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Theorem. Let $\mathcal{K}\subseteq 2^{X}$ a set family such that $\emptyset \in\mathcal{K}$ and be given an application $\nu\colon\mathcal{K}\to[0,+\infty]$ such that $\nu(\emptyset)=0$, we define $$\mu^*(E)=\inf\left\{\sum_{n\in\mathbb{N}}\nu(I_n)\;\middle|\;E\subseteq\bigcup_{n\in\mathbb{N}}I_n\;\text{and}\;\{I_n\}_{n\in\mathbb{N}}\subseteq\mathcal{K}\right\},$$ then $\mu^*$ is an outer measure.

Proof.(With Fubini's Theorem) We prove that $\mu^*$ is $\sigma$-subadditive, the other properties are trivial. Let $\{E_n\}_{n\in\mathbb{N}}\subseteq 2^{X}$ a countable set family. We suppose that $\mu^*(E_n)<+\infty$ for all $n\in\mathbb{N}$, that is for all $n\in\mathbb{N}$ exists $\{I_{n,k}\}_{k\in\mathbb{N}}\subseteq \mathcal{K}$ such that $E_n\subseteq\bigcup_{k\in\mathbb{N}}I_{n,k}$.

Be fixed $\varepsilon> 0$. For what has been said above and for the properties of the infimum, for all $n\in\mathbb{N}$ exists $\{I_{n,k}\}\subseteq\mathcal{K}$ such that $$E_n\subseteq\bigcup_{k\in\mathbb{N}}I_{n,k}\quad\text{and}\quad\mu^*(E_n)+\frac{\varepsilon}{2^n}>\sum_{k\in\mathbb{N}}\nu(I_{n,k}).$$ We observe that $$\bigcup_{n\in\mathbb{N}}E_n\subseteq\bigcup_{n\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}I_{n,k}=\bigcup_{(n,k)\in\mathbb{N}\times\mathbb{N}}I_{n,k}\quad\text{and}\quad\{I_{n,k}\}_{n,k\in\mathbb{N}}\subseteq\mathcal{K}.$$ At this point, in order to apply the definition of $\mu^*$, it is necessary to specify that it is equivalent to $$\mu^*(E)=\inf\left\{\sum_{(n,k)\in\mathbb{N}\times\mathbb{N}}\nu(I_{n,k})\;\middle|\;E\subseteq\bigcup_{(n,k)\in\mathbb{N}\times\mathbb{N}}I_{n,k}\;\text{and}\;\{I_{n,k}\}_{(n,k)\in\mathbb{N}\times\mathbb{N}}\subseteq\mathcal{K}\right\}.$$ In general, what is important is that the set $E$ can be covered by a family $\{I_\gamma\}_{\gamma\in\Gamma}\subseteq\mathcal{K}$, where $\Gamma$ is a countable set of indices. By definition of $\mu^*$ we have \begin{equation} \mu^*\bigg(\bigcup_{n\in\mathbb{N}}E_n\bigg)\le\sum_{(n,k)\in\mathbb{N}\times\mathbb{N}}\nu(I_{n,k})\color{BLUE}{=}\sum_{n\in\mathbb{N}}\sum_{k\in\mathbb{N}}\nu(I_{n,k})<\sum_{n\in\mathbb{N}}\bigg[\nu(I_{n,k})+\frac{\varepsilon}{2^n}\bigg]=\sum_{n\in\mathbb{N}}\mu^*(E_n)+\varepsilon, \end{equation} where the blue equal is the Fubini's Theorem.$\hspace{9cm}\square$

Proof.(Without Fubini's Theorem) Let $\{E_n\}_{n\in\mathbb{N}}\subseteq 2^{X}$ a countable set family. We suppose that $\mu^*(E_n)<+\infty$ for all $n\in\mathbb{N}$, that is for all $n\in\mathbb{N}$ exists $\{I_{nk}\}_{k\in\mathbb{N}}\subseteq \mathcal{K}$ such that $E_n\subseteq\bigcup_{k\in\mathbb{N}}I_{nk}$.

