The problems are from Lebesgue measure theory. Which are true and which are false?

1. E is measurable if and only if m*(A)>=m*(A int E)+m*(A int E^C) for any bounded set A in R.

2. Let v be a nonnegative, extended real-valued function on X. Also let countable additivity hold for any finite collection of disjoint sets in X. Then for any infinite sequence of sets, countable subadditivity holds.

3. E is measurable if and only if given e>0, there is a sequence {I_n} of disjoint intervals such that E is included in the union of the I_n's and m*(union of I_n\E's)<e.

Please explain the false ones if there are any with counter examples!

1. E is measurable if and only if m*(A)>=m*(A int E)+m*(A int E^C) for any bounded set A in R.

2. Let v be a nonnegative, extended real-valued function on X. Also let countable additivity hold for any finite collection of disjoint sets in X. Then for any infinite sequence of sets, countable subadditivity holds.

3. E is measurable if and only if given e>0, there is a sequence {I_n} of disjoint intervals such that E is included in the union of the I_n's and m*(union of I_n\E's)<e.

Please explain the false ones if there are any with counter examples!

2. What is X? If it is any old collection of subsets, you don't get countable subadditivity. Here's a really silly counterexample. Let X = {{},{1,2},{2,3},{1,2,3}} with v({})=0, v({1,2})=1, v({2,3})=1, v({1,2,3})=5. v is a measure on X. Countable additivity is trivially true since there are no two non-empty disjoint sets in X, but countable subadditivity fails: v({1,2}union{1,3}) = v({1,2,3}) = 5 > v({1,2}) + v({2,3}) = 2.

3. This is false for any set E of positive measure. Or did the question state, "E has measure zero if ..."?

3. This is false for any set E of positive measure. Or did the question state, "E has measure zero if ..."?

Oh!! That makes it clear for 2!! Thank you very much.

Why is 3 false? I can't understand your point...

Could you give some more details? (->) is true isn't it?

What happens to 1?

Thank you so much

Why is 3 false? I can't understand your point...

Could you give some more details? (->) is true isn't it?

What happens to 1?

Thank you so much

For 3, (<-) Let A be any set and given e>0 let I denote the union of the collection of disjoint intervals satisfying the condition.

Then m*(A INT E)<=m*(A INT I)

m*(A INT E^c)=m*(A INT E^c INT I)+m*(A INT E^c INT I^c)=m*(A INT E^c INT I)+m*(A INT I^c)

<e+m*(A INT I^c)

So m*(A INT E)+m*(A INT E^c)<=m*(A INT I)+m*(A INT I^c)+e=m*(A)+e?

Since e was arbitrary, m*(A INT E)+m*(A INT E^c)<=m*(A)??

Isn't this correct?

What am I missing??

Thanks again!!

Then m*(A INT E)<=m*(A INT I)

m*(A INT E^c)=m*(A INT E^c INT I)+m*(A INT E^c INT I^c)=m*(A INT E^c INT I)+m*(A INT I^c)

<e+m*(A INT I^c)

So m*(A INT E)+m*(A INT E^c)<=m*(A INT I)+m*(A INT I^c)+e=m*(A)+e?

Since e was arbitrary, m*(A INT E)+m*(A INT E^c)<=m*(A)??

Isn't this correct?

What am I missing??

Thanks again!!

For 3, E is certainly measurable if it can be covered by a countable union of disjoint intervals of arbitrarily small length (in fact, E would have measure zero). However, there are plenty of measurable sets that cannot be covered in this way. Just take the closed interval [0,1]. Whichever way you cover it with intervals, the intervals will always have combined length at least 1.

Owlpride, you missed "union of I_n\E's". cuhbl89, why are you sure that m*(A INT E^c)=m*(A INT E^c INT I)+m*(A INT E^c INT I^c) is true? I think that the additivity holds for measurable sets and you haven't proved that these are such.

About 2. -> I guess we are talking about sigma algebras. But even then, I'm confused why we use "countable additivity" for "finite collections" (maybe just additivity?). And even if we have sigma algebra with additivity on finite collections, I think that it's not true.

X = {1, 2, 3, 4.... }. We can take v({n,n+1...})=100+1/n and v({n})=1/n-1/(n+1). Then the additivity is obvious, but we still don't have subadditivity, since v({1,2,3...})=101>v(1)+v(2)+...=1.

I haven't studied real analysis and Lebesgue measure for many years and probably I'm talking very stupid things, though...

About 2. -> I guess we are talking about sigma algebras. But even then, I'm confused why we use "countable additivity" for "finite collections" (maybe just additivity?). And even if we have sigma algebra with additivity on finite collections, I think that it's not true.

X = {1, 2, 3, 4.... }. We can take v({n,n+1...})=100+1/n and v({n})=1/n-1/(n+1). Then the additivity is obvious, but we still don't have subadditivity, since v({1,2,3...})=101>v(1)+v(2)+...=1.

I haven't studied real analysis and Lebesgue measure for many years and probably I'm talking very stupid things, though...

