Proposition. Let $X$ a set, $\mathcal{C}\subseteq 2^X$ an elementary family such that $\emptyset\in\mathcal{C}$ and $\nu$ a measure on $\mathcal{C}$. Then exists a unique measure $\mu$ on the minimal algebra $\mathcal{A}_0(\mathcal{C})$ such that $\mu_{|\mathcal{C}}=\nu.$
Proof. Denote with $\mathcal{U}_0(\mathcal{C})$ the family of the finite disjoint unions of $\mathcal{C}$ elements. I've already proved that $$\mathcal{U}_0(\mathcal{C})=\mathcal{A}_0(\mathcal{C}).$$ Hence for each $E\in\mathcal{A}_0(\mathcal{C})$ exists $E_1,\dots, E_n\in\mathcal{C}$ disjoint such that $E=\cup_{k=1}^n E_k.$
We define an application $\mu\colon\mathcal{A}_0(\mathcal{C})\to\overline{\mathbb{R}}_+$ in the following way $$\mu(E):=\sum_{k=1}^n\nu(E_k)\quad\text{for each}\quad E\in\mathcal{A}_0(\mathcal{C}).\tag1$$
Obviously $\mu_{|\mathcal{C}}=\nu.$
First step. Definition $(1)$ is well placed. The proof is clear.
Second step. $\big[$$\mu$ is a measure on $\mathcal{A}_0(\mathcal{C})$$\big]$.
Let $\{F_l\}\subseteq\mathcal{A}_0(\mathcal{C})$ a sequence of disjoint subsets such that $F:=\bigcup_{l=1}^{\infty} F_l\in\mathcal{A}_0(\mathcal{C}).$ We must prove that $$\mu(F)=\sum_{l=1}^{\infty} \mu(F_l).\tag2$$
Since $F\in\mathcal{A}_0(\mathcal{C})$, exists $B_1,\dots, B_m\in\mathcal{C}$ disjoint such that $F=\sum_{k=1}^{m} B_k.$ Therefore, $$B_k=B_k\cap F=\bigcup_{l=1}^\infty (B_k\cap F_l),\quad k=1,\dots, m.$$
Similarly, for each $F_l\in\mathcal{A}_0(\mathcal{C})$ exists a disjoint finite family $\{E_{jl}\}\subseteq \mathcal{C}\quad(j=1,\dots, n_l)$ such that $F_l=\bigcup_{j=1}^{n_l} E_{jl},\;l\in\mathbb{N}.$ Then
$$B_k=\bigcup_{l=1}^{\infty} (B_k\cap F_l)=\bigcup_{l=1}^{\infty}\bigcup_{j=1}^{n_l}(B_k\cap E_{jl}).$$ Moreover, $$F_l=F\cap F_l=\bigcup_{k=1}^m\bigcup_{j=1}^{n_l} (B_k\cap E_{jl}),$$ hence $$\mu(F_l)=\sum_{k=1}^m\sum_{j=1}^{n_l}\nu(B_k\cap E_{jl}),$$ for $(1).$
Now, $$\nu(B_k)=\nu\bigg(\bigcup_{l=1}^{\infty}\bigcup_{j=1}^{n_l}(B_k\cap E_{jl})\bigg)\color{RED}{=} \sum_{l=1}^{\infty}\sum_{j=1}^{n_l}\nu(B_k\cap E_{jl})$$
Question Is the equality in red true?
I know that $B_k\in\mathcal{C}$ and $\nu$ is a measure on it, therefore $\nu$ is $\sigma$– additive but I'm used to having only one union in the argument of $\nu$, in this case how does it work?
How I feel about it
I think that equality is correct, because fix $l=1, l=2$ and consider $A_{n_1}:=\bigcup_{j=1}^{n_1} (B_k\cap E_{j1})$ and $A_{n_2}:=\bigcup_{j=1}^{n_2} (B_k\cap E_{j2})$, we must prove that $A_{n_1}\cap A_{n_2}=\emptyset$.
We know that $\{E_{j1}\}_{j=1}^{n_1}$ is disjoint and that $F_1=\cup_{j=1}^{n_1} E_{j1}$, moreover $\{E_{j2}\}_{j=1}^{n_2}$ is disjoint and that $F_2=\cup_{j=1}^{n_2} E_{j2}$. Therefore, since $F_1\cap F_2=\emptyset$, we have that $$\bigg[\bigcup_{j=1}^{n_1} (B_k\cap E_{j1})\bigg]\cap\bigg[ \bigcup_{j=1}^{n_2} (B_k\cap E_{j2})\bigg]=\emptyset,$$ hence
$$\nu\bigg(\bigcup_{l=1}^{\infty}\bigcup_{j=1}^{n_l}(B_k\cap E_{jl})\bigg)=\sum_{l=1}^{\infty}\nu\bigg(\bigcup_{j=1}^{n_l} (B_k\cap E_{jl})\bigg).$$
At this point we can use the finite additivity of $\nu$, since the sequence $\{E_{jl}\}_{j=1}^{n_l}$ for each fixed $l\in\mathbb{N}$ is disjoint. Thus obtaining the thesis.
It's correct?
Thanks!
Best Answer
Yes this is correct. An alternative way to see it, although your method is perfectly fine I add it just for exposition, is to just verify that $$B_k\cap E_{jl}$$ are each disjoint. Indeed, the double union really is just a single countable union (the double union is a countable union of finite sets and so is countable) and thus you can just check disjointness of the individual terms rather than unions of the said terms. But then the disjointness is trivial, simply from the construction of the given sets.