A $\sigma$-finite measure on a semialgebra Consider the following family of sets $$\mathcal{I}_0=\{(a,b]\;|\;-\infty\le a\le b< +\infty\}\cup\{(a,+\infty)\,|\; a\in\mathbb{R}\}.$$
I have shown that this family is a semialgebra. Let $F\in\mathcal{I}_0.$
The map $\lambda_0\colon \mathcal{I}_0\to[0,\infty]$ is defined as follows \begin{split}
\lambda_0(\emptyset):=&0\\
\lambda_0((a,b]):=&b-a\quad (-\infty\le a<b\le \infty)\\
\lambda_0((a,+\infty)):=&+\infty\quad a\in\mathbb{R}
\end{split}

Question 0. Since it is a semialgebra, how can I prove that $\lambda_0$ is additive and monotone?



Proposition. The application $\lambda_0$ is a $\sigma-$ finite measure on $\mathcal{I}_0$

Proof. Let a sequence $\{F_l\}\subseteq\mathcal{I}_0$ disjoint such that $F:=\bigcup_{l=1}^\infty F_l\in\mathcal{I}_0$ We prove that $$\lambda_0(F)=\sum_{l=1}^\infty\lambda_0(F_l).$$
It is not limitative to suppose $$\bigcup_{l=1}^n F_l\in\mathcal{I}_0\quad\text{for all}\quad n\in\mathbb{N}.$$

Question 1. Why?

Since $\bigcup_{l=1}^n F_l\subseteq F$ for all $n\in\mathbb{N}$ and $\lambda_0$ is additive and monotone we have: $$\lambda_0\big(\bigcup_{l=1}^n F_l\big)=\sum_{l=1}^n\lambda_0(F_l)\le \lambda_0(F)$$ for all $n\in\mathbb{N}$, then $$\sum_{l=1}^\infty\lambda_0(F_l)\le\lambda_0(F)$$
the second inequality is easy once the question below is solved.

I don't understand why the following statements:
Since $F\in\mathcal{I}_0$ exists a sequence $\{H_n\}\subseteq \mathcal{I}_0$ such that
$(a)\;$ The interval $H_n$ is bounded and its closure $\text{cl}(H_n)\subseteq F$ for all $n\in\mathbb{N}.$
$(b)\;$ $\lambda_0(H_n)\rightarrow\lambda_0(F)$ for $n\rightarrow+\infty$

My attempt.
Since $F\in\mathcal{I}_0$, then $F$ can be of two types.
If $F=(a,b]$, we can consider $H_n=\big(a+\frac{1}{n}, b-\frac{1}{n}\big]$, then
$$cl(H_n)\subseteq F,$$ and $$\lambda_0(H_n)=b-\frac{2}{n}-a\to\lambda_0(F)\quad\text{for}\quad n\to\infty.$$ If $F=(a,\infty)$, we can consider $H_n=(a-n,a]$, then $$cl(H_n)\subseteq F,$$ and $$\lambda_0(H_n)=n\to\infty=\lambda_0(F)\quad\text{for}\quad n\to\infty.$$
And therefore we can conclude that since $F\in\mathcal{I}_0$, exists a sequence $\{H_n\}\subseteq\mathcal{I}_0$ with the aforementioned properties.

Question it's correct?

Thanks!
 A: Question 0.
Monotone: Suppose $F_1 \subseteq F_2$ for $F_1,F_2 \in \mathcal{I}_0$. We wish to show $\lambda_0(F_1) \le \lambda_0(F_2)$. This is trivial if $F_2 = (a,+\infty)$ for some $a \in \mathbb{R}$, so suppose otherwise: $F_2 = (a,b]$ for some $-\infty \le a \le b <+\infty$. Since $F_1 \subseteq F_2$, $F_1 = (c,d]$ for some $c \ge a$ and $d \le b$. But then $\lambda_0(F_1) = d-c \le b-a = \lambda_0(F_2)$.
Additive: Suppose $F_1,F_2 \in \mathcal{I}_0$ with $F_1\cup F_2 \in \mathcal{I}_0$ and $F_1 \cap F_2 = \emptyset$. We wish to show $\lambda_0(F_1\cup F_2) = \lambda_0(F_1)+\lambda_0(F_2)$. Since we have shown $\lambda_0$ is monotone, the equality is trivial if $F_1 = (a,+\infty)$ for some $a \in \mathbb{R}$ or $F_2 = (b,+\infty)$ for some $b \in \mathbb{R}$. So suppose $F_1 = (a,b]$ and $F_2 = (c,d]$ for some $-\infty \le a,b,c,d < + \infty$, $a \le b$, $c \le d$. Without loss of generality, say $a \le c$. Since $F_1 \cup F_2 \in \mathcal{I}_0$, it must be that $c \le b$. And since $F_1\cap F_2 = \emptyset$, it must be that $c=b$. Then $\lambda_0((a,b]\cup(c,d]) = \lambda_0((a,d]) = d-a = (d-c)+(b-a) = \lambda_0(F_2)+\lambda_0(F_1)$, as desired.
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Question 1. This is a good question. The answer to it is very much dependent on the $\mathcal{I}_0$ we are working with presently. The answer is that we may (re)order the $F_l$'s so that $\cup_{l=1}^n F_l \in \mathcal{I}_0$ for each $n$. More precisely, there are $(\tilde{F_l})_{l=1}^\infty$ such that $\{\tilde{F_l}\} = \{F_l\}$ and $\cup_{l=1}^n \tilde{F_l} \in \mathcal{I}_0$ for each $n \ge 1$ and Below is a proof.
Each $F_l$ is of the form $(a_l,b_l]$ or $(a_l,+\infty)$ and the union $\cup_{l=1}^\infty F_l$ is of the form $(a,b]$ or $(a,+\infty)$. Let's first suppose that $\cup_{l=1}^\infty F_l = (a,b]$. This means each $F_l = (a_l,b_l]$ for some $-\infty \le a_l \le b_l < +\infty$. Since $b \in \cup_{l=1}^\infty F_{l}$, $b \in F_{l_1}$ for some $l_1$. Let $\tilde{F_1} = F_{l_1}$. Note $\tilde{F_1} \in \mathcal{I}_0$. Now $(\cup_{l=1}^\infty F_l)\setminus \tilde{F_1} = (a,a_{l_1}]$, so there is some $l_2$ with $a_{l_1} \in F_{l_2}$. We must have $b_{l_2} = a_{l_1}$, so if we let $\tilde{F_2} = F_{l_2}$, we then have $(\cup_{l=1}^\infty F_l)\setminus(\tilde{F_1}\cup \tilde{F_2}) = (a,a_{l_2}]$. Note that $\cup_{l=1}^2 \tilde{F_l} = (a_{l_2},b_{l_1}] \in \mathcal{I}_0$. We can keep constructing $\tilde{F_l}$ in this manner. It is clear that we will have $\{\tilde{F_l}\} = \{F_l\}$ and $\cup_{l=1}^n \tilde{F_l} \in \mathcal{I}_0$ for each $n \ge 1$. I leave the case of $\cup_{l=1}^\infty F_l = (a,+\infty)$ to you.
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Final question.
The case of $F = (a,b]$ is correct (and well-written). The case of $F = (a,+\infty)$ is not correct. The $H$ you defined does not have $cl(H_n) \subseteq F$ since we don't even have $H_n \subseteq F$ since $a-\frac{n}{2} \not \in F$ (for example). You should instead look at, for example, $H_n = (a+\frac{1}{n},n]$.
