I would like to re-propose an issue that opened in another topic.
Let $\mathcal{F}$ an algebra of sets and let $\mu_{0}$ a finite and $\sigma$- additive measure.
We denote with $\mathcal{F_\sigma}$ the family all countable unions of sets of $\mathcal{F}$ and with $\mathcal{F_\delta}$ the family all countable intersection of sets of $\mathcal{F}$.
We extend $\mu_{0}$ to two family $\mathcal{F_\sigma}$ and $\mathcal{F}_{\delta}$ in the following way:
if $A\in\mathcal{F_{\sigma}}$ \begin{equation} \mu_{1}(A)=\sup\{\mu_{0}(A'), A'\subset A,\; A'\in\mathcal{F}\} \end{equation} else if $B\in\mathcal{F_\delta}$ \begin{equation} \mu_{2}(B)=\inf\{\mu_{0}(B'), B'\supset B,\; B'\in\mathcal{F}\}. \end{equation}
We know that a measure is monotone, but in this case $\mu_{1}$ and $\mu_{2}$ they are different, furthermore they are not really measures.
If $B\subset A$, I think that $\mu_2(B)<\mu_1(A)$, but my doubt resides here:
Let $B\subset A$, where $B\in\mathcal{F_\delta}$ and $A\in\mathcal{F_\sigma}$
if $A'$, $B'\in\mathcal{F}$ such that $B'\subset A'$ for all $A'\subset A$ and $B'\supset B$, then $\mu_0(B')<\mu_0(A')$, therefore \begin{equation} inf_{B'\supset B} \{\mu_0 (B')\}<sup_{A'\subset A} \{\mu_0 (A')\}, \end{equation} then,
\begin{equation} \mu_2(B)<\mu_1(A). \end{equation}
Now, from the definition of $\mu_2$, $B'$ must contain $B$, therefore $B'$ it could contain $A$, therefore $A'$, in this case exists a set of $\mathcal{F}$ for which $\mu_0(B')>\mu_0(A')$.
Same thing if $A'\subset B$.
So how do we prove that, if $B\subset A$, then $\mu_2(B)<\mu_1(A)$, without assuming that $B'\subset A'$ for all $B'\supset B$ and $A'\subset A$?
Thanks!
I assume the symbol $\subset$ allows set equality. (See note below.) So I will write $\subseteq$ to avoid confusing myself.
We are given: $\mu_0$ is finite and sigma-additive. So: if $A_n \in \mathcal F$, $A_1 \supseteq A_2 \supseteq \cdots$ and $\bigcap A_n = A \in \mathcal F$, then $\lim \mu_0(A_n) = \mu_0(A)$. And similarly for an increasing sequence.
Claim. Let $B \in \cal F_\delta$, $A \in \cal F_\sigma$, $B \subseteq A$. Then: $\mu_2(B) \le \mu_1(A)$.
Note: we cannot prove strict inequality $\mu_2(B) < \mu_1(A)$, contrary to the claim in the OP.
By definition of $\mu_1$, there is a sequence $A_n \in \cal F$, $A_n \subseteq A$ with $\lim \mu_0(A_n) = \mu_1(A)$. Since $A \in \cal F_\sigma$, we may assume $\bigcup A_n = A$. By taking finite unions, we may assume $A_1 \subseteq A_2 \subseteq \cdots$.
By definition of $\mu_2$, there is a sequence $B_n \in \cal F$, $B_n \supseteq B$ with $\lim \mu_0(B_n) = \mu_2(B)$. Since $B \in \cal F_\delta$, we may assume $\bigcap B_n = B$. By taking finite intersections, we may assume $B_1 \supseteq B_2 \supseteq \cdots$.
Then we have $B_n \setminus A_n \in \cal F$, $B_1 \setminus A_1 \supseteq B_2 \setminus A_2 \supseteq \cdots$, and $\bigcap (B_n\setminus A_n) = B \setminus A = \varnothing \in \cal F$. So $\lim \mu_0(B_n \setminus A_n) = 0$.
Compute $$ \mu_0(B_n) = \mu_0(B_n \cap A_n)+\mu_0(B_n \setminus A_n) \le \mu_0(A_n)+\mu_0(B_n\setminus A_n) . $$ Then let $n \to \infty$ to get $$ \mu_2(B) \le \mu_1(A)+0 = \mu_1(A), $$ as required.
note
This does not work if the symbol $\subset$ is strict inclusion in the definitions $\mu_1, \mu_2$.
Example. In set $\Omega = \{1,2,3,4,5\}$ let $\mathcal F$ be the power set, let $\mu_0$ be counting measure. Take $B=\{1,2\}$, $A=\{1,2,3\}$. Then:
Any $A' \subset A$ has at most two elements, so $\mu_1(A) = 2$.
Any $B' \supset B$ has at least $3$ elements, so $\mu_2(B) = 3$.
So we have $B \subset A$ but not $\mu_2(B) \le \mu_1(A)$.