Uniqueness of product measure (non $\sigma$-finite case)

$\begingroup$ I don't have a counterexample on hand (and I'm at home). The Wikipedia page refers to Ash's book for the statement that if the measures are $\sigma$-finite, then the extension is unique, but I don't know if it provides an example of non-uniqueness when the measures are not. The Wikipedia page on the Hahn-Kolmogorov Theorem gives an example in which the extension of a finitely-additively function to a measure is not unique, so perhaps it can be adapted. I'll think about it, and check my books on Monday. (Sorry!) $\endgroup$

$\begingroup$ OK, thanks! The uniqueness in the $\sigma$-finite case can be proved using Dynkin's $\pi -\lambda$ theorem (easily when the space is finite; otherwise write the space as union of finite subspaces). $\endgroup$

$\begingroup$ As Arturo said, it's not true in general. As usual, counterexamples should be obtained by considering $[0,1] \times [0,1]$ with Lebesgue measure on the first coordinate and counting measure on the second coordinate. Let $\mu = \lambda \times \#$ be the usual product measure. Let $\nu$ be the measure (!) $\nu(G) = \sup\<\mu(G \cap (E \times F))\,:\,\lambda(E) \lt \infty, \# F \lt \infty\>$. Then I think this should give a counterexample. $\endgroup$

2 Answers 2

Uniqueness does not hold in general.

Besides the usual product measure of two measure spaces $(X,\mathfrak,\mu)$ and $(Y,\mathfrak,\nu)$, there is a different version of the product measure, called the complete locally determined product. For products of non-$\sigma$-finite measure spaces it has many desirable properties that the usual product measure lacks. In his Measure Theory, Fremlin goes so far as to call the usual product measure the primitive product measure, see chapter 5 of volume 2 for more details on the construction and its basic properties.

What follows below is taken more or less directly from Fremlin's exposition.

The construction of the complete locally determined version of the product measure is as follows: Let $\pi = \mu \times \nu$ be the usual product measure on the full $\sigma$-algebra $\mathfrak = \mathfrak \otimes \mathfrak$ obtained from performing the Carathéodory construction on the measurable rectangles. It seems more natural to do this than to restrict to the $\sigma$-algebra $\mathfrak \times \mathfrak \subset \mathfrak \otimes \mathfrak$ generated by the measurable rectangles from the beginning. The complete locally determined product measure $p$ is an “inner regularization” of the usual product measure $\pi$. Its definition is that for $P \in \mathfrak$ one puts

If you're willing to believe that $p$ is a measure, you can skip the next section of this answer.

Let us check that $p$ is indeed a measure: Clearly, $p(\emptyset) = 0$. If $P_n \in \mathfrak$ is a sequence of pairwise disjoint sets, we can estimate for every pair $M \in \mathfrak,N \in \mathfrak$ with $\mu(M) \lt \infty$ and $\nu(N) \lt \infty$ that $$ \pi \left( \bigcup_^ <\infty>P_n \cap (M \times N)\right) = \sum_^<\infty>\; \pi\left(P_n \cap (M \times N)\right) \leq \sum_^<\infty>\;p(P_n), $$ hence $p\left(\bigcup_^<\infty>P_n\right) \leq \sum_^<\infty>\;p(P_n),$ so $p$ is $\sigma$-subadditive. To prove $\sigma$-additivity, note that we can choose for every $t \lt \sum p(P_n)$ a large enough $k$ such that there are $t_1 \lt p(P_1), \ldots, t_k \lt p(P_k)$ with $t \lt t_1 + \cdots + t_k$. Then we can choose $M_i \in \mathfrak,N_i \in \mathfrak$ with $\mu(M_i)\lt \infty$ and $\nu(N_i) \lt \infty$ such that $t_i \leq \pi(P_i \cap (M_i \times N_i))$. Let $M = M_1 \cup \cdots \cup M_k$, $N = N_1 \cup \cdots \cup N_k$. Then, by definition, and the fact that $\mu(M) \lt \infty$ and $\nu(N)\lt \infty$, we have $$ p\left(\bigcup_^\infty P_n\right) \geq \pi\left(\bigcup_^\infty P_n \cap (M \times N)\right) \geq \sum_^k \; \pi (P_n \cap (M \times N)) \geq \sum_^k \; \pi(P_n \cap (M_n \times N_n)) \gt t. $$ Thus $t \lt p(\bigcup P_n) \leq \sum p(P_n)$ for every $t < \sum p(P_n)$. It follows that $p$ is $\sigma$-additive on disjoint sequences sets in $\mathfrak$, hence $p$ is a measure on $\mathfrak$.

Exercise 1: It is clear that $p(M \times N) = \mu(M)\,\nu(N)$ for $M$ and $N$ of finite measure. To ensure this equality for all measurable sets $M$ and $N$, we need to suppose that $(X,\mathfrak,\mu)$ and $(Y,\mathfrak,\nu)$ are semi-finite in the sense that every set of infinite measure contains a measurable set of finite positive measure. This implies that $\mu(M) = \sup<\<\mu(E)\,:\,E \subset M, \mu(E) \lt \infty\>>$ and similarly for $\nu$. Using this, check that the equality $p(M \times N) = \mu(M)\, \nu(N)$ holds for all $M \in \mathfrak$ and $N \in \mathfrak$.

Exercise 2: Let $(X,\mathfrak,\mu)$ and $(Y,\mathfrak,\nu)$ be $\sigma$-finite spaces. Prove that $p = \pi$.

Exercise 3: Let $(X,\mathfrak,\mu)$ and $(Y,\mathfrak,\nu)$ be arbitrary measure spaces. If $\lambda$ is any measure on $\mathfrak$ satisfying $\lambda(M \times N) = \mu(M)\nu(N)$ then $p(P) \leq \lambda(P) \leq \pi(P)$.

We are finally ready to concoct our counterexample—most of the basic counterexamples in connection with product measures are variations of this theme.

Let $(X,\mathfrak,\mu) = ([0,1],\Sigma,\lambda)$ be the unit interval with Lebesgue measure. Let $(Y,\mathfrak,\nu) = ([0,1],\mathcal([0,1]),\#)$ be the unit interval equipped with counting measure. Let $\pi$ be the usual product measure and $p$ its locally determined version. Since both $\mu$ and $\nu$ are semi-finite, we have by exercise 1 above that $p(M \times N) = \mu(M) \nu(N)$ for all measurable $M \subset X$ and $N \subset Y$.

Notice that the diagonal $\Delta = \$ is in the $\sigma$-algebra $\mathfrak$ generated by the measurable rectangles, because $$\Delta = \bigcap_^<\infty>\; \bigcup_^\; \left[\frac,\frac\right] \times \left[\frac,\frac\right].$$

Exercise 4: $\pi(\Delta) = \infty$.

Exercise 5: $p(\Delta) = 0$.

Thus, $\pi$ and $p$ are distinct, while agreeing on the measurable rectangles themselves. Since $\Delta \in \mathfrak \times \mathfrak$ this also holds for the restrictions of $\pi$ and $p$ from $\mathfrak$ to $\mathfrak$.