Infinite Union of Non-disjoint Sets

In summary, the conversation discusses using the definition of external measure to prove the inequality \mu(\bigcup^{\infty}_{k=1}A_{k})\leq \sum^{\infty}_{k=1}\mu(A_{k}) where μ is the Lebesgue measure and the A's are a countable set of Borel sets. The approach involves considering elements that are only present in one set and using induction to show that the sums of the measures of these elements are less than or equal to the infinite sum of the measures of all the sets.
  • #1
Yagoda
46
0

Homework Statement


To give some context, I'm trying to show that [itex]\mu(\bigcup^{\infty}_{k=1}A_{k})\leq \sum^{\infty}_{k=1}\mu(A_{k})[/itex] where μ is the Lebesgue measure and the A's are a countable set of Borel sets.

Since the A's may not be disjoint, I'm trying to rewrite the left side of the equation to somehow show that the elements that are present in more than one set are not being counted in the same way that they are on the right, which causes the left side to be smaller (or equal if all sets are disjoint).

Homework Equations





The Attempt at a Solution


In class we were given the hint to consider elements that are only in one set, present in 2 sets, in 3 sets, etc. and use this to rewrite the inequality.

The elements that are only in one set particular set [itex]A_{k} = A_{k} \setminus \bigcup^{k-1}_{i=1} A_{i}[/itex]. When I union all of these, it looks pretty messy.
I'm stuck on figuring out how to write elements that are present in more than one set. Does this approach look like it's on the right track?

(From what I've read it seems to common to build this up using the outer Lebesgue measure, but we haven't covered that. It seem like the way we're going about this is kind of unconventional)
 
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  • #2
use the definition of external measure as infimum.
 
  • #3
Yagoda said:

Homework Statement


To give some context, I'm trying to show that [itex]\mu(\bigcup^{\infty}_{k=1}A_{k})\leq \sum^{\infty}_{k=1}\mu(A_{k})[/itex] where μ is the Lebesgue measure and the A's are a countable set of Borel sets.

Since the A's may not be disjoint, I'm trying to rewrite the left side of the equation to somehow show that the elements that are present in more than one set are not being counted in the same way that they are on the right, which causes the left side to be smaller (or equal if all sets are disjoint).

Homework Equations





The Attempt at a Solution


In class we were given the hint to consider elements that are only in one set, present in 2 sets, in 3 sets, etc. and use this to rewrite the inequality.

The elements that are only in one set particular set [itex]A_{k} = A_{k} \setminus \bigcup^{k-1}_{i=1} A_{i}[/itex]. When I union all of these, it looks pretty messy.
I'm stuck on figuring out how to write elements that are present in more than one set. Does this approach look like it's on the right track?

(From what I've read it seems to common to build this up using the outer Lebesgue measure, but we haven't covered that. It seem like the way we're going about this is kind of unconventional)

Can you show [itex] \mu(A \cup B) \leq \mu(A) + \mu(B)? [/itex] If so, you can get by induction that [tex] V_n \equiv \mu \left( \cup_{i=1}^n A_i \right) \leq \sum_{i=1}^n \mu(A_i) \leq \sum_{i=1}^{\infty} \mu(A_i).[/tex] The numbers Vn are non-negative, increasing in n and are all <= the infinite sum of μ(Ai). What does that tell you?

RGV
 
  • #4
Yes, I think I can show that μ(A∪B)≤μ(A)+μ(B).

What is Vn, though?
 
  • #5
Yagoda said:
Yes, I think I can show that μ(A∪B)≤μ(A)+μ(B).

What is Vn, though?

I *defined* Vn to be [itex] \mu \left( \cup_{i=1}^n A_i \right),[/itex]; did you miss the [itex] \equiv [/itex] sign?

RGV
 

FAQ: Infinite Union of Non-disjoint Sets

What is an "Infinite Union of Non-disjoint Sets"?

An infinite union of non-disjoint sets refers to a mathematical concept where an infinite number of sets are combined together, and some of the sets may share elements with each other. This results in a larger set that includes all the elements from the individual sets.

How is an "Infinite Union of Non-disjoint Sets" different from a regular union of sets?

The main difference is that in a regular union, the sets being combined must be disjoint, meaning they have no elements in common. In an infinite union of non-disjoint sets, the sets can have overlapping elements.

What are some real-world applications of "Infinite Union of Non-disjoint Sets"?

This concept is commonly used in statistics and probability to calculate the likelihood of events happening. It is also used in computer science for data clustering and in linguistics for analyzing language patterns.

Can an infinite union of non-disjoint sets have a finite result?

Yes, it is possible for an infinite union of non-disjoint sets to result in a finite set. This can happen when the sets being combined have a finite number of elements, but there is still an infinite number of sets being combined.

Are there any limitations to using an infinite union of non-disjoint sets?

One limitation is that it may not be possible to accurately represent an infinite number of sets in real-world situations. Additionally, the calculations involved in this concept can become complex and computationally intensive for large sets of data.

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