Let $(X,\mathcal{A}, \mu)$ be a measure space, where $\mathcal{A}$ is a $\sigma$-algebra over $X$ and $\mu:\mathcal{A} \to [0,\infty]$ is a measure. It is equivalent to say that $\mu$ satisfies the following conditions:
1) $\mu(\emptyset) = 0$
2) $\mu$ is mono: $A \subseteq B \Rightarrow \mu(A) \leq \mu(B)$, for all $A,B \in \mathcal{A}$
3) $\mu(A) + \mu(B) = \mu(A \cup B)+ \