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Topic Title: Sigma-field question - very elementary
Created On Fri Sep 02, 05 11:04 AM

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bhutes
Senior Member

Posts: 515
Joined: May 2005

Fri Sep 02, 05 11:04 AM

Let Omega = {1, 2, 3,4, 5, 6} (Omega is the sample space)
Let E = {2, 4, 6}
Let F = {1, 2, 3, 4}

I would like to find the sigma-algebra generated by E and F.

Do the following constitute all the elements of the sigma-field generated by E and F:
null set
E = {2, 4, 6}
E? = {1, 3, 5}
F={1,2,3,4}
F? ={5,6}
EF = {2, 4}
EF? = {6}
E?F = {1, 3}
E?F? = {5}
Omega

OR

Do the following elements also constitute the elements of the sigma-field generated by E and F?
(in addtion to the ones listed above):
{1,2,3,4,6} (generated by union of E and F)
{1,2,3,4,5} (generated by union of F and E'F')
{1,3,5,6} (generated by union of E and EF')
{2,4,5} (genearted by union of EF and E'F')
{1,3,6} (generated by union of E'F and EF')
{2,4,5,6} (generated by union of E and E'F')

-----------------------------------------------------------
In other words, sigma-field generated by E and F is
(i) all possible combinations (unions and intersections) of E and F and their complements only, or ....
(ii) we should consider the further union and intersections of the sets created in the initial intersections and unions of E, F, as well. (so to say, given any two sets in the sigma field, their union and intersection also must be included in the sigma-field)

Just let me know, if the answer is (i) or (ii)
(I know it's elementary .... but I need a firm confirmation from some who's absolutely sure.)

edit: I was nearly sure all along ... but the example on the first page of this article created doubts. that's why.

Edited: Fri Sep 02, 05 at 11:10 AM by bhutes

Posts: 107
Joined: Oct 2004

Fri Sep 02, 05 12:30 PM

The first set you give satisfies the conditions to be a sigma algelbra.
Also the second set satisfies the conditions but the first is included in the second (by construction) and hence it is smaller.
both contains the sets {E,F}.
so, the smallest sigma algebra is the first one.
Once you get a sigma algebra that contains your set, why to bother making it larger when you are looking for the smallest one?.

hope it helps

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bhutes
Senior Member

Posts: 515
Joined: May 2005

Fri Sep 02, 05 01:06 PM

Quote

null set
E = {2, 4, 6}
E? = {1, 3, 5}
F={1,2,3,4}
F? ={5,6}
EF = {2, 4}
EF? = {6}
E?F = {1, 3}
E?F? = {5}
Omega

But, I don't think this is a sigma-algebra at all... eg. {5} is an element but it's complement is not !

I was expecting someone to confirm to me that the pdf attachment is wrong .... and the answer should be (ii) (not (i) )

Am I missing something basic ... for example, check this link
The above set fails to satisfy the conditions to be a sigma algebra. Isn't it?

Posts: 107
Joined: Oct 2004

Fri Sep 02, 05 01:31 PM

You are absolutely right. (I didnt realized it wasn't yet a sigma algebra).

Let's do it well now. (H is the sigma algebra we are looking for)
we compute the intersection and all that as for (i)
({2,4,6},{1,3,5},{1,2,3,4},{5,6},{2,4},{6},{1,3},{5})
and then we complete this set to a sigma algebra using the definition.
(a) we add the missing complements ({1,5,6},{1,2,3,4,5},{1,2,4,5,6},{1,2,3,4,6})
(b) and missing unions ({1,3,5,6},...
(c) We check if it is sigma algebra and go (a) if not.

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bhutes
Senior Member

Posts: 515
Joined: May 2005

Mon Sep 19, 05 01:42 PM

Have yet another elementary question:

Question 1:
What are the constituents of the Borel algebra on the set of real numbers?

Answer: (my indicative attempts are in BOLD - please correct them, if they are wrong)
1. All intervals (a,b) where a,b belong to R eg. open interval (1,5) belongs to B
Yes?
2. All unions (or intersections) of intervals (a,b) where a,b belong to R eg. (1,5) U (7,10) belongs to B
No?-disjoint unions are not allowed.
3. Individual points on the real number line. {5} belongs to B ??
Yes?-but Borel measure would be zero for points.
4. Countably infinite sets belong to B?? eg. {2,4,6,8,10 ..... } belong to B ??
No?-disjoint unions not allowed ( ..... but all sigma algebra allow unions of all subsets. I'm confused !)

