maths.freeMeasure Theory › Uses › Product measure

Product measure

In mathematics, given two measurable spaces and measures on them, one can obtain a product measurable space and a product measure on that space.

Product measure

In mathematics, given two measurable spaces and measures on them, one can obtain a product measurable space and a product measure on that space. Conceptually, this is similar to defining the Cartesian product of sets and the product topology of two topological spaces, except that there can be many natural choices for the product measure.

Let \((X_1, \Sigma_1)\) and \((X_2, \Sigma_2)\) be two measurable spaces, that is, \(\Sigma_1\) and \(\Sigma_2\) are sigma algebras on \(X_1\) and \(X_2\) respectively, and let \(\mu_1\) and \(\mu_2\) be measures on these spaces. Denote by \(\Sigma_1 \otimes \Sigma_2\) the sigma algebra on the Cartesian product \(X_1 \times X_2\) generated by subsets of the form \(B_1 \times B_2\), where \(B_1 \in \Sigma_1\) and \(B_2 \in \Sigma_2\):

\(\Sigma_1 \otimes \Sigma_2 = \sigma\left( \lbrace B_1 \times B_2 \mid B_1 \in \Sigma_1, B_2 \in \Sigma_2 \rbrace \right)\)

This sigma algebra is called the product σ-algebra on the product space.

A product measure \(\mu_1 \times \mu_2\) (also denoted by \(\mu_1 \otimes \mu_2\) by many authors) is defined to be a measure on the measurable space \((X_1 \times X_2, \Sigma_1 \otimes \Sigma_2)\) satisfying the property

\((\mu_1 \times \mu_2)(B_1 \times B_2) = \mu_1(B_1) \mu_2(B_2) \qquad (B_1 \in \Sigma_1, B_2 \in \Sigma_2)\).

(In multiplying measures, some of which are infinite, we define the product to be zero if any factor is zero.)

In fact, when the spaces are \(\sigma\)-finite, the product measure is uniquely defined, and for every measurable set E,

\((\mu_1 \times \mu_2)(E) = \int_{X_2} \mu_1(E^y)\,d\mu_2(y) = \int_{X_1} \mu_2(E_{x})\,d\mu_1(x),\)

Condensed: the full section is in Wikipedia.

Examples

  • Given two measure spaces, there is always a unique maximal product measure μmax on their product, with the property that if μmax(A) is finite for some measurable set A, then μmax(A) = μ(A) for any product measure μ. In particular its value on any measurable set is at least that of any other product measure. This is the measure produced by the Carathéodory extension theorem.
  • Sometimes there is also a unique minimal product measure μmin, given by μmin(S) = supAS, μmax(A) finite μmax(A), where A and S are assumed to be measurable.
  • Here is an example where a product has more than one product measure. Take the product X×Y, where X is the unit interval with Lebesgue measure, and Y is the unit interval with counting measure and all sets are measurable. Then, for the minimal product measure the measure of a set is the sum of the measures of its horizontal sections, while for the maximal product measure a set has measure infinity unless it is contained in the union of a countable number of sets of the form A×B, where either A has Lebesgue measure 0 or B is a single point. (In this case the measure may be finite or infinite.) In particular, the diagonal has measure 0 for the minimal product measure and measure infinity for the maximal product measure.

İndi siz Heç bir hesablama proqramı bu ədədi həll edə bilməz, lakin onun hissələri hesablana bilər. Aşağıdakılardan birini sınayın ya da özünüz yazın.

Özün işlə

Pulsuz hesab hər dərsdə qeydlər əlavə edir, bitirdiyiniz işlərin qeydini saxlayır, həll etdiyiniz problemləri bir yerdə saxlayır və bu səhifə haqqında soruşa biləcəyiniz müəllimi də əlavə edir. Riyaziyyat özü daxil olub-olmadığınıza baxmayaraq hər kəsə açıqdır.

Qeyd Et Giriş

Burada istifadə edilən simvollar

Tam təsviri, şəkli və hər hərfin nə demək olduğunu görmək üçün hər hansı simvolu vurun.

İnsanlar soruşurlar

What is wrong with the Riemann integral?

It fails on functions that oscillate too much, and it does not interact well with limits: the limit of integrable functions need not be integrable. Lebesgue's integral fixes both.

Bu səhifənin bəzi hissələri Wikipedia (CC BY-SA 4.0). Bu, bizim üçün çox böyük bir uğurdur, biz bunu çox istəyirik.

Daha çox Measure Theory