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Let $(X,\mathscr{A},\mu), (Y,\mathscr{B},\nu)$ be two measure spaces, then we have the product measurable space $(X\times Y, \mathscr{A}\times\mathscr{B})$ where $\mathscr{A}\times\mathscr{B}$ is the $\sigma$-algebra generated by $\{A\times B, A\in \mathscr{A}, B\in \mathscr{B}\}$. Applying Caratheodory extension we can always construct a measure $m$ on $\mathscr{A}\times\mathscr{B}$ so that $m(A\times B)=\mu(A)\nu(B)$. Further, if $\mu$ and $\nu$ are $\sigma$-finite, by monotone class theorem such a measure $m$ is uniquely determined. Now my question is:

When $\mu$ and $\nu$ are not $\sigma$-finite, is product measure still unique?

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Not in general. –  Arturo Magidin Oct 8 '11 at 20:09
    
@Arturo, could you please give me a counter example (or a link to a counterexample)? –  Syang Chen Oct 8 '11 at 21:12
    
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!) –  Arturo Magidin Oct 8 '11 at 21:17
    
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). –  Syang Chen Oct 8 '11 at 22:24
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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. –  t.b. Oct 9 '11 at 17:29
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2 Answers 2

up vote 14 down vote accepted

Uniqueness does not hold in general.

Besides the usual product measure of two measure spaces $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$, 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 $\mathfrak{P} = \mathfrak{M} \otimes \mathfrak{N}$ obtained from performing the Carathéodory construction on the measurable rectangles. It seems more natural to do this than to restrict to the $\sigma$-algebra $\mathfrak{M} \times \mathfrak{N} \subset \mathfrak{M} \otimes \mathfrak{N}$ 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 $P \in \mathfrak{P}$ one puts

$$p(P) = \sup{\{\pi(P \cap (M \times N))\,:\,\mu(M) \lt \infty, \nu(N) \lt \infty\}}.$$

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 $P_n \in \mathfrak{P}$ is a sequence of pairwise disjoint sets, we can estimate for every pair $M \in \mathfrak{M},N \in \mathfrak{N}$ with $\mu(M) \lt \infty$ and $\nu(N) \lt \infty$ that $$ \pi \left( \bigcup_{n=1}^{\infty} P_n \cap (M \times N)\right) = \sum_{n=1}^{\infty}\; \pi\left(P_n \cap (M \times N)\right) \leq \sum_{n=1}^{\infty}\;p(P_n), $$ hence $p\left(\bigcup_{n=1}^{\infty}P_n\right) \leq \sum_{n=1}^{\infty}\;p(P_n),$ 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 $M_i \in \mathfrak{M},N_i \in \mathfrak{N}$ 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 $$ p\left(\bigcup_{n=1}^\infty P_n\right) \geq \pi\left(\bigcup_{n=1}^\infty P_n \cap (M \times N)\right) \geq \sum_{n = 1}^k \; \pi (P_n \cap (M \times N)) \geq \sum_{n=1}^k \; \pi(P_n \cap (M_n \times N_n)) \gt t. $$ 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 $\mathfrak{P}$, hence $p$ is a measure on $\mathfrak{P}$.

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 $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ 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 $M \in \mathfrak{M}$ and $N \in \mathfrak{N}$.

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

Exercise 3: Let $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ be arbitrary measure spaces. If $\lambda$ is any measure on $\mathfrak{P}$ 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 $(X,\mathfrak{M},\mu) = ([0,1],\Sigma,\lambda)$ be the unit interval with Lebesgue measure. Let $(Y,\mathfrak{N},\nu) = ([0,1],\mathcal{P}([0,1]),\#)$ 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 $\Delta = \{(x,x) \,:\,x \in [0,1]\}$ is in the $\sigma$-algebra $\mathfrak{M \times N}$ generated by the measurable rectangles, because $$\Delta = \bigcap_{n =1}^{\infty}\; \bigcup_{k=0}^{n-1}\; \left[\frac{k}{n},\frac{k+1}{n}\right] \times \left[\frac{k}{n},\frac{k+1}{n}\right].$$

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

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

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

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Hi, t.b.. Would you say that Fremlin's books are better than Bogachev? I'm looking for some nice books on measure theory. Too bad Fremlin's books are not published with a regular publisher. –  Jonas Teuwen Oct 10 '11 at 12:10
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I don't have a real preference. Both have their drawbacks and virtues, but both are excellent and reliable in general. I find Bogachev's book very readable, and I like it a lot. His treatment of topological measure spaces doesn't really suit my tastes and I'm not so interested in measures on topological vector spaces. Fremlin's books are somewhat too dry and sometimes push a bit too much in direction of generality for my taste, but precisely because of that they contain a wealth that is of immediate use for my work, more so than Bogachev, that's why I think I like Fremlin a bit more... –  t.b. Oct 10 '11 at 12:20
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I give more simple example. Let $\mu =\lambda \times \#$ be usual product of $\lambda$ and $\#$. Let $\mu_1$ be defined as follows: $$\mu_1(X)=\sum_{y \in [0,1]}\lambda_y(([0,1]\times \{y\})\cap X)$$for all Lebesgue measurable subset $X$ of $[0,1]\times [0,1]$, where $\lambda_y$ is Lebesgue measure defined on $[0,1]\times \{y\}$. Then measures $\mu$ and $\mu_1$ are agreeing on measurable rectangles but $\mu(\{(x,x):x \in [0,1]\})=+\infty$ and $\mu_1( (\{(x,x):x \in [0,1]\})=0$.

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This is actually the same example: your $\mu_1$ is $p$ in the other answer, but I guess it is good to have the explicit expression... –  Martin Jan 2 '13 at 11:56
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