Outer Lebesgue Measure limit's proof

In summary: In the infinite case, I think you should be able to come up with a series of sets to make the inequality ##\displaystyle \bigcup_{k=1}^{\infty} E_{k} \leq \lim_{k \to \infty} |E_{k}|_{e}## true.In summary, we need to prove that if \{ E_{k} \}_{k \in \mathbb{N}} is an increasing sequence of subsets of R^{p}, then the outer Lebesgue measure of their union is equal to the limit of the outer Lebesgue measure of each individual set. We have proven the inequality | \displaystyle \bigcup_{k=1}
  • #1
SqueeSpleen
141
5
If [itex]\{ E_{k} \}_{k \in \mathbb{N}}[/itex] is an increasing sequence of subsets of [itex]R^{p}[/itex], then:
[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} = \lim_{k \to \infty} |E_{k}|_{e}[/itex]
I proved:
[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} \geq \lim_{k \to \infty} |E_{k}|_{e}[/itex]
But I don't know how to prove the other inequality.

[itex]E_{n} = \displaystyle \bigcup_{k=1}^{n} E_{k} \subseteq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]
So, for monotony of the outer Lebesgue measure we have:
\newline
[itex]| E_{n} |_{e} = | \displaystyle \bigcup_{k=1}^{n} E_{k} |_{e} \leq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]
\newline
So the limit has also to be equal or lesser.
My problem is that I have not idea how to link the limit of the sets with the other limit in the other way, I could decompose:
[itex]\displaystyle \bigcup_{k=1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle \bigcup_{k=n+1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle (\bigcup_{k=n+1}^{\infty} E_{k} - \bigcup_{k=1}^{n} E_{k})[/itex] but the outer measure is subaditive, so the decomposition of the measure I could do to try to get something arbitrary small would be greater than my previous set and I have no guarantee it's close than the measure of the other set (vitali's could basically double the external measure I'm trying to delimite).
If they're measurables it's very easy, but in general I don't know how to prove it.
 
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  • #2
SqueeSpleen said:
If [itex]\{ E_{k} \}_{k \in \mathbb{N}}[/itex] is an increasing sequence of subsets of [itex]R^{p}[/itex], then:



[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} = \lim_{k \to \infty} |E_{k}|_{e}[/itex]
I proved:
[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} \geq \lim_{k \to \infty} |E_{k}|_{e}[/itex]
But I don't know how to prove the other inequality.

[itex]E_{n} = \displaystyle \bigcup_{k=1}^{n} E_{k} \subseteq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]
So, for monotony of the outer Lebesgue measure we have:
\newline
[itex]| E_{n} |_{e} = | \displaystyle \bigcup_{k=1}^{n} E_{k} |_{e} \leq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]
\newline
So the limit has also to be equal or lesser.
My problem is that I have not idea how to link the limit of the sets with the other limit in the other way, I could decompose:
[itex]\displaystyle \bigcup_{k=1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle \bigcup_{k=n+1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle (\bigcup_{k=n+1}^{\infty} E_{k} - \bigcup_{k=1}^{n} E_{k})[/itex] but the outer measure is subaditive, so the decomposition of the measure I could do to try to get something arbitrary small would be greater than my previous set and I have no guarantee it's close than the measure of the other set (vitali's could basically double the external measure I'm trying to delimite).
If they're measurables it's very easy, but in general I don't know how to prove it.

Well the first thing I would do is split the problem into two cases; one where ##|\bigcup_{k=1}^{\infty} E_{k} |_{e}## is finite, and one where it is not.

For the harder (finite) case, I think you might just need to roll up your sleeves, dust off the old ##\epsilon##s and ##\delta##s (or I guess in this case ##\epsilon##s and ##n##s), and play around in the muck for a bit. Maybe draw a picture/cartoon of a "nice-looking" finite case, come up with an heuristic argument you think works there, and then see if you can give that argument some rigor that makes it work in the general case.
 

Related to Outer Lebesgue Measure limit's proof

1. What is the Outer Lebesgue Measure limit's proof?

The Outer Lebesgue Measure limit's proof is a mathematical proof used to calculate the outer Lebesgue measure of a set, which is a measure of the size or volume of a set in n-dimensional space.

2. Why is the Outer Lebesgue Measure limit's proof important?

The Outer Lebesgue Measure limit's proof is important because it is used to define and calculate the Lebesgue measure, which is a more general and useful measure than the traditional notion of volume in Euclidean space. It is also used in various areas of mathematics, including analysis and probability theory.

3. How is the Outer Lebesgue Measure limit's proof different from other measures?

The Outer Lebesgue Measure limit's proof is different from other measures because it uses a different approach to defining and calculating the measure of a set. It takes into account the "outer" points of a set, rather than just the "inner" points, which allows for a more accurate and flexible measure.

4. What are the main steps in the Outer Lebesgue Measure limit's proof?

The main steps in the Outer Lebesgue Measure limit's proof include defining the outer measure, proving that it is a measure, and then using it to calculate the Lebesgue measure of a given set. This involves constructing a sequence of simple sets that approximate the given set, and then taking the limit of the measure of these sets.

5. Are there any applications of the Outer Lebesgue Measure limit's proof?

Yes, there are many applications of the Outer Lebesgue Measure limit's proof in various areas of mathematics, including analysis, topology, and probability theory. It is also used in the construction of fractals and in the study of geometric measure theory.

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