Measure defined on Borel sets that it is finite on compact sets

In summary, the problem statement introduces a measure ##\mu## defined on the Borel sets of ##\mathbb R^n## that is finite on compact sets. It then defines a class ##\mathcal H## of Borel sets satisfying certain conditions involving the measure ##\mu##. The task is to prove that open and compact sets belong to ##\mathcal H##, that ##\mathcal H## is a ##\sigma-##algebra if ##\mu## is finite, and that ##\mathcal H## coincides with the ##\sigma-##algebra of Borel. To prove these, one needs to use the properties of open and compact sets, as well as the
  • #1
mahler1
222
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The problem statement

Let ##\mu## be a measure defined on the Borel sets of ##\mathbb R^n## such that ##\mu## is finite on the compact sets. Let ##\mathcal H## be the class of Borel sets ##E## such that:

a)##\mu(E)=inf\{\mu(G), E \subset G\}##, where ##G## is open.

b)##\mu(E)=sup\{\mu(K), K \subset E\}##, where ##K## is compact.

Prove the following:

i. The open and compact sets belong to ##\mathcal H##.

ii. If ##\mu## is finite, ##\mathcal H## is a ##\sigma-##algebra.

iii. ##\mathcal H## coincides with the ##\sigma-##algebra of Borel.

The attempt at a solution

For i., maybe I could find an increasing sequence of compact sets ##\{K_n\}_{n \in \mathbb N}##(##K_n \subset K_{n+1}##) such that all are contained in ##E##, the problem is that I don't know how to construct this sequence; I suppose that, in an analogous way, I can construct a decreasing sequence of open sets ##\{G_n\}_{n \in \mathbb N}## (##G_{n+1} \subset G_n##) such that ##E## is contained in all of them.

For ii., it's easy to verify that ##\emptyset \in \mathcal H##, it remains to prove that if ##E \in \mathcal H##, then ##E^c \in \mathcal H##, and that if ##E_n \in \mathcal H## for a sequence of sets, then ##\bigcup_{n \in \mathbb N} E_n \in \mathcal H##. I couldn't prove that the complement of ##E## must be in ##\mathcal H##, I'll write what I did for countable unions:

Suppose ##E_n \in \mathcal H## for a sequence of sets, call ##E=\bigcup_{n \in \mathbb N} E_n##. By hypothesis, we have that ##\mu(E_n)## is finite for each ##n##. Given ##\epsilon>0##, we can choose for each ##n##, an open set ##G_n## : ##\mu(G_n)\leq \mu(E_n)+\dfrac{\epsilon}{2^n}##, if I call ##G=\bigcup_{n \in \mathbb N} G_n##, then ##E \subset G## and ##\mu(G) \leq \mu(E)+ \epsilon##. This means that ##\mu(E)## satisfies a). Analogously, we can show that ##\mu(E)## satisfies b), from here it follows ##E \in \mathcal H##.

For iii., assuming I could prove i., I can say that ##B \subset \mathcal H## since the open sets are contained in ##\mathcal H##, it remains to show that ##\mathcal H \subset B##.

I am pretty stuck in all three items, I would appreciate some help with this exercise and if someone could tell me if what I did for countable unions in ii. is correct.
 
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  • #2
mahler1 said:
The problem statement

Let ##\mu## be a measure defined on the Borel sets of ##\mathbb R^n## such that ##\mu## is finite on the compact sets. Let ##\mathcal H## be the class of Borel sets ##E## such that:

a)##\mu(E)=inf\{\mu(G), E \subset G\}##, where ##G## is open.

b)##\mu(E)=sup\{\mu(K), K \subset E\}##, where ##K## is compact.

Prove the following:

i. The open and compact sets belong to ##\mathcal H##.

ii. If ##\mu## is finite, ##\mathcal H## is a ##\sigma-##algebra.

iii. ##\mathcal H## coincides with the ##\sigma-##algebra of Borel.

The attempt at a solution

For i., maybe I could find an increasing sequence of compact sets ##\{K_n\}_{n \in \mathbb N}##(##K_n \subset K_{n+1}##) such that all are contained in ##E##, the problem is that I don't know how to construct this sequence; I suppose that, in an analogous way, I can construct a decreasing sequence of open sets ##\{G_n\}_{n \in \mathbb N}## (##G_{n+1} \subset G_n##) such that ##E## is contained in all of them.

You need to use that ##E## is actually compact (or open). So take ##E## compact. Then (b) shouldn't be too difficult. For ##A##, perhaps you should think of

[tex]\{x\in \mathbb{R}^n~\vert~d(x,E)<1/n\}[/tex]

For ii., it's easy to verify that ##\emptyset \in \mathcal H##, it remains to prove that if ##E \in \mathcal H##, then ##E^c \in \mathcal H##, and that if ##E_n \in \mathcal H## for a sequence of sets, then ##\bigcup_{n \in \mathbb N} E_n \in \mathcal H##. I couldn't prove that the complement of ##E## must be in ##\mathcal H##,

Take an ##E\in \mathcal{H}##. Let's prove ##(a)## for ##E^c##. You must find an open set ##G## such that ##E^c\subseteq G## and such that ##\mu(G) - \varepsilon<\mu(E^c)##. Take complements, then you need to find a certain closed subset of ##E##. Use that ##E\in \mathcal{H}## to find this.

