\)-finite measure \(\mu\) defined over the product space \((X \times Y, \mathcal{A} \ox \mathcal{B})\). Is it characterized by its value on measurable rectangles? This is true if \(\mu\) is assumed to be finite, as can be shown via the \(\pi\)-\(\lambda\) theorem. It is also true if \(\mu\) is the product of two \(\sigma\)-finite measures, as discussed above. However, it turns out that it is not true in general. The goal of this note is to provide a counter example. This counter example was presented to me by Sébastien Gouëzel, and is based on problem 14E in Problems for Mathematicians, Young and Old by Paul R. Halmos.
We will build a function \(B : \RR^2 \to \RR_+\) such that for any \(E, F \in \mathcal{L}(\RR)\) with positive measure, $B$ has an infinite integral over \(E \times F\) against the Lebesgue measure on \(\RR^2\). Therefore, using the same trick as before, it will not be possible to distinguish the measure with density \(B\) from the same measure to which we add the Lebesgue measure over the diagonal of \(\RR^2\) by simply looking at their values over measurable rectangles. On the other hand, because \(B\) is finite everywhere, the measure with density \(B\) will be \(\sigma\)-finite (take \(X_n := \{x \mid B(x) \leqslant n\} \cap [-n, n]^2\) to get a sequence of spanning sets with finite measure). In what follows, if \(E \in \mathcal{L}(\RR)\), we will denote by \(\abs{E}\) the Lebesgue measure of $E$.
To construct \(B\), consider \((r_n)_{n \in \NN}\) a dense sequence in \(\RR\), and let \(I_n\) be the interval centered at $r_n$ and with length \(1/2^n\). We set
\[\begin{align*} A : \RR & \to \RR \\ x & \mapsto \sum_{n \geqslant 0} 4^n \mathbb{I}_{I_n}(x). \end{align*}\]Because \(\sum_{n \geqslant 0} \abs{I_n} < \infty\), the Borel-Cantelli lemma implies that for \(\mathrm{Leb}\)-almost every \(x\), the sum defining \(A(x)\) contains only a finite number of non-zero terms, and is therefore finite. We redefine \(A\) by setting it to \(0\) on the zero-measure set where it is infinite.
We now set \(B(x, y) := A(x - y)\). Let \(E, F \in \mathcal{L}(\RR)\) with positive measure. Doing the change of variable \(u = x-y\) and \(v = y\), we have
\[\begin{align*} \int_{E \times F} B(x, y) \mathrm dx \mathrm dy & = \int_{\RR^2} A(x-y) \mathbb{I}_{E \times F}(x, y) \mathrm dx \mathrm dy \\ & = \int_{\RR^2} A(u) \mathbb{I}_{E \times F}(u+v, v) \mathrm du \mathrm dv \\ & = \int_\RR A(u) \int_\RR \mathbb{I}_{E \times F} (u + v, v) \mathrm dv \mathrm du \\ & = \int_\RR A(u) \int_\RR \mathbb{I}_{(E - u) \cap F}(v) \mathrm dv \mathrm du \\ & = \int_\RR A(u) |(E - u) \cap F| \mathrm du. \end{align*}\]We will now prove that, for some constant \(c > 0\), there exist arbitrarily large $n$ such that for every \(u \in I_n, \abs{(E - u) \cap F} \geqslant c\). This will imply that
\[\int_{E \times F} B(x, y) \mathrm dx \mathrm dy \geqslant c \int_{I_n} A(x) \mathrm dx \geqslant c 4^n |I_n| = c 2^n,\]which in turn implies that \(\int_{E \times F} B(x, y) \mathrm dx \mathrm dy = \infty\). To do that, recall that because \(E\) and \(F\) both have positive measure, the Lebesgue’s density theorem implies that there exist \(x \in E\), \(y \in F\) and \(\epsilon > 0\) such that
\[|E \cap [x-\epsilon, x+\epsilon]| > (4 \epsilon)/3 \quad \text{and} \quad |F \cap [y-\epsilon, y+\epsilon]| > (4 \epsilon)/3.\]Let \(u \in [x - y - \epsilon/7, x - y + \epsilon/7]\). Then
\[|(E - u) \cap [x - u - \epsilon, x - u + \epsilon]| = |E \cap [x - \epsilon, x + \epsilon]| \geqslant (4 \epsilon)/3.\]Moreover, \(x - u \in [y - \epsilon/7, y + \epsilon/7]\), which means that
\[\begin{align*} |(E - u) \cap [y - \epsilon, y + \epsilon]| & \geqslant |(E - u) \cap [x - u - \epsilon, x - u + \epsilon] \cap [y - \epsilon, y + \epsilon]| \\ & = |((E - u) \cap [x - u - \epsilon, x - u + \epsilon]) \backslash \\ & \qquad ([y - \epsilon, y + \epsilon]^c \cap [x - u - \epsilon, x - u + \epsilon])| \\ & \geqslant |(E - u) \cap [x - u - \epsilon, x - u + \epsilon]| - \\ & \qquad |[y - \epsilon, y + \epsilon]^c \cap [x - u - \epsilon, x - u + \epsilon]| \\ & \geqslant (4\epsilon)/3 - (2 \epsilon)/7 > \epsilon. \end{align*}\]We also know that \(\abs{F \cap [y - \epsilon, y + \epsilon]} > (4 \epsilon)/3\). Because \(\abs{[y - \epsilon, y + \epsilon]} = 2 \epsilon\), we deduce that
\[|(E - u) \cap F| \geqslant |(E - u) \cap F \cap [y - \epsilon, y + \epsilon]| > (7 \epsilon)/3 - 2\epsilon = \epsilon/3.\]Take \(c := \epsilon/3\). To conclude, we notice that because the sequence \((r_n)_{n \in \NN}\) is dense, the interval \([x - y - \epsilon/8, x - y + \epsilon/8]\) contains infinitely many terms of the sequence, which implies that there exist arbitrarily large \(n\) such that \([x - y - \epsilon/7, x - y + \epsilon/7]\) contains \(I_n\). \(\square\)