# Product Sigma Algebra (generated)

1. May 29, 2007

### MWG@berlin

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- S:{E in XxY; E=finite union of AixBi pairwise disjoint,Ai in A Bi in B}

The question is if we allow 'countably infinite union to replace finite union in the above Ring/Algebra, I conjecture (my presumptious inspection with the properties) that this set is already 'a sigma algebra' hence contains the smallest sigma algebra generated as in the literature. (Since S contains {E1xE2inXxY; E1 in A, E2 in B})

Consequence (BIG PROBLEM): As a result, we inevitably arrive at a very useful Theorem (which cannot be right):

let E be an element in the product sigma algebra (generated as in the literature), it follows E=countably union of AixBi such that Ai in A, Bi in B

Remarks:
Note that with the Theorem in the consequence, many related proofs can be further simplified. That is why I think I am missing something. I believe that when I release the restriction of "finite" union, it "still" cannot possibly be a sigma algebra.

Is S really a sigma algebra if the finite union is replaced by countably instead?

Last edited: May 29, 2007
2. May 29, 2007

### cliowa

One very important thing in a measure space is the measure, agreed? This measure then induces the sigma algebra of all measurable subsets of the space.

You first need to specify your new measure on $X\times Y$, then we can talk about sigma algebras.

Maybe I'm missing the point here, but wouldn't you agree that if any sigma algebra which contains S (defined as in the above) contains all countable unions of elements of S? Wouldn't this sigma algebra therefore contain also all sets of the form {E in XxY; E=countable union of AixBi, Ai in A, Bi in B}?

Have you had a look at "Evans, Gariepy: Measure and fine properties of functions"? They do a pretty good job at explaining all those things; at least that's the way I feel about the book.

3. May 29, 2007

### MWG@berlin

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 the product measure, we first establish the product sigma algebra, your answer is correct, smallest sigma algebra that generate S is the standard procedure the literatures (almost all including Rudin, Babarian, few others).

I also missed a point set AixBi in S is pairwise disjoint for all i.
Key is..if we let finite union become countable infinite union, it is indeed?a sigma algebra
(which I verify a few times..unless countable union of countable union is not countable)

Then the problem is....my theroem will be established every E in the product sigma algebra in any book can be represented by UAixBi where Ai in A, Bi in B
...this is a problem...this "new" thereom will siimply the proof of fubini's theorem and free us from the use of the "monotonic class".

Thats y this "new" thereom "oughted" to be wrong...but i cannot disproof it.

Sorry * "standard generated product sigma algebra will be a subset of S (countably union)
since S(countably union) contains S.

Last edited: May 29, 2007
4. May 30, 2007

### cliowa

I don't really see the problem with your theorem here.
Basically, what you're dealing with is a sigma algebra generated by some sets, let's call them Di (where i has as range some indexing set, not necessarily countable), and the collection of all those Di we'll call DD. Then you define the sigma algebra containing DD as the family of all sets in DD plus those you get when you take countable unions, intersections and complements. This clearly establishes all the properties of a sigma algebra. Now, does it really surprise you that any element in that sigma algebra can be expressed as a countable union of the "generating sets" Di?

I don't understand a word. Could you try to be more clear? Maybe just write everything down using set notation.

5. May 30, 2007

### MWG@berlin

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 theorem:

the mapping x to v(Ex) is A-measurable where Ex = x section of E
the mapping y to u(Ey) is B-measurable Ey = y section of E

Standard literatures use monotone class, which is a more complicated
indirect tools to proof the above two statements (full proof could be of 20 lines). But with the "new" theorem, the "new" proof of the 2 statements could be finished in 2 lines.

Because then the mapping x to v(Ex) can be explicity represented as:
Countable summation of 1[Ai]V(Bi) where 1[Ai] is the charateristics (01) function of Ai, it then follows that the mapping is measurable. The proof is finished without using "monotone class".

6. May 30, 2007

### cliowa

I'm not really sure what's going on here (haven't thought about it enough), but don't you ignore that the characteristic function of a countable union of some sets is NOT the countable summation of the characteristic function of the sets themselves?

Say some sets $A_i\times B_i$ and $A_j\times B_j$ overlap, aren't you then summing up too much?

7. May 30, 2007

### MWG@berlin

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 doesnt matter whether they overlap. it is still a well defined
measurable function "let x be in Ai^Aj, then the summand = 1[Ai](x)+ 1[Aj](x)=2
as for x in Ai\Aj the summand simply =1[Ai](x)=1
4th fact: AixBi and AjxBj are disjoint sets in the countable version of S.
implies {Ai} disjoints hence 1{UAi} = Summand of 1{Ai} which is a 1 0 function.

Last edited: May 30, 2007
8. May 30, 2007

### cliowa

I think you don't see what I'm trying to say. Let's take an example. Let X=Y=[0,1], take the Lebesgue measure and it's induced sigma algebras A=B. Then have a look at the product sigma algebra. Take for example the square Q:=[0,1]x[0,1]. Then clearly this can be expressed as $([0,3/4]\times [0,1])\cup ([1/4,1]\times [0,1])$, where those sets are in the sigma algebras. Now the value of v(Ex)=1 for all x, but your calculation would give a different answer, depending on the current the actual x.

Do you understand what I mean? How about the "facts" you posted? Still feeling good about all of them?

Last edited: May 30, 2007
9. May 30, 2007

### MWG@berlin

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?

Last edited: May 30, 2007
10. May 30, 2007

### cliowa

Would you not agree that one can write $Q=([0,3/4]\times [0,1])\cup ([1/4,1]\times [0,1])$? All those intervals are in the respective sigma algebras, therefore in the product sigma algebra.

I don't acknowledge the truth of fact 4! What am I supposed to get?

11. May 30, 2007

### MWG@berlin

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}.

p.s. sorry about the typo...I meant S...look at the 4th fact again

Last edited: May 30, 2007
12. May 30, 2007

### cliowa

Yes, I still claim that the countable union version gives rise to the same sigma algebra. I'm not so sure about the pairwise disjoint thing, but wouldn't you agree on this: If I first do not allow for countable, but only for finite unions of sets Ai, Bi and then as the second step look at the smallest sigma algebra containing those finite unions, I still get all countable unions, precisely because I create a sigma algebra. It doesn't matter whether I take finite unions then allow them to be countably infinite, or whether I allow them to be infinite right from the start. The only difference is that when I compare the resulting product sigma algebra with the initial family of sets S (with finite unions) it contains far more sets then the family of sets S (with finite unions), whereas if I allow infinite unions I get the whole sigma algebra.

Actually, for the definition of the product sigma it suffices to require this: The product sigma algebra is the smallest sigma algebra containing every set of the form $R\times T$, where $R\in A, T \in B$. Agreed?

Now, in my last post I was trying to give an example for the fact that your way of writing the function $x \mapsto \nu(E_x)$ is flawed.

13. May 30, 2007

### MWG@berlin

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 sets
in the product sigma algebra, but if S is not a sigma algebra, then there exists set in the product sigma algebra which is not in S.

One way to see this is that the x----V(Ey) statement could be proven in 2 lines which I did it, if your claim (and what I am trying to as well) is rite.

I write x----V(Ey) as a "direct consequence of your claim. Set E in S means
E=UAixBi p.d. MEANING......A1xB1 ^ A2xB2 ^........are all empty...they are disjoint...thats y the 1 function has no problem it will only take 1 and 0
get it?

14. May 30, 2007

### cliowa

First point: In the end, you need to have a sigma algebra, the product sigma algebra. Which special set you choose to "enlarge" in order to get the sigma algebra doesn't really matter, as I pointed out.

But you're wrong on one important point: Not all the elements of the resulting sigma algebra are disjoint! Even if you take as a start the set S (as defined in your first post, finite unions, pairwise disjoint sets), when generating the sigma algebra you're allowed to take countable unions and countable intersections and complements. If there is any other set G in your sigma algebras besides the empty set or whole space $X\times Y$, then the sets in the sigma algebra cannot be disjoint, because this set G is not disjoint with the whole space: $(X\times Y) \cap G\neq\emptyset$!

Give me an example of any sigma algebra which is not trivial, where all sets in the sigma algebra are disjoint! You can't!

15. May 30, 2007

### MWG@berlin

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 explcitly defined in F the set. You are contradicting your first claim.

16. May 30, 2007

### cliowa

Where did I claim that? I did not!

This is not a valid criterion. Just because the set S consists of unions of p.d. AixBi does NOT mean the whole sigma algebra does! Do you understand my words?

17. May 30, 2007

### MWG@berlin

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. .This is how F is defined.

You dunt even need to generate anything since YOU claimed F is a product sigma algebra already.

DO YOU UNDERSTAND, FINALLY?

Last edited: May 30, 2007
18. May 31, 2007

### cliowa

Please, do yourself and me a favor and read precisely what I write! I said that the thing you call F gives rise, which means induces and not is, a sigma algebra, and I further claimed that the sigma algebra generated that way gives you the same sigma algebra as you would get if you were to start from the set S.

Obviously, you do not have that specific capacity. Maybe we should use some translation machine. Else I suggest you start again and read paying attention to the exact phrasing.

I keep repeating myself: Just because you use disjoint sets to generate the sigma algebra does NOT mean the sigma algebra will consist only of disjoint sets.

19. May 31, 2007

### MWG@berlin

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 whole thing if you generate F, every one knows the sun comes up from the east and sets at the west!!!

Did I every use the disjoint thing to generate F? I never say I generate F, THE QUESTION
always is F a sigma algebra???...I think you ve been missing out quite a lot. Dunt write again
if you dunt EVEN understand the question.

Last edited: May 31, 2007
20. May 31, 2007

### cliowa

Ok, I was only trying to interpret the somewhat diffuse question you have. Let's stop the accusations for a moment and get to the math.

First of all, if you start with two measurable sets X,Y 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 defined. In precisely what manner do you build up the AixBi, which are supposed to be pairwise disjoint? Say, for example, I start with A1=X and B1=Y. Then the only other thing I can generate, which is disjoint, is the empty set. Is this a sigma algebra? Yes, not a very interesting one, though. Or do you want the Ai to be pairwise disjoint?