0
$\begingroup$

Using the framework of this question Is this a measure on the sigma algebra of countable and cocountable subsets of R?. Where $\Omega=\mathbb R$ and $A$ is the collection of subsets of $Ω$ such that A $∈ A$ if and only if either A is countable or $A^C$ is countable. We can define a Caratheodory outer measure $μ^*$ associated to μ by

$μ^*(E)=\inf${$\sum_{n=1}^∞μ(A_n):A_n∈ A, E \subset \cup_{n=1}^∞A_n$}, for all $E \subset Ω$.

I am having problems in showing that this $μ^*$ is not a content on $P(Ω)$, I have tried to show that $μ^*(E)=0$ if E is countable and equals 1 if $E^C$ is countable, but how can I prove it? And moreover, I have to show that $A$ is the collection of $μ^*$-measurable subsets of $Ω$. Does anyone know how to do it?

  • 0
    Your question is sloppy. 1) How is $A$ defined? 2) Don't you mean $E\subseteq\bigcup_{n=1}^{\infty}A_n$? 3) Don't you mean $E\subseteq\Omega$ (instead of $E\in\Omega$) 4) What do you mean by "..is not a content on..." (I don't know that expression, but that could be a lack of myself)? 5) Do we have $\Omega=\mathbb R$ here (as in the linked question)? – 2017-02-13
  • 0
    Yes I edited the question. And to prove that it is not a content I have to show that $μ(\cup_{i=1}^nA_i) = \sum_{i=1}^n(μ(A_i))$ is not satisfied. – 2017-02-13

2 Answers 2

1

Let it be that $\Omega$ is an uncountable set (e.g. $\Omega=\mathbb R)$.

I denote: $\mathcal A:=\{A\in\wp(\mathbb\Omega)\mid A\text{ is countable or is cocountable}\}$

Let it be that $E$ is not countable.

If $E\subseteq\bigcup_{n=1}^{\infty}A_n$ for $A_n\in\mathcal A$ then some integer $m$ must exist such that $A_m$ is not countable. Then $A_m$ must be cocountable so that $\sum_{n=1}^{\infty}\mu(A_n)\geq\mu(A_m)=1$. This observation justifies the conclusion that $\mu^*(E)\geq1$. We can take $A_1=\Omega$ and $A_n=\varnothing$ for $n\geq2$ and doing so we find that $\mu^*(E)\leq1+0+0+\cdots=1$. Proved is now that $\mu^*(E)=1$ whenever $E$ is not countable.

Let it be that $E$ is countable.

We have $E\subseteq\bigcup_{n=1}^{\infty}A_n$ for $A_1=E$ and $A_n=\varnothing$ for $n\geq2$, and if $E$ is countable then this is a union of sets in $\mathcal A$. This results in $\mu^*(E)\leq 0+0+\cdots=0$. From the definition of $E$ it follows that also $\mu^*(E)\geq0$ so we conclude that $\mu^*(E)=0$ whenever $E$ is countable.

If $E\notin\mathcal A$, i.e. $E$ is not countable and not cocountable then:$$\mu^*(\Omega)=1<1+1=\mu^*(E)+\mu^*(E^c)=\mu^*(\Omega\cap E)+\mu^*(\Omega\cap E^c)$$ This reveals that measurable sets are necessarily elements of $\mathcal A$, and - if I understand well - that $\mu^*$ is not a content on $\Omega$.

Let it be that $E$ is countable.

If $A$ is countable then:$$\mu^*(A)=0=0+0=\mu^*(A\cap E)+\mu^*(A\cap E^c)$$

If $A$ is not countable then:$$\mu^*(A)=1=0+1=\mu^*(A\cap E)+\mu^*(A\cap E^c)$$

Let it be that $E$ is cocountable.

If $A$ is countable then:$$\mu^*(A)=0=0+0=\mu^*(A\cap E)+\mu^*(A\cap E^c)$$

If $A$ is not countable then:$$\mu^*(A)=1=1+0=\mu^*(A\cap E)+\mu^*(A\cap E^c)$$

This because at least one of the sets $A\cap E$ and $A\cap E^c$ is not countable and $A\cap E^c$ is countable.

Proved is now that elements of $\mathcal A$ are measurable, and combined with the former conclusion that justifies the conclusion that $\mathcal A$ is exactly the collection of measurable sets.

  • 0
    Thank you very much for your answer. Sorry that I made a mistake: In order to prove that is not a content we need to show that $μ(\cup_{i=1}^nA_i) = \sum_{i=1}^n(μ(A_i))$ is not satisfied for some finitely pairwise disjoint sets $A_1, ..., A_n$ that $\in$ to the sigma-algebra $A$. A content is a ring that satisfies finite-additivity and not sigma-additivity. – 2017-02-13
  • 0
    That cannot be proved for $\mu$, since $\mu$ is a measure on $\mathcal A$. However it can be proved for $\mu^*$ on $\wp(\Omega)$. In my answer you find $\mu^*(\Omega)<\mu^*(E)+\mu^*(E^c)$ if $E\notin\mathcal A$. Note that $E$ and $E^c$ are disjoint and $E\cup E^c=\Omega$. Exactly what you need. – 2017-02-13
  • 0
    Yes you are right, I need to prove it for $\mu^*$ but $E\notin\mathcal A$, the disjoint sets you are using don't belong to $\mathcal A$ and they should in order to prove it. – 2017-02-13
  • 0
    Again that's not possible since $\mu^*$ restricted to $\mathcal A$ equals $\mu$ and is a measure on $\mathcal A$. – 2017-02-13
1

In this post, I will try to motivate why it's no surprise that in order to prove that $\mu^*$ is not a content on $\mathfrak p(\Omega)$ one has to consider sets $E \not \in \mathcal A$.

We have the following measure-theoretic result:

Let $\mu \colon \mathfrak h \to \overline{\mathbb{R}}$ be a $\sigma$-finite pre-measure on the semiring $\mathfrak h \subseteq \mathfrak p(\Omega)$ and $\mu^*$ the outer measure associated to $\mu$. Let $A_{\mu^*}$ denote the $\sigma$-algebra of $\mu^*$-measurable sets. Then $\mu^*_{|A_{\mu^*}}$ is the completion of $\mu_{|\sigma(\mathfrak{h})}$, where $\sigma(\mathfrak h)$ denotes the $\sigma$-algebra generated by $\mathfrak h$.

In our case

  • $\mu$ is not only a $\sigma$-finite pre-measure, it's a measure.
  • $\mathcal{A}$ is not only a semi-ring, it's a $\sigma$-algebra.
  • the measure space $(\Omega, \mathcal A, \mu)$ is already complete.

In this case, constructing the outer measure $\mu^*$ won't yield anything new. As @drhab already wrote: The set of $\mu^*$-measurable sets is precisely $\mathcal A$, i.e. in order to look for non-measurable sets, i.e. sets for which the content-property of finite additivity is violated, you have to consider sets $E \not \in \mathcal{A}$.