# Homework Help: Measure theory

1. Aug 15, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
Consider a measure f mapping from a family of sets A to [0,infinity]

Prove that f is countably additive on A.

3. The attempt at a solution
To show equality from an inequality we do ie.

a<=b>=a so a=b

I tried this strategy with measures and sets but couldn't see it through. I might need to construct another set but don't see what to construct.

2. Aug 15, 2007

### matt grime

Isn't this just the fact that a countable sum is the limit of its finite subsums (by definition)?

Or, in the thing you're doing with a<=b<=a (which is what you should have written, rather than writing b>=a twice), a is the infinite sum of the measures, and b is the measure of the infinite union of the sets. What you need to do is show b=>a-e for any e, which is just saying something about sums being limits of partial subsums.

Last edited: Aug 15, 2007
3. Aug 17, 2007

### pivoxa15

I prefer to show reverse inequality.

4. Aug 17, 2007

### matt grime

This would show the reverse inequality, if by that you mean it would show that a<=b<=a: one is clear b<=a, and the other would show that b>a-e for any e. thus b=>a.

5. Aug 17, 2007

### pivoxa15

subtractions are avoided when dealing with measures because the element of infinity is in the set which cannot be subtracted. So equivalently show b+e>a for any e? To show b>=a

6. Aug 17, 2007

### matt grime

Uh, what on earth does that mean? Just treat the case of a collection of sets having infinite measure separately if you must, but it is not important - the sequence either converges or diverges.

Look, I have a sequence of measurable (i.e. of finite measure) sets

S_1, S_2, S_3,...

And I know that m(S_1 u S_2 u S_3 .....) => m(S_1 u S_2 ... u S_n)

for any n. Now do you understand why this is just a statement about the sum of a series? I am *not* subtracting any measures, the -e is purely from what it means for an infinite sum to be the limit of its partial subsums.

OK, let's put it this way: if any of the S_i has infinite measure, so does the union and there is nothing to prove. So suppose each S_i has finite measure.

Let X_n be the union of S_1,..,S_n, and let X be the union of all the S_i

Then m(X_n)=m(S_1)+..+m(S_n) by hypothesis, and

What do we know? m(X) is less than or equal to the limit of m(X_n) by assumption, and m(x) is greater than or equal to m(X_n) for all n - it's just a question about sequences:

suppose that a_n tends to a and that b satisfies a<=b and b=>a_n for all n, then b=a (note this even includes the case when a is infinity.

Last edited: Aug 17, 2007
7. Aug 17, 2007

### pivoxa15

I am unsure about the statement 'b=>a_n for all n'. Do you mean b>a_n for any finite n?

I am a bit rusty with proof of infinite sequences and series.

8. Aug 18, 2007

### pivoxa15

I think I see where you are getting at. You have laid out most of the solution in post 6 and recquires one more step or so to complete it. All I have to prove is that m(x) is greater than or equal to m(X_n) for all n.

9. Aug 18, 2007

### pivoxa15

In fact to prove 'm(X) is greater than or equal to m(X_n) for all n.' All one needs is the monatonicity of measure since X_n is a subset of X for all n so m(X_n)<=m(X) for all n. => lim(n->infinity) m(X_n)<=m(X)

but m(X)<= lim(n->infinity) m(X_n) as given so m(X)=lim(n->infinity) m(X_n) = infinite series as each partial sum is a finite series. Take n-> infinity and you get an infinite series.

The assumption in this is that all sets S_i in X are disjoint and rightly so they should for equality. It seems like this is a nice problem to test your knowledge of infinite series and sequences.

10. Aug 18, 2007

### matt grime

I meant what I wrote. And how can there be anything other than a 'finite n' since n is an element of the natural numbers?

That is a bit worrying if you're doing measure theory.

11. Aug 19, 2007

### pivoxa15

This statement is correct?

'since X_n is a subset of X for all n so m(X_n)<=m(X) for all n implies lim(n->infinity) m(X_n)<=m(X)'

12. Aug 20, 2007

### pivoxa15

Matt, just an observation. I never recall you make a post telling me I am right. Is not replying your way of saying that there is nothing wrong with the post?

13. Aug 20, 2007

### matt grime

What do you want me to say?

"since X_n is a subset of X for all n so m(X_n)<=m(X) for all n"

is obviously true by the definition of measure.

m(X_n) is an increasing sequence of real numbers bounded above be m(X), so of course the limit of m(X_n) is less than m(X), again by the basic definitions of analysis.