properties for measure


Theorem [1, 2, 3, 4] Let (E,,μ) be a measure spaceMathworldPlanetmath, i.e., let E be a set, let be a σ-algebra of setsMathworldPlanetmath in E, and let μ be a measureMathworldPlanetmath on . Then the following properties hold:

  1. 1.

    Monotonicity: If A,B, and AB, then μ(A)μ(B).

  2. 2.

    If A,B in , AB, and μ(A)<, then

    μ(BA)=μ(B)-μ(A).
  3. 3.

    For any A,B in , we have

    μ(AB)+μ(AB)=μ(A)+μ(B).
  4. 4.

    Subadditivity: If {Ai}i=1 is a collectionMathworldPlanetmath of sets from , then

    μ(i=1Ai)i=1μ(Ai).
  5. 5.

    Continuity from below: If {Ai}i=1 is a collection of sets from such that AiAi+1 for all i, then

    μ(i=1Ai)=limiμ(Ai).
  6. 6.

    Continuity from above: If {Ai}i=1 is a collection of sets from such that μ(A1)<, and AiAi+1 for all i, then

    μ(i=1Ai)=limiμ(Ai).

Remarks In (2), the assumptionPlanetmathPlanetmath μ(A)< assures that the right hand side is always well defined, i.e., not of the form -. Without the assumption we can prove that μ(B)=μ(A)+μ(BA) (see below). In (3), it is tempting to move the term μ(AB) to the other side for aesthetic reasons. However, this is only possible if the term is finite.

Proof. For (1), suppose AB. We can then write B as the disjoint unionMathworldPlanetmathPlanetmath B=A(BA), whence

μ(B)=μ(A(BA))=μ(A)+μ(BA).

Since μ(BA)0, the claim follows. Property (2) follows from the above equation; since μ(A)<, we can subtract this quantity from both sides. For property (3), we can write AB=A(BA), whence

μ(AB) = μ(A)+μ(BA)
μ(A)+μ(B).

If μ(AB) is infiniteMathworldPlanetmathPlanetmath, the last inequalityMathworldPlanetmath must be equality, and either of μ(A) or μ(B) must be infinite. Together with (1), we obtain that if any of the quantities μ(A),μ(B),μ(AB) or μ(AB) is infinite, both sides in the equation are infinite and the claim holds. We can therefore without loss of generality assume that all quantities are finite. From AB=B(AB), we have

μ(AB)=μ(B)+μ(AB)

and thus

2μ(AB)=μ(A)+μ(B)+μ(AB)+μ(BA).

For the last two terms we have

μ(AB)+μ(BA) = μ((AB)(BA))
= μ((AB)(AB))
= μ(AB)-μ(AB),

where, in the second equality we have used properties for the symmetric set difference (http://planetmath.org/SymmetricDifference), and the last equality follows from property (2). This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof of property (3). For property (4), let us define the sequencePlanetmathPlanetmath {Di}i=1 as

D1=A1,Di=Aik=1i-1Ak.

Now DiDj= for i<j, so {Di} is a sequence of disjoint sets. Since i=1Di=i=1Ai, and since DiAi, we have

μ(i=1Ai) = μ(i=1Di)
= i=1μ(Di)
i=1μ(Ai),

and property (4) follows.

TODO: proofs for (5)-(6).

References

Title properties for measure
Canonical name PropertiesForMeasure
Date of creation 2013-03-22 13:45:28
Last modified on 2013-03-22 13:45:28
Owner matte (1858)
Last modified by matte (1858)
Numerical id 8
Author matte (1858)
Entry type Theorem
Classification msc 60A10
Classification msc 28A10