I'll write what I did for countable unions:

Suppose ##E_n \in \mathcal H## for a sequence of sets, call ##E=\bigcup_{n \in \mathbb N} E_n##. By hypothesis, we have that ##\mu(E_n)## is finite for each ##n##.

I don't see why ##\mu(E_n)## is finite.

Given ##\epsilon>0##, we can choose for each ##n##, an open set ##G_n## : ##\mu(G_n)\leq \mu(E_n)+\dfrac{\epsilon}{2^n}##, if I call ##G=\bigcup_{n \in \mathbb N} G_n##, then ##E \subset G## and ##\mu(G) \leq \mu(E)+ \epsilon##.

Not sure if this is easy to see. We don't have necessarily that the ##G_n## and ##E_n## are pairswise disjoint. So you might not be able to use ##\sigma##-additivity.

For iii., assuming I could prove i., I can say that ##B \subset \mathcal H## since the open sets are contained in ##\mathcal H##, it remains to show that ##\mathcal H \subset B##.

But ##\mu## is a Borel measure, so it's only defined on the Borel sets.
 
  • #3
mahler1 said:
The problem statement

Let ##\mu## be a measure defined on the Borel sets of ##\mathbb R^n## such that ##\mu## is finite on the compact sets. Let ##\mathcal H## be the class of Borel sets ##E## such that:

a)##\mu(E)=inf\{\mu(G), E \subset G\}##, where ##G## is open.

b)##\mu(E)=sup\{\mu(K), K \subset E\}##, where ##K## is compact.

Prove the following:

i. The open and compact sets belong to ##\mathcal H##.

ii. If ##\mu## is finite, ##\mathcal H## is a ##\sigma-##algebra.

iii. ##\mathcal H## coincides with the ##\sigma-##algebra of Borel.

The attempt at a solution

For i., maybe I could find an increasing sequence of compact sets ##\{K_n\}_{n \in \mathbb N}##(##K_n \subset K_{n+1}##) such that all are contained in ##E##, the problem is that I don't know how to construct this sequence; I suppose that, in an analogous way, I can construct a decreasing sequence of open sets ##\{G_n\}_{n \in \mathbb N}## (##G_{n+1} \subset G_n##) such that ##E## is contained in all of them.

For ii., it's easy to verify that ##\emptyset \in \mathcal H##, it remains to prove that if ##E \in \mathcal H##, then ##E^c \in \mathcal H##, and that if ##E_n \in \mathcal H## for a sequence of sets, then ##\bigcup_{n \in \mathbb N} E_n \in \mathcal H##. I couldn't prove that the complement of ##E## must be in ##\mathcal H##, I'll write what I did for countable unions:

Suppose ##E_n \in \mathcal H## for a sequence of sets, call ##E=\bigcup_{n \in \mathbb N} E_n##. By hypothesis, we have that ##\mu(E_n)## is finite for each ##n##. Given ##\epsilon>0##, we can choose for each ##n##, an open set ##G_n## : ##\mu(G_n)\leq \mu(E_n)+\dfrac{\epsilon}{2^n}##, if I call ##G=\bigcup_{n \in \mathbb N} G_n##, then ##E \subset G## and ##\mu(G) \leq \mu(E)+ \epsilon##. This means that ##\mu(E)## satisfies a). Analogously, we can show that ##\mu(E)## satisfies b), from here it follows ##E \in \mathcal H##.

For iii., assuming I could prove i., I can say that ##B \subset \mathcal H## since the open sets are contained in ##\mathcal H##, it remains to show that ##\mathcal H \subset B##.

I am pretty stuck in all three items, I would appreciate some help with this exercise and if someone could tell me if what I did for countable unions in ii. is correct.

I think you are trying to go too fast here. Let's just stick with i) for a while. You are dealing with ##\mathbb R^n##. You probably know a lot about ##\mathbb R^n##. You probably know Heine-Borel. Use that to attack i).
 

What is a measure defined on Borel sets?

A measure defined on Borel sets is a mathematical concept used to assign a numerical value to certain subsets of a given set, known as Borel sets. It is a way of quantifying the size or extent of these sets.

What does it mean for a measure to be finite on compact sets?

A measure being finite on compact sets means that the numerical value assigned by the measure to any compact set is a finite number. This is important because it allows for the measure to accurately quantify the size of these sets, which are often used in mathematical analysis.

Why is it important for a measure to be finite on compact sets?

In mathematical analysis, compact sets are often used to represent physical objects or systems. Having a measure that is finite on these sets allows for more accurate and meaningful calculations, as the measure can provide a numerical representation of the size or extent of these objects or systems.

How is a measure defined on Borel sets used in practical applications?

Measures defined on Borel sets have a wide range of practical applications, particularly in physics, engineering, and economics. They are used to quantify physical quantities such as length, volume, or mass, and to analyze and predict the behavior of complex systems.

What are some examples of measures defined on Borel sets?

Some common examples of measures defined on Borel sets include the Lebesgue measure, which is used to measure length, area, and volume in Euclidean space, and the Haar measure, which is used in probability theory and harmonic analysis. Other examples include the probability measure and the Hausdorff measure.

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