Supremum with respect to the order of measures on $(X,A)$ Suppose that $(X,\leq )$ is an ordered set, we can define the maximum and the infimum of this set,now let $(X,A)$ be a measurable space and let $M(X,A)$ be the set of all measures on $(X,A)$, we now define an order $\leq$ by declaring that $\mu _{1}\leqslant \mu _{2}$ iff $\mu _{1}\left ( A_{0} \right )\leq \mu _{2}\left ( A_{0} \right )$ for each $A_{0}\in A$ my questions are as follow.
(a) Let $\phi \neq Y\subseteq M(X,A)$ and for each $B$ in $A$, define $$\mu \left ( B \right )=\sup\left \{ \sum_{n\in\mathbb{N} }^{}\varphi \left ( n \right )\left ( B_{n} \right ):\varphi :\mathbb{N}\rightarrow Y,B_{n}\mbox{ is a partition of }B \mbox{ with } B_{n}\in A\mbox{ for each }n\right \}$$My question is that how to show that $\mu$ is a measure and it is a supremum of $Y$.
(b) is it true that every $\phi \neq Y\subseteq M(X,A)$ admits a lower bound?
 A: $\newcommand\C{\mathscr C}\newcommand\ep{\varepsilon}\newcommand\A{\mathscr A}\newcommand\N{\mathbb N}$Yes, $\mu$ is a measure.
Indeed, let $\A:=A$. Say that a partition of a set $B\in\A$ is measurable if all its pieces are in $\A$.
For any $B\in\A$, let $P(B)$ denote the set of all countable measurable partitions of $B$. Then
$$\mu(B)=\mu_Y(B):=\sup\Big\{\sum_{n\in\N}f(n)(C_n)\colon f\colon\N\to Y, (C_n)_{n\in\N}\in P(B)\Big\}. \tag{0}$$
Take any pairwise disjoint $B_1,B_2,\dots$ in $\A$ and let $B:=B_1\cup B_2\cup\cdots$.
Take any $(C_n)_{n\in\N}\in P(B)$ and any $f\colon\N\to Y$. Then $(C_n\cap B_j)_{n\in\N}\in P(B_j)$ for each $j\in\N$. So,
$$\sum_{n\in\N}f(n)(C_n)=\sum_{n\in\N}\sum_{j\in\N}f(n)(C_n\cap B_j)
=\sum_{j\in\N}\sum_{n\in\N}f(n)(C_n\cap B_j)
\le\sum_{j\in\N}\mu(B_j),$$
whence
$$\mu(B)\le\sum_{j\in\N}\mu(B_j). \tag{1}$$
Let us now show that
$$\mu(B)\ge\sum_{j\in\N}\mu(B_j). \tag{2}$$
Since $\mu\ge0$, here we may and will assume that $\mu(B_j)<\infty$ for all $j\in\N$ (otherwise, (2) is trivial). Take now any real $\ep>0$. Then for each $j\in\N$ there is a partition $(C_{j,n})_{n\in\N}\in P(B_j)$ and a function $f_j\colon\N\to Y$ such that $\sum_{n\in\N}f_j(n)(C_{j,n})\ge\mu(B_j)-\ep/2^j$. Moreover, $(C_{j,n})_{(j,n)\in\N^2}\in P(B)$. So,
$$\mu(B)\ge\sum_{j\in\N}\sum_{n\in\N}f_j(n)(C_{j,n})
\ge\sum_{j\in\N}(\mu(B_j)-\ep/2^j)
=\sum_{j\in\N}\mu(B_j)-\ep.$$
So, (2) follows. By (1) and (2), $\mu$ is a measure.

Moreover, for any $B\in\A$, we have $(B,\emptyset,\emptyset,\dots)\in P(B)$. Taking now any measure $y\in Y$ and, say, letting $f(n)=y$ for all $n\in\N$, we see that
$\mu\ge y$ for any $y\in Y$. However, it is easy to see that in general $\mu$ will not be in $Y$.
It is also clear that any $y\in Y$ is a lower bound on $\mu$. Moreover, for any $Y_0\subseteq Y$, the measure $\mu_{Y_0}$ is a lower bound on $\mu_Y$.
