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

In summary: E^{c-1} = \cap_{k=1}^{\infty} E_k.In Step 1 you did not follow through all the way: if ##E = \cap_k E_k,## then E^c = \bigcup_k... E^{c-1} = \cap_{k=1}^{\infty} E_k.
  • #1
ChemEng1
52
0

Homework Statement


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

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.
 
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  • #2
ChemEng1 said:

Homework Statement


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

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##?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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, [itex]C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}[/itex]

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: [itex]m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right][/itex]. [itex]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[/itex]
 
  • #8
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?
 
  • #9
ChemEng1 said:
By DeMorgan, [itex]C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}[/itex]


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

Lower Bound:
Consider: [itex]m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right][/itex]. [itex]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[/itex]

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

By DeMorgan, [itex]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}[/itex]

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

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

By DeMorgan, [itex]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}[/itex]

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

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

Easier: from ##E = \cap_k E_k## we have ##E^c = \cup_k E_k^c## (no need for all those intermediate steps). Then [tex] 0 \leq m(E^c) \leq \sum_k m(E_k^c) = \sum_k 0 =0,[/tex] 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.
 

1. What is the Lebesgue measure?

The Lebesgue measure is a mathematical concept used to measure the size or volume of measurable subsets of a given space, typically a subset of the real numbers. It is a generalization of the more familiar concept of length, area, and volume in one, two, and three dimensions, respectively. It was introduced by French mathematician Henri Lebesgue in the early 20th century.

2. How is the Lebesgue measure different from the traditional notion of measure?

The traditional notion of measure, also known as the Riemann integral, relies on partitioning a space into smaller and smaller pieces and approximating the size of a subset by summing the sizes of these pieces. The Lebesgue measure, on the other hand, is defined in terms of a more abstract concept called a measure space, which allows for a more general and powerful approach to measuring subsets.

3. What are measurable subsets and why are they important in the Lebesgue measure?

A measurable subset is a subset of a space that can be assigned a numerical measure using the Lebesgue measure. In order to use the Lebesgue measure, a subset must be measurable. This is important because it allows for the calculation of the size or volume of subsets that may not have a traditional notion of measure, such as fractals or irregularly shaped objects.

4. How is the Lebesgue measure used in real-world applications?

The Lebesgue measure has a wide range of applications in fields such as physics, engineering, and economics. It is used to calculate probabilities in probability theory, to measure the length of curves in differential geometry, and to analyze the distribution of data in statistics, among others.

5. What are some of the key properties of the Lebesgue measure?

The Lebesgue measure has several important properties, including countable additivity, translation invariance, and monotonicity. Countable additivity means that the measure of the union of countably many disjoint subsets is equal to the sum of their individual measures. Translation invariance means that the measure of a subset is the same regardless of its position in the space. Monotonicity means that the measure of a subset is always greater than or equal to the measure of its subset. These properties make the Lebesgue measure a powerful and versatile tool in mathematical analysis.

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