Sequence of measurable subsets of [0,1] (Lebesgue measure, Measurable)

ChemEng1
Messages
52
Reaction score
0

Homework Statement


Let \left\{E_{k}\right\}_{k\in N} be a sequence of measurable subsets of [0,1] satisfying m\left(E_{k}\right)=1. Then m\left(\bigcap^{\infty}_{k=1}E_{k}\right)=1.

Homework Equations


m denotes the Lebesgue measure.
"Measurable" is short for Lebesgue-measurable.

The Attempt at a Solution


My intuition was to construct 2 disjoint and uncountable subsets of an uncountable set that all have measure equal to 1.

After much struggling, I don't think this is possible. To be able to construct such a counterexample would mean there is a mapping of each subset to the natural numbers which would make them both countable.

Are there non-constructive approaches to building this counterexample? My attempts haven't yielded anything meaningful. Each example (rational and irrational, transcendental and algebraic) I've considered ends up with measure 1 and measure 0 subsets instead of both measure 1.

I am not sure if my counterexample building skills are lacking or I just need to adjust my intuition. Any help would be appreciated.
 
Physics news on Phys.org
ChemEng1 said:

Homework Statement


Let \left\{E_{k}\right\}_{k\in N} be a sequence of measurable subsets of [0,1] satisfying m\left(E_{k}\right)=1. Then m\left(\bigcap^{\infty}_{k=1}E_{k}\right)=1.

Homework Equations


m denotes the Lebesgue measure.
"Measurable" is short for Lebesgue-measurable.

The Attempt at a Solution


My intuition was to construct 2 disjoint and uncountable subsets of an uncountable set that all have measure equal to 1.

After much struggling, I don't think this is possible. To be able to construct such a counterexample would mean there is a mapping of each subset to the natural numbers which would make them both countable.

Are there non-constructive approaches to building this counterexample? My attempts haven't yielded anything meaningful. Each example (rational and irrational, transcendental and algebraic) I've considered ends up with measure 1 and measure 0 subsets instead of both measure 1.

I am not sure if my counterexample building skills are lacking or I just need to adjust my intuition. Any help would be appreciated.

What is the measure of the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##?
 
Ray Vickson said:
What is the measure of the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##?

If I can construct a counterexample of 2 disjoint and noncountable subsets on [0,1] of measure 1, then the measure of the complement of intersections would be 1. However, I have not been able to find such a counterexample. The answer based on what I have considered is 0.
 
ChemEng1 said:
If I can construct a counterexample of 2 disjoint and noncountable subsets on [0,1] of measure 1, then the measure of the complement of intersections would be 1. However, I have not been able to find such a counterexample. The answer based on what I have considered is 0.

So, you do not intend to answer my question. Fair enough, but that's it for me.
 
Ray Vickson said:
So, you do not intend to answer my question. Fair enough, but that's it for me.
I answered your question the best I could.

Your question is closely related to the question I am posing. If I could simply answer it, then I wouldn't've started the thread.
 
ChemEng1 said:
I answered your question the best I could.
Try again, but this time answer the question. Start by answering this question: What is the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##? When you have answered this, can you find an upper bound on the measure of that complement?
 
D H said:
Try again, but this time answer the question. Start by answering this question: What is the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##? When you have answered this, can you find an upper bound on the measure of that complement?

By DeMorgan, C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}

D H said:
When you have answered this, can you find an upper bound on the measure of that complement?
A lower bound immediately comes to mind. I'm still stewing on how to get an upper bound to pop out.

Lower Bound:
Consider: m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]. m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]= m\left[0,1\right]-m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\geq m\left[0,1\right]-\Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=1-0=1
 
You have a set which is a subset of [0,1] and has measure bigger than or equal to 1, and you're not sure what its measure is?
 
ChemEng1 said:
By DeMorgan, C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}


A lower bound immediately comes to mind. I'm still stewing on how to get an upper bound to pop out.

Lower Bound:
Consider: m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]. m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]= m\left[0,1\right]-m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\geq m\left[0,1\right]-\Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=1-0=1

In Step 1 you did not follow through all the way: if ##E = \cap_k E_k,## then
E^c = \bigcup_k E_k^c
 
  • #10
Ray Vickson said:
In Step 1 you did not follow through all the way: if ##E = \cap_k E_k,## then
E^c = \bigcup_k E_k^c
Yep. I messed that up too. One more time.

By DeMorgan, C\left(\cap_{k=1}^{\infty} E_k\right)=C\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]=\bigcup_{k=1}^{\infty}E^{c}_{k}

Consider: m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right). m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\leq \Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=0

Therefore, m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)=0\Rightarrow m\left(\cap_{k=1}^{\infty}E_{k}\right)=1
 
  • #11
ChemEng1 said:
Yep. I messed that up too. One more time.

By DeMorgan, C\left(\cap_{k=1}^{\infty} E_k\right)=C\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]=\bigcup_{k=1}^{\infty}E^{c}_{k}

Consider: m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right). m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\leq \Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=0

Therefore, m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)=0\Rightarrow m\left(\cap_{k=1}^{\infty}E_{k}\right)=1

Easier: from ##E = \cap_k E_k## we have ##E^c = \cup_k E_k^c## (no need for all those intermediate steps). Then 0 \leq m(E^c) \leq \sum_k m(E_k^c) = \sum_k 0 =0, where the ##`\leq '## is from some standard property or result, depending on exactly what approach was used in the course/notes/textbook.
 
Last edited:
  • #12
I like the sandwhich approach better than mine.

Thanks for the pointers. I really lost the forest from the trees on this problem.

I got it stuck in my head that perhaps the irrationals were "dense enough" to create 2 subsets with the same measure. It didn't dawn on me to use L.measure properties to disprove the existence of that set.
 
Back
Top