# Measure Theory Question

1. Aug 20, 2006

### Oxymoron

If I have a sigma-algebra, A, consisting of subsets of X where X = {1,2,3,4}, and I also have a measure on A such that

m({1,2}) = 1
m({1,2,3}) = 2
m({1,2,3,4}) = 3

Then my question is this:

Is the set E = {3} a member of the sigma-algebra?

I figured that since a subset E of X is in the s-algebra A if X\E is in A. Then since X\E = {1,2,4} and {1,2,4} is certainly a subset of X, E is in A? Is this correct?

2. Aug 20, 2006

### Hurkyl

Staff Emeritus
Why not just say {3} is a subset of X, and thus is in A?

Wait... when you say "A consists of subsets of X"... do you mean all subsets?

3. Aug 20, 2006

### Oxymoron

Hmmm...the question says "...A is a s-algebra of subsets of X", so I guess that could mean

1. "...A is a s-algebra of all subsets of X"
2. "...A is a s-algebra of any collection of, but not necessarily all subsets of X"

Im tempted to go with (1.) until I find out for certain. I hope it's not going to make much of a difference?

4. Aug 20, 2006

### matt grime

It means precisely that A is an s algebra of subsets. (Which is 2. in your list.) The problem is that your orginal post contains the pleonasm of 'sigma algebra... consisting of subsets of' (by definition a sigma algebra is made up of subsets) and leads to some slight ambiguity. You would have been better posting the question as it is stated directly. We do not know if {3} is in E since you have not said what E is.

5. Aug 20, 2006

### Oxymoron

Duly noted.

E is a set, it is the set containing the element 3: E = {3}. Also, Im asking whether E is in A, not whether {3} is in E.

6. Aug 20, 2006

### matt grime

My mistake. Let me correct it:

We do not know if {3} is in A since you have not said what A is.

7. Aug 21, 2006

### Oxymoron

All Im told is that A is a sigma-algebra of subsets of X = {1,2,3,4} and I have to work out if such sets as {3} or {3,4} or perhaps even {1,4} are in A, and additionally if I have enough information to compute the measure of each of these sets.

8. Aug 21, 2006

### matt grime

I said before that it helps to write out the question in full, verbatim, which you duly noted. I don't think that is a literal rendition of the text.

There are many sigma algebras on a given set, not least the trivial sigma algebra, just like there are many topologies. There is nowhere near enough information to answer the question.

In your first post you mentioned that the sets {1}, {1,2}, {1,2,3} were all measurable (information that hasn't appeared again), for instance, implying they are in A, and thus so are {4},{4,3},{4,3,2}, and hence {1,4}, {1,3,4}, {1,2,4} and all others that can be obtained by intersections, unions and complements.

Last edited: Aug 21, 2006
9. Aug 21, 2006

### Oxymoron

Full Question:

Suppose A is a sigma-algebra of subsets of X = {1,2,3,4} and m is a measure on A such that

m({1,2}) = 1, m({1,2,3}) = 2, m(X) = 3

Is the set E={3} in A and do you have enough information to computer m(E)?

Now A is the collection of all measurable subset of X. So we already know that {1,2},{1,2,3}, and {1,2,3,4} are in A. So X\{1,2} = {3,4} and X\{1,2,3} = {4}, X\{1,2,3,4} = {} are in A, and, as you said, so are all the other sets obtained by intersections, unions and complements.

Last edited: Aug 21, 2006
10. Aug 21, 2006

### matt grime

Well, I can see the answer to the first part (if E is in A) just by looking at the information there, indeed I stopped writing out sets that are in A exactly where I did and said the things about intersections etc for a good reason.... (remember you're supposed to do the legwork here.)

11. Aug 21, 2006

### Oxymoron

I'm not sure, but I don't think I said {1} was measurable. Perhaps you meant {1,2,3,4}?

12. Aug 21, 2006

### Oxymoron

...and I'm getting there!

Ok, let me get one thing straight: If I was not given the information about the measures (m({1,2})=1, etc...) then I would not have been able to decide whether or not {3} (or any set for that matter) is in the sigma algebra? Because from that information we now know which sets are measurable and from that we can work out which other sets are measurable and hence in A?

[So it is like topology, we cannot possibly know which sets are open unless we are given a specific topology]

If so, I do not think that {3} is in the sigma-algebra because I cannot construct the set {3} out of intersections, unions and complements of {1,2}, {1,2,3}, and X (which we know from the information given are in the sigma-algebra).

However, a set such as {4} would be because {4} is X\{1,2,3} (and {1,2,3} is in the s-algebra). Also, {3,4} is in because {3,4} is X\{1,2}.

The set {1,4} is not in the sigma algebra, and neither is {1,3} for the same reason.

EDIT: X\{1,2,3} = {4} and {4} U {1,2} = {1,2,4}. So since {4} in A and {1,2} in A, then is {4} U {1,2} = {1,2,4} in A? But then is would {3} be a measurable set since X\{1,2,4} = {3}?? Hmm, could I generate every subset of {1,2,3,4} in this way? Let me think...

Last edited: Aug 22, 2006
13. Aug 22, 2006

### Oxymoron

Is that "... and all others that can be obtained by intersections, unions and complements of {1,2}, {1,2,3}, {1,2,3,4} with themselves." or is that "... and all others that can be obtained by intersections, unions and complements of every possible subset of X.

Because if it is the latter, then I reckon I could generate the power set of X using as many intersections, unions, and complements as I wanted with {1,2}, {1,2,3}, X, with {3,4}, {4}, {}, and then Int., Uni, and Comp. again with those 6 sets, and then again, and again.

However, if it were the former, Then I can only generate 6 sets, namely:

{1,2}, {1,2,3}, {1,2,3,4}, {3,4}, {4}, {}

14. Aug 22, 2006

### Oxymoron

Would I be safe in concluding that {4} is in A because it is X\{1,2,3} and {1,2,3} is measurable with m({1,2,3}) = 2. Then

m({4}) = m(X\{1,2,3}) = m(X) - m({1,2,3}) = 3 - 2 = 1

15. Aug 22, 2006

### matt grime

Yes, obviously: those are the only ones we know for sure are measurable. But this of course is assuming that we are just saying A is some sigma algebra. Your book may have asserted something like 'through out this book any sigma algebra on a finite set is to be assumed to be the maximal sigma algebra' or something.

The sigma algebra generated by {1,2}, {1,2,3} does not just have 6 elements, or if they are they aren't the 6 you wrote. The things you wrote aren't a sigma algebra. They are not closed under intersections. (HINT {3} is the intersection of two of the sets you have written down!!!)

16. Aug 22, 2006

### Oxymoron

Yes! {3} = {1,2,3} n {3,4} and {1,2,3} is measurable and {3,4} is measurable because {3,4} = X\{1,2}, hence the intersection {3} is measurable. This is what I was asking! - could I use a measurable set {1,2}, take the complement with another measurable set X, and make the measurable set {3,4}, then take IT and intersect it with another measurable set {1,2,3} to make the measurable set {3}? Obviously I must be able to.

Then is X\{3} = {1,2,4} a measurable set? I think it is.

So I spent a few minutes working out all the measurable sets I could come up with:

{1,2}
{1,2,3}
X
{3,4} (=X\{1,2})
{4} (=X\{1,2,3})
{0} (=X\X)
{3} (={3,4} n {1,2,3} = {1,2,3}\{1,2})
{1,2,4} (=X\{3})

And there are no more.