Recent content by MWG@berlin

If you still dutn understand what rudin means, here it is a text suitable for you, Chapter 6 product measure page 116 section34 Theorem 1 Measure and Integration Sterling K. Berberian 1966 second print. Borrow it and read before you write again. Thereom 1: S is a Ring (Algebra you just need...

The class contains all measurable rectangles is precisly S which is an algebra...rudin generate this set...Q=E in S... (class of all eleementary sets) exactly the same thing...elementary set means E=finite UAixBi meaning finite union of measurable rectangle.

CONTAINS EVERY MEASURABLE RECTANGLES MEANS CONTAINING S ALREADY LOOK AT WHAT IS Q=R1U...URm...om page 160... it means E=A1xBiU...UAmxBm where they are disjoint...RiintersectRj =empty Ri=AixBi, Rj=AjxBj. The class conatins all measurable retangles is precisely S. you do not understand this...

First of all, if you start with two measurable sets X,Y Ans1.[X and Y are raw sets. (X,A) and (X,B) are sigma finite measurable space] and two sigma algebras A, B respectively, and "define" S:={E in XxY; E=finite union of AixBi pairwise disjoint,Ai in A Bi in B}, this set is not well...

Please pay attention and run thru the thread again, My question is ALWAYS w/o generating S, is F being a countable version of S a sigma algebra. Then you answer something non relevant sayign that generating S is the same as generating F, who ask you to generate F? Thats no point for the...

You claimed set F is a product sigma algebra, dude! LOOK What you wrote: "Yes, I still claim that the countable union version gives rise to the same sigma algebra." WHO CAN UNDERSTAND WHAT YOU MEANT THEN? Let E be an element of the product sigma algebra F it follows E=UAixBi (AixBi)i p.d...

You are missing something: THINK it over before you write again. You claimed this set countable version of S...(lets call it F)={E in XxY;E=UAixBi, (AixBi)i p.d. Ai in A, Bi in B} IS A PRODUCT SIGMA ALGEBRA Look, is any set in this product sigma algebra expressible as p.d. set. This is...

1st point: "whereas if I allow infinite unions I get the whole sigma algebra" whether once the uncoutable union replace the finite union wil turn the set into a sigma algebra is my main question, which u said it is. After you replace S with a coutable version, it will in deed contain some...

I meant Set S We are talking about whether a countable union version of set S (as in the first post above) is a product sigma algebra, and you claimed that it is. Let me put it here again S={E in X x Y such that E = countable UAixBi AixBi pairwise disjoint (p.d.) with Ai in A, Bi in B}...

Then clearly this can be expressed as ([0,3/4]\times [0,1])\cup ([1/4,1]\times [0,1])? They are not even disjoint. look at the 4th fact, and the set S with finite union replaced by countable union. Get it?

Thx trust this: 1st fact:Ai is in A 1[Ai] is always A measurable for all i 2nd fact: countable summation of 1[Ai] = limit of finite summand of 1[Ai] limit function of measurable functions is again measurable. 3rd fact: it doesn't matter whether they overlap. it is still a well defined...

Lets say the theorem has no problem, we then do not need to even consider "the monotone class" to proof many properties and theorem in product sigma algebra. For example, let E be an element in the product sigma algebra, u,v, the sigma finite measure in A and B respectively. then we have this...

Thx Cliowa: (the way you establish is the construction of an outer measure, but that is not neccessary, the way goes like this...we have a set X...we then find a "suitable" sigma algebra A then we define a measure on A. that is the triple (X,A,u) Let me clarify a bit, before we define...

http://integrals.wolfram.com/index.jsp

Dear mates, I run into a difficulty on the captioned area and am looking forward for enlightenments. Consider two measurable spaces (X,A) and (Y,B). The standard way to generate the product sigma algebra (XxY,AxB) is to consider the smallest sigma algebra that contains: Ring/Algebra-...