@logaritym: I used the fact that intervals are measurable. So, a countable union of disjoint intervals I is measurable, isn't it??

Sorry for misusing "countable additivity". Just finite additivity was what I wanted to say. I didn't know the appropriate term

I think you got what I wanted to say now!! Thanks!!

@owlpride: Why are you talking about intervals with arbitrarily small length?

Also, we are taking E away from the union of the collection of intervals so we aren't concerned with the sum of the lengths of the intervals aren't we? Thanks!!

Do I have a mistake in the argument above?

Sorry for misusing "countable additivity". Just finite additivity was what I wanted to say. I didn't know the appropriate term

I think you got what I wanted to say now!! Thanks!!

@owlpride: Why are you talking about intervals with arbitrarily small length?

Also, we are taking E away from the union of the collection of intervals so we aren't concerned with the sum of the lengths of the intervals aren't we? Thanks!!

Do I have a mistake in the argument above?

I claim that item 3 is true:

Assume first that E is measurable. If E has finite measure, then for any e > 0, we can find a countable sequence of intervals In of length Ln covering E such that m*(Uln) - m*(E) < e by definition. Let O denote the union of the In's. By measurability of E, m*(O \ E) = m*(O) - m*(E) < e. Since any open subset of the reals is a countable union of disjoint intervals and O is open, one direction is done.

(If E has infinite measure, apply the above procedure to E intersect [-n,n] with e/2^n).

Now for the converse. For each integer k, there is an open set Ok (namely, a union of ln's) containing E such that m*(Ok \ E) < 1/k. Let A be the intersection of all the Ok; then it is clear that A contains E and m*(A \ E) < 1/k for all k, so m*(A \ E) = 0. Since a set of measure zero is measurable, we have A\E measurable, and A is clearly measurable (it's Borel, in fact). It follows that E = A \ (A\E) is measurable, completing the proof.

Interestingly, we didn't even use disjointness in the second part of the proof. This assertion can be made into something a bit more useful as well: for any measurable set, there's a finite disjoint collection of intervals such that the outer measure of the symmetric difference of the sets is small. One can use this to prove the density of step functions in L^p, among other things.

Hope this clarifies things!

Assume first that E is measurable. If E has finite measure, then for any e > 0, we can find a countable sequence of intervals In of length Ln covering E such that m*(Uln) - m*(E) < e by definition. Let O denote the union of the In's. By measurability of E, m*(O \ E) = m*(O) - m*(E) < e. Since any open subset of the reals is a countable union of disjoint intervals and O is open, one direction is done.

(If E has infinite measure, apply the above procedure to E intersect [-n,n] with e/2^n).

Now for the converse. For each integer k, there is an open set Ok (namely, a union of ln's) containing E such that m*(Ok \ E) < 1/k. Let A be the intersection of all the Ok; then it is clear that A contains E and m*(A \ E) < 1/k for all k, so m*(A \ E) = 0. Since a set of measure zero is measurable, we have A\E measurable, and A is clearly measurable (it's Borel, in fact). It follows that E = A \ (A\E) is measurable, completing the proof.

Interestingly, we didn't even use disjointness in the second part of the proof. This assertion can be made into something a bit more useful as well: for any measurable set, there's a finite disjoint collection of intervals such that the outer measure of the symmetric difference of the sets is small. One can use this to prove the density of step functions in L^p, among other things.

Hope this clarifies things!

Last edited by Flow on Wed Feb 16, 2011 6:42 pm, edited 1 time in total.

Sorry, I misread question 3. I read "m*(union of I_n's)<e" instead of "m*(union of I_n\E's)<e". My bad!

Question 1 is true. Its called Caratheodory's criteria for measurability. I've seen treatments where this is taken as the definition of measurable.

@Flow: I know that -> is easily proved using open sets but why do you use open sets for the opposite direction?

It just says disjoint intervals not open sets, doesn't it?

Your proof for <- seems to be the proof of the existence of a G_delta set containing E and differing from it by zero.

I think its not what the question is asking. What do you think? Thanks!!

@blerg: I think the definition of a measurable set due to Caratheodory says any sets not just bounded sets. Thanks!!

It just says disjoint intervals not open sets, doesn't it?

Your proof for <- seems to be the proof of the existence of a G_delta set containing E and differing from it by zero.

I think its not what the question is asking. What do you think? Thanks!!

@blerg: I think the definition of a measurable set due to Caratheodory says any sets not just bounded sets. Thanks!!

@cuhbl89-

You are right, did produce a G delta set "A" containing E, but the key idea was not that m*(A)=m*(E) (this is always possible), but that m*(A\E) was zero. The reason is that A is certainly measurable (being a Borel set), and A\E is certainly measurable (having outer measure zero), so since the measurable sets form a sigma algebra, E = A \ (A\E) must be measurable!

As I remarked, we do not even need to use the disjointness of the initial open intervals to create our G delta set. What we needed was that we could find collections of intervals covering E in the sense that m*(union of ints \ E) gets small, not just that m*(union of ints)-m*(E) gets small; this was how I used the hypothesis. The open sets I referred to were just the unions of these intervals. Hope this makes sense!