Just let me know Yes / No against each of the above four items.

----------------------------------
Question 2:
Is Borel algebra defined on n-dimensitonal spaces i.e. R^n ?
Yes?-should be. No reason to the contrary.
Is this a valid constituent of a Borel Algebra on R^3 : ( (1,5) , {5}, {2,4,6,8,10 ... } ) ?
No?-disallowed because the third dimension is disjoint.
A Borel measure must always be a subset of [0,infinity) ... for all Borel sets on all R^n (irrespective of whether n=1 or higher)
i.e. the measure has to be a single real value (in other words, a scalar. Measure can't be a vector).

Just a Yes / No against each is fine. (I'm embarassed by the low-end question ... but resort to this as last resort)

Posts: 107
Joined: Oct 2004

Mon Sep 19, 05 02:07 PM

Quote
Originally posted by: bhutes
Have yet another elementary question:

Question 1:
What are the constituents of the Borel algebra on the set of real numbers?

Answer: (my indicative attempts are in BOLD - please correct them, if they are wrong)
1. All intervals (a,b) where a,b belong to R eg. open interval (1,5) belongs to B
Yes?

Yes

Quote

2. All unions (or intersections) of intervals (a,b) where a,b belong to R eg. (1,5) U (7,10) belongs to B
No?-disjoint unions are not allowed.

Yes countable disjoint unions are allowed.

Quote

3. Individual points on the real number line. {5} belongs to B ??
Yes?-but Borel measure would be zero for points.

Yes

Quote

4. Countably infinite sets belong to B?? eg. {2,4,6,8,10 ..... } belong to B ??
No?-disjoint unions not allowed ( ..... but all sigma algebra allow unions of all subsets. I'm confused !)

Yes countable disjoint unions are allowed.

Quote

----------------------------------
Question 2:
Is Borel algebra defined on n-dimensitonal spaces i.e. R^n ?
Yes?-should be. No reason to the contrary.

Yes

Quote

Is this a valid constituent of a Borel Algebra on R^3 : ( (1,5) , {5}, {2,4,6,8,10 ... } ) ?
No?-disallowed because the third dimension is disjoint.

Yes, it is

Quote

A Borel measure must always be a subset of [0,infinity) ... for all Borel sets on all R^n (irrespective of whether n=1 or higher)
i.e. the measure has to be a single real value (in other words, a scalar. Measure can't be a vector).

Yes but a correction in the expression. A Borel measure is not a subset of [0,infinity) but a non negative real function (scalar) from a delta-ring.

Hope this helped.

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bhutes
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Posts: 515
Joined: May 2005

Mon Sep 19, 05 02:09 PM

Gerry
Junior Member

Posts: 6
Joined: Jul 2003

Tue Oct 04, 05 06:46 PM

For what its worth, its actually somewhat difficult to construct sets that aren't borel measurable (in the borel sigma algebra on the reals.) An example is if you break the reals into equivalence classes where two numbers are equivalent if they differ by a rational number. So all the rationals are in the [0] equivalence class, but there are lots of other equivalence classes. Now let the set A be one element from equivalent class (this requires the axiom of choice of possibily something a little weaker, I'm not a logician). A isn't measurable, and the reason is that the real numbers are a countable union of sets which look like A translated by a rational number. When you try and figure out what measure lebesgue assigns to this set you run into some problems. The point is, descriptive set theory is hard. There are Borel measurable sets, and analytic sets, and Lebesgue measurable sets, however, you are probably not going to construct a pathological example by accident. If you're construction doesn't involve the axiom of choice, its probably borel measurable.
Gerard

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bhutes
Senior Member

Posts: 515
Joined: May 2005

Mon Dec 05, 05 07:13 AM

I have another question in the same region of maths -

We know that all Euclidean spaces are Hilbert spaces (eg. R^2 fullfills all requirements of a 2-dimensional Hilbert Space (i.e. L^2) ).

What is the additional constraint which Euclidean Space has, but Hilbert space does not?

To make the question clearer: We have the definition of a vector space .... impose some additional constraints on it, we get to innner-product space ..... impose some additional constraints on inner-product space, we get the definition of a hilbert space. In that sequence, what is the flexibility that Hilbert space offers, but euclidean spaces do not offer.

