Let $([0,1],\mathcal{B},m)$ be the Borel sigma algebra with lebesgue measure and $([0,1],\mathcal{P},\mu)$ be the power set with counting measure. Consider the product $\sigma$-algebra on $[0,1]^2$ and product measure $m \times \mu$.

(1) Is $D=\{(x,x)\in[0,1]^2\}$ measurable?

(2) If so, what is $m \times \mu(D)$?

Edit: The product measure is defined on a rectangle by $m \times \mu(A\times B)=m(A)\mu(B)$ and on general set taking infimum over union of rectangles containing $D$. (Maybe we should use $0 \cdot \infty=0$)

  • $\begingroup$ Are you sure that $m\times\mu$ is well-defined? I have only seen product measures been defined between two $\sigma$-finite measures. $\endgroup$ – Stefan Hansen Apr 18 '13 at 7:28
  • $\begingroup$ @StefanHansen Well, the sigma finiteness gives the unique product measure. So it may not be uniquely defined. $\endgroup$ – Gobi Apr 18 '13 at 7:47
  • $\begingroup$ Then what do you mean by $m\times\mu(D)$? A list of possible values? What is the purpose then? $\endgroup$ – Stefan Hansen Apr 18 '13 at 7:49
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    $\begingroup$ @StefanHansen: Usually people intend the product measure $\pi$ to be what you get by applying Carathéodory's method: define an outer measure by taking $\pi(A)$ to be the infimum of $\sum_n \mu(E_n)\nu(F_n)$ over all countable unions of measurable rectangles covering $A$; then restrict to the $\sigma$-algebra of $\pi$-measurable sets. This is an honest measure defined on a $\sigma$-algebra containing the measurable rectangles. The issue is with uniqueness (Hahn-Kolmogorov's uniqueness theorem does not apply without $\sigma$-finiteness). $\endgroup$ – Martin Apr 18 '13 at 8:01
  • $\begingroup$ And the measure is ugly: The diagonal in the OP has infinite measure, but it contains no subset of finite positive measure. $\endgroup$ – Martin Apr 18 '13 at 8:01

I am assuming that the product measure is the one induced by the product outer measure. (This avoids any issue with ambiguity of definition.)

Any set of the form $B \times A$, with $B$ Borel is measurable ($A$ is arbitrary).

Take the sets $S_n = \{(1,1)\} \cup\left( \cup_{k=0}^{n-1} [\frac{k}{n},\frac{k+1}{n})^2 \right)$. Clearly each $S_n$ is measurable, and hence $D = \cap_{n \ge 0} S_n $ is measurable.

We must have $(m \times \mu) D = \infty$.

To see this, suppose $D \subset \bigcup_{n \ge 0} B_n \times A_n$, where $B_n$ is Borel, and $A_n$ is arbitrary. Let $D_n = D \cap (B_n \times A_n)$, and let $\pi_x((x,y)) = x$ and similarly $\pi_y((x,y)) = y$. Since $[0,1] \subset \cup_n \pi_x D_n$, we must have $m^* (\pi_x D_{n'} ) >0$ for some $n'$, where $m^*$ is the Lebesgue outer measure (using $m^*$ avoids having to worry about the measurability of $\pi_x D_{n'}$).

Hence $m B_{n'} \ge m^* (\pi_x D_{n'} ) > 0$. In addition, $\pi_x D_{n'}$ must be uncountable (otherwise $m^* (\pi_x D_{n'} ) $ would be zero). Furthermore, $\pi_y D_{n'} = \pi_x D_{n'}$, $\pi_y D_{n'} \subset A_{n'}$, hence $A_{n'}$ is also uncountable. Hence $m(B_{n'}) \mu(A_{n'}) = \infty$, and so $\sum_{n \ge 0} m(B_{n}) \mu(A_{n}) = \infty$ for any cover of $D$ by measurable rectangles. Hence $(m \times \mu) D = (m \times \mu)^* D =\infty$.

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    $\begingroup$ The ambiguity is only in the fact that there is no unique measure satisfying $(m \times \mu)(A \times B) = m(A) \cdot \mu(B)$. For example you can check that $\nu(E) = \sum_y m(E\cap ([0,1] \times \{y\}))$ is a measure and has that property for measurable rectangles, but for this measure the diagonal has measure zero. $\endgroup$ – Martin Apr 18 '13 at 18:49
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    $\begingroup$ Sure, the measure defined by the outer measure is unique. The formula for $\nu$ is supposed to mean: take the one-dimensional Lebesgue measure of every horizontal section $E_y = \{x \in [0,1] \mid (x,y) \in E\}$ with $y \in [0,1]$. Take the sum $\nu(E) = \sum_{y \in [0,1]} m(E_y)$ (in other words I take $\nu(E) = \int \int \chi_E(x,y) \,dm(x)\,d\mu(y)$). For example $[0,1] \times \{0\}$ will have measure $1$. This is a measure defined on the Borel sets of the interval times the discrete interval and one can check that it has the property that $\nu(A \times B) = m(A) \mu(B)$. $\endgroup$ – Martin Apr 18 '13 at 19:16
  • $\begingroup$ @Martin: I figured out what you meant. Thanks. $\endgroup$ – copper.hat Apr 18 '13 at 19:18

Since $$ \mathcal{B}(\mathbb{R}^2)=\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})\subseteq \mathcal{B}(\mathbb{R})\otimes \mathcal{P} $$ it is enough to show that $D\in\mathcal{B}(\mathbb{R}^2)$. For example you could show that $D$ is closed in $\mathbb{R}^2$.

For the second part: Is $m\times \mu$ even a well-defined measure on $\mathcal{B}\otimes \mathcal{P}$?

The setup I'm used to is the following: Let $(X,\mathcal{E},\mu)$ and $(Y,\mathcal{F},\nu)$ be two $\sigma$-finite measure spaces. Then there exists a unique measure $\pi$ on $(X\times Y,\mathcal{E}\otimes\mathcal{F})$ such that $$ \pi(A\times B)=\mu(A)\nu(B),\quad A\in\mathcal{E},\,B\in\mathcal{F}. $$ The measure $\pi$ is called the product-measure and it is explicitly given by the formula $$ \pi(U)=\int_X\nu(U_x)\,\mu(\mathrm dx)=\int_Y\mu(U^y)\,\nu(\mathrm dy),\quad U\in\mathcal{E}\otimes\mathcal{F},\tag{1} $$ where $U_x=\{y\in Y\mid (x,y)\in U\}$ and $U^y=\{x\in X\mid (x,y)\in U\}$ are the sections.

Now, $([0,1],\mathcal{P},\mu)$ is clearly not $\sigma$-finite, so I doubt that we can talk about the product measure. My point is that we can't take $(1)$ as the definition of $m\times\mu$ because the two integrals aren't equal.


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