Be fixed $\varepsilon> 0$. For what has been said above and for the properties of the infimum, for all $n\in\mathbb{N}$ exists $\{I_{nk}\}\subseteq\mathcal{K}$ such that $$E_n\subseteq\bigcup_{k\in\mathbb{N}}I_{nk}\quad\text{and}\quad\mu^*(E_n)+\frac{\varepsilon}{2^n}>\sum_{k\in\mathbb{N}}\nu(I_{nk}).$$ Let $f\colon\mathbb{N}\to\mathbb{N}\times\mathbb{N}$ be a bijection and we place $I_{n,k}:=I_{nk}$ for all $(n,k)\in\mathbb{N}\times\mathbb{N}$. We consider the sequence $\{I_{f(r)}\}_{r\in\mathbb{N}}$. We prove that $\{I_{f(r)}\}_{r\in\mathbb{N}}\subseteq\mathcal{K}$ and that is a covering of $\bigcup_{n\in\mathbb{N}} E_n$.We observe that $$\bigcup_{r\in\mathbb{N}}I_{f(r)}= \bigcup_{(n,k)\in\mathbb{N}\times\mathbb{N}}I_{nk}=\bigcup_{n\in\mathbb{N}}\bigg[\bigcup_{k\in\mathbb{N}}I_{nk}\bigg]\supseteq\bigcup_{n\in\mathbb{N}}E_n.$$ By definition of $\mu^*$ we have $$\mu^*\bigg(\bigcup_{n\in\mathbb{N}}E_n\bigg)\le\sum_{r\in\mathbb{N}}\nu(I_{f(r)}).$$ Now we place $$f(r):=(n_r,k_r)\quad\text{and}\quad K_r=\max\{k_1,\dots, k_r\},$$ and we consider the partial sum s-th. Therefore \begin{equation} \begin{split} \sum_{r=1}^s \nu(I_{f(r)})=&\nu(I_{f(1)})+\cdots+\nu(I_{f(s)})\\ =&\nu(I_{n_1k_1})+\cdots+\nu(I_{n_sk_s})\quad\text{We remember that}\quad I_{nk}:=I_{n,k}\\ \color{RED}{\le}& \sum_{n=1}^{K_r}\color{BLUE}{\sum_{k\in\mathbb{N}}\nu(I_{nk})}\\ <&\sum_{n=1}^{K_r}\bigg[\mu^*(E_n)+\frac{\varepsilon}{2^n}\bigg]\\ <&\sum_{n\in\mathbb{N}}\bigg[\mu^*(E_n)+\frac{\varepsilon}{2^n}\bigg]\\ =&\sum_{n\in\mathbb{N}}\mu^*(E_n)+\varepsilon. \end{split} \end{equation} For $\varepsilon\to 0$ we have $$\sum_{r=1}^s \nu(I_{f(r)})\le\sum_{n\in\mathbb{N}}\mu^*(E_n).$$

Question 1. Why is the inequality in red true?

$$$$

Question 2. Could it be that some blue series is divergent?

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Therefore $$\sum_{r\in\mathbb{N}}\nu(I_{f(r)}):=\lim_{s\to+\infty}\sum_{r=1}^s \nu(I_{f(r)})\color{BLUE}{\le}\sum_{n\in\mathbb{N}}\mu^*(E_n)$$

Question 3. Why is the inequality in blue true? In general if $\{a_n\}$ and $\{b_n\}$ are two real number sequence such that $\lim a_n=a\in\mathbb{R}$ and $\lim b_n=b\in\mathbb{R}$, then if $a_n\le b_n$ for all $n\in\mathbb{N}$, then $a<b$. Is this the case? That is, the series $\sum_{n\in\mathbb{N}}\mu^*(E_n)$ is convergent?

Thanks for your patience!

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Question 1. Why is the inequality in red true?

Because if $a_n\geq0$ for all $n $, then for any $k $ you have $a_k\leq\sum_na_n $.

Question 2. Could it be that some blue series is divergent?

No, you assumed that it was bounded above by $\mu^*(E_n)+\varepsilon/2^n $ and that $\mu^*(E_n)<\infty $.

Question 3. Why is the inequality in blue true? In general if $\{a_n\}$ and $\{b_n\}$ are two real number sequence such that $\lim a_n=a\in\mathbb{R}$ and $\lim b_n=b\in\mathbb{R}$, then if $a_n\le b_n$ for all $n\in\mathbb{N}$, then $a<b$. Is this the case? That is, the series $\sum_{n\in\mathbb{N}}\mu^*(E_n)$ is convergent?

Almost. If $a_n\leq b_n $ for all $n $, then $a\leq b $.

And no, bounding a series below tells you nothing about its convergence. And it could very well be the case that $\mu^*(\bigcup_nE_n)=\infty $ and $\sum_r\nu (I_{f (r)}=\infty $.

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  • $\begingroup$ Thanks for your answer. So, if we place $a_s:=\sum_{r=1}^s \nu(I_{f(r)})$, and we have already proved that $a_s\le\sum_{n\in\mathbb{N}}\mu^*(E_n):=\lim_{k\to+\infty}\sum_{n=1}^k\mu^*(E_k)$ for all $s\in\mathbb{N}$, then can we conclude that $\lim_{s\to+\infty} a_s\le\lim_{k\to+\infty}\sum_{n=1}^k\mu^*(E_k)$? Even if we do not know if the term $\lim_{k\to+\infty}\sum_{n=1}^k\mu^*(E_k)$ is finite? $\endgroup$ – Jack J. Dec 26 '18 at 15:04
  • $\begingroup$ I think that I have understand: With the notations of the above comment. The sequence $\{a_s\}_{s\in\mathbb{N}}$ is increasing, then $\lim_{s\to+\infty}a_s=\sup_s\{a_s\}$. Now, if the right term is finite we have that $a_s\le R$ for all $s\in\mathbb{N}$, then $\sup_s\{a_s\}\le R$, where $R\in\mathbb{R}$. If the right term is $+\infty$, with the order relation in $\bar{\mathbb{R}}_+$, we have that $a_s\le+\infty$ for all $s\in\mathbb{N}$, then $\sup_s\{a_s\}=+\infty.$ Correct? $\endgroup$ – Jack J. Dec 26 '18 at 15:29
  • $\begingroup$ Yes and yes, but you cannot conclude that $\sup\{a_s\}=\infty $. It may be finite, for all you know. $\endgroup$ – Martin Argerami Dec 26 '18 at 15:41
  • $\begingroup$ Sorry, obviously "$\le$". Thanks! $\endgroup$ – Jack J. Dec 26 '18 at 15:51

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