I am not able to understand why hilbert spaces are used in so many models of the real world .... where my natural intuition would say, "euclidean spaces are a good-enough generalization" eg. probability spaces, spacetime in theory of relativity (physics) etc. etc.

Edited: Mon Dec 05, 05 at 09:25 AM by bhutes

Posts: 107
Joined: Oct 2004

Mon Dec 05, 05 11:08 AM

The main difference is that in infinity dimensions, you do not have a Euclidean space anymore, and what you need is its generalization, Hilbert spaces.

Iin finite dimensions, hilbert spaces dont add anything to the picture, in infinity spaces they are the way to work.

examples of infinity dimensions, :a space of functions, a espace of linear operators, etc.

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mkeller
Junior Member

Posts: 2
Joined: Sep 2005

Wed Dec 07, 05 08:53 AM

Hi bhutes, actually L^2 is not necessarily a 2-dim. space as you write. The 2 superscript denotes the norm/inner product that is to be used. E.g. L^2(R) is defined as all functions R -> R for which

and the inner product is

. This space L^2(R) is actually infinite-dimensional and I cannot think of a way to write is as a Euclidian space.

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bhutes
Senior Member

Posts: 515
Joined: May 2005

Wed Dec 07, 05 09:06 AM

Thanks for clarifying ... to be frank I have reading about all this here and there (but many terms go over my head).

Yeah, I came across a parallel description of hilbert space : The set of all measurable functions that are square-integrable.

The alternate one is (from what I understood): It's a complete vector space, with norm defined (i.e. inner product space) + it's complete with respect to the norm + all cauchy sequences converge.

Are these two descriptions of hilbert space equivalent?

(Also could you confirm that L^n space is also defined ... and would simply mean that the nth-integrable functions are finite in that space?)

Thanks.

edit: I was fine as far a vector space (which all hilbert spaces are) are defined in terms of constraints on the elements of the set (norm etc.)
But defining a space in terms of functions ??? (or constraints on them) ... is very uncomfortable and non-intuitive.

Edited: Wed Dec 07, 05 at 09:13 AM by bhutes

Posts: 107
Joined: Oct 2004

Wed Dec 07, 05 10:16 AM

Bhutes,

The spaces of square integrable functions is an EXAMPLE of hilbert space., and it shouldn'tbe taken as a definition. There are many other different examples.

p-integrable spaces also exists (L^p spaces) and are also different examples of banach spaces. (every hilbert space is a Banach space, but the inverse is not true)

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bhutes
Senior Member

Posts: 515
Joined: May 2005

Wed Dec 07, 05 10:35 AM

Quote
Originally posted by: vplanas
Bhutes,

The spaces of square integrable functions is an EXAMPLE of hilbert space., and it shouldn'tbe taken as a definition. There are many other different examples.

Oh, thanks !! This is very reassuring.

I am quite comfortable with the other definition (.. or as you say, the only definition).
Thanks a lot, Victor, for your help .... but I do hope to revisit this thread after reading up more.

Posts: 121
Joined: Dec 2005

Wed Dec 07, 05 04:36 PM

That is correct but you have to be careful because equality of elements in L^2 is equality "almost everywhere" and it not quite the same as regular equality. In Hilbert space, ||f||=0 <=> f=0. In L^2 ||f||=0 <=> f=0 almost everywhere.

Quote
Originally posted by: vplanas
Bhutes,

The spaces of square integrable functions is an EXAMPLE of hilbert space., and it shouldn'tbe taken as a definition. There are many other different examples.

p-integrable spaces also exists (L^p spaces) and are also different examples of banach spaces. (every hilbert space is a Banach space, but the inverse is not true)

Edited: Wed Dec 07, 05 at 04:37 PM by saimqn

Posts: 107
Joined: Oct 2004

Wed Dec 07, 05 08:40 PM

Yes and not, saigmqn

the spaces l2 (with lower case L) consisting on the square integrable functions is the one that we are descrbing. As you sais, there are functions that for which their distance (in the l2 norm) is zero.
This defines an equivalence relation in l2 (agan lower case L).
By taking the quotient of the space by this equivaence relation we obtain the space L2 (now upper case L), in which the norm is a proper one, that is, the distance of two "object" (they are not functions anymore but quotients) is zero if and only if the onject is the same.

however, it's always more intuitive to continue thinking that this objects ar ethe squeare integrable functions, even if we are just working with some representations of the quotient space.

i do not if i was very clear here, but it's late.

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