1
$\begingroup$

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 $0 \cdot \infty=0$)

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
3
$\begingroup$

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 $(m \times \mu) D = (m \times \mu)^* D =\infty$.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.