You are right, did produce a G delta set "A" containing E, but the key idea was not that m*(A)=m*(E) (this is always possible), but that m*(A\E) was zero. The reason is that A is certainly measurable (being a Borel set), and A\E is certainly measurable (having outer measure zero), so since the measurable sets form a sigma algebra, E = A \ (A\E) must be measurable!

As I remarked, we do not even need to use the disjointness of the initial open intervals to create our G delta set. What we needed was that we could find collections of intervals covering E in the sense that m*(union of ints \ E) gets small, not just that m*(union of ints)-m*(E) gets small; this was how I used the hypothesis. The open sets I referred to were just the unions of these intervals. Hope this makes sense!

For the first problem.

m*(A)<=m*(A int E)+m*(A int E^C), because every union of coverages of A int E and A int E^C also covers A.

So we have m*(A)=m*(A int E)+m*(A int E^C) for every bounded set A. Now we just have to prove that this is true for unbounded sets.

Let B - unbounded set. We split the real line into segments with length 1. Let's take the intervals Jk=(k,k+1) where k in Z. Let Bk = B int Jk. We have that m*(Bk)=m*(Bk int E)+m*(Bk int E^C) for every k, since Bk are bounded. Now we just have to prove that m*(C)=sum_{k in Z} m*(C int (k,k+1)). If we take a coverage of C, we can easily produce coverages of C int (k,k+1) just by intersecting. If we take coverages of C int (k,k+1), we can easily produce a coverage of C just by adding sufficiently small open intervals covering k in Z, such that their sum is < e for arbitrary e. We should also notice that in every case we have countable coverages.

Isn't this right?

Also, I still don't understand problem 2.

m*(A)<=m*(A int E)+m*(A int E^C), because every union of coverages of A int E and A int E^C also covers A.

So we have m*(A)=m*(A int E)+m*(A int E^C) for every bounded set A. Now we just have to prove that this is true for unbounded sets.

Let B - unbounded set. We split the real line into segments with length 1. Let's take the intervals Jk=(k,k+1) where k in Z. Let Bk = B int Jk. We have that m*(Bk)=m*(Bk int E)+m*(Bk int E^C) for every k, since Bk are bounded. Now we just have to prove that m*(C)=sum_{k in Z} m*(C int (k,k+1)). If we take a coverage of C, we can easily produce coverages of C int (k,k+1) just by intersecting. If we take coverages of C int (k,k+1), we can easily produce a coverage of C just by adding sufficiently small open intervals covering k in Z, such that their sum is < e for arbitrary e. We should also notice that in every case we have countable coverages.

Isn't this right?

Also, I still don't understand problem 2.

I would guess 2 means to say the following:

Let X be a σ-algebra. Let v be a function from X into {non-negative reals, positive infinity} such that for a sequence of any disjoint sets A_i in X, we have v(∪A_i) = Σv(A_i). Prove that for any sequence of sets B_i och X (not necessarily disjoint), v(∪B_i) ≤ Σv(B_i).

The proof could then go:

Suppose B_i ⊂ X. Let A_i = B_i \ {B_1 ∪ ... ∪ B_{i-1}}. Since B_i can be written as the disjoint union of A_i and B_i ∩ (B_1 ∪ ... ∪ B_{i-1}), we have

v(B_i) = v(A_i) + v(B_i ∩ (B_1 ∪ ... ∪ B_{i-1}))

and it follows that

v(B_i) ≥ v(A_i).

Now, the A_i are disjoint, and ∪A_i = ∪B_i, so

v(∪B_i) = v(∪A_i) = Σv(A_i) ≤ Σv(B_i)

as was to be proven.

(The proof goes through even if X is merely a ring, provided we take the countable (sub)additivity properties to only be required when the unions ∪A_i or ∪B_i are in X.)

Let X be a σ-algebra. Let v be a function from X into {non-negative reals, positive infinity} such that for a sequence of any disjoint sets A_i in X, we have v(∪A_i) = Σv(A_i). Prove that for any sequence of sets B_i och X (not necessarily disjoint), v(∪B_i) ≤ Σv(B_i).

The proof could then go:

Suppose B_i ⊂ X. Let A_i = B_i \ {B_1 ∪ ... ∪ B_{i-1}}. Since B_i can be written as the disjoint union of A_i and B_i ∩ (B_1 ∪ ... ∪ B_{i-1}), we have

v(B_i) = v(A_i) + v(B_i ∩ (B_1 ∪ ... ∪ B_{i-1}))

and it follows that

v(B_i) ≥ v(A_i).

Now, the A_i are disjoint, and ∪A_i = ∪B_i, so

v(∪B_i) = v(∪A_i) = Σv(A_i) ≤ Σv(B_i)

as was to be proven.

(The proof goes through even if X is merely a ring, provided we take the countable (sub)additivity properties to only be required when the unions ∪A_i or ∪B_i are in X.)

Haha, I see... The "bad word" is "finite", not "countable"

Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”

Users browsing this forum: cater99 and 4 guests