Krdytkyu

Theorem. Let $mathcal{K}subseteq 2^{X}$ a set family such that $emptyset inmathcal{K}$ and be given an application $nucolonmathcal{K}to[0,+infty]$ such that $nu(emptyset)=0$, we define $$mu^*(E)=infleft{sum_{ninmathbb{N}}nu(I_n);middle|;Esubseteqbigcup_{ninmathbb{N}}I_n;text{and};{I_n}_{ninmathbb{N}}subseteqmathcal{K}right},$$
then $mu^*$ is an outer measure.

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

Be fixed $varepsilon> 0$. For what has been said above and for the properties of the infimum, for all $ninmathbb{N}$ exists ${I_{n,k}}subseteqmathcal{K}$ such that $$E_nsubseteqbigcup_{kinmathbb{N}}I_{n,k}quadtext{and}quadmu^*(E_n)+frac{varepsilon}{2^n}>sum_{kinmathbb{N}}nu(I_{n,k}).$$
We observe that $$bigcup_{ninmathbb{N}}E_nsubseteqbigcup_{ninmathbb{N}}bigcup_{kinmathbb{N}}I_{n,k}=bigcup_{(n,k)inmathbb{N}timesmathbb{N}}I_{n,k}quadtext{and}quad{I_{n,k}}_{n,kinmathbb{N}}subseteqmathcal{K}.$$ At this point, in order to apply the definition of $mu^*$, it is necessary to specify that it is equivalent to $$mu^*(E)=infleft{sum_{(n,k)inmathbb{N}timesmathbb{N}}nu(I_{n,k});middle|;Esubseteqbigcup_{(n,k)inmathbb{N}timesmathbb{N}}I_{n,k};text{and};{I_{n,k}}_{(n,k)inmathbb{N}timesmathbb{N}}subseteqmathcal{K}right}.$$ In general, what is important is that the set $E$ can be covered by a family ${I_gamma}_{gammainGamma}subseteqmathcal{K}$, where $Gamma$ is a countable set of indices. By definition of $mu^*$ we have
begin{equation}
mu^*bigg(bigcup_{ninmathbb{N}}E_nbigg)lesum_{(n,k)inmathbb{N}timesmathbb{N}}nu(I_{n,k})color{BLUE}{=}sum_{ninmathbb{N}}sum_{kinmathbb{N}}nu(I_{n,k})<sum_{ninmathbb{N}}bigg[nu(I_{n,k})+frac{varepsilon}{2^n}bigg]=sum_{ninmathbb{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 ${E_n}_{ninmathbb{N}}subseteq 2^{X}$ a countable set family. We suppose that $mu^*(E_n)<+infty$ for all $ninmathbb{N}$, that is for all $ninmathbb{N}$ exists ${I_{nk}}_{kinmathbb{N}}subseteq mathcal{K}$ such that $E_nsubseteqbigcup_{kinmathbb{N}}I_{nk}$.

Be fixed $varepsilon> 0$. For what has been said above and for the properties of the infimum, for all $ninmathbb{N}$ exists ${I_{nk}}subseteqmathcal{K}$ such that $$E_nsubseteqbigcup_{kinmathbb{N}}I_{nk}quadtext{and}quadmu^*(E_n)+frac{varepsilon}{2^n}>sum_{kinmathbb{N}}nu(I_{nk}).$$ Let $fcolonmathbb{N}tomathbb{N}timesmathbb{N}$ be a bijection and we place $I_{n,k}:=I_{nk}$ for all $(n,k)inmathbb{N}timesmathbb{N}$. We consider the sequence ${I_{f(r)}}_{rinmathbb{N}}$. We prove that ${I_{f(r)}}_{rinmathbb{N}}subseteqmathcal{K}$ and that is a covering of $bigcup_{ninmathbb{N}} E_n$.We observe that $$bigcup_{rinmathbb{N}}I_{f(r)}= bigcup_{(n,k)inmathbb{N}timesmathbb{N}}I_{nk}=bigcup_{ninmathbb{N}}bigg[bigcup_{kinmathbb{N}}I_{nk}bigg]supseteqbigcup_{ninmathbb{N}}E_n.$$ By definition of $mu^*$ we have $$mu^*bigg(bigcup_{ninmathbb{N}}E_nbigg)lesum_{rinmathbb{N}}nu(I_{f(r)}).$$ Now we place $$f(r):=(n_r,k_r)quadtext{and}quad K_r=max{k_1,dots, k_r},$$ 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})quadtext{We remember that}quad I_{nk}:=I_{n,k}\
color{RED}{le}& sum_{n=1}^{K_r}color{BLUE}{sum_{kinmathbb{N}}nu(I_{nk})}\
<&sum_{n=1}^{K_r}bigg[mu^*(E_n)+frac{varepsilon}{2^n}bigg]\
<&sum_{ninmathbb{N}}bigg[mu^*(E_n)+frac{varepsilon}{2^n}bigg]\
=&sum_{ninmathbb{N}}mu^*(E_n)+varepsilon.
end{split}
end{equation}
For $varepsilonto 0$ we have $$sum_{r=1}^s nu(I_{f(r)})lesum_{ninmathbb{N}}mu^*(E_n).$$

Question 1. Why is the inequality in red true?

$$$$

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

$$$$

Therefore $$sum_{rinmathbb{N}}nu(I_{f(r)}):=lim_{sto+infty}sum_{r=1}^s nu(I_{f(r)})color{BLUE}{le}sum_{ninmathbb{N}}mu^*(E_n)$$

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

Thanks for your patience!

Question 1. Why is the inequality in red true?

Because if $a_ngeq0$ for all $n $, then for any $k $ you have $a_kleqsum_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 $lim a_n=ainmathbb{R}$ and $lim b_n=binmathbb{R}$, then if $a_nle b_n$ for all $ninmathbb{N}$, then $a<b$. Is this the case? That is, the series $sum_{ninmathbb{N}}mu^*(E_n)$ is convergent?

Almost. If $a_nleq b_n $ for all $n $, then $aleq 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_rnu (I_{f (r)}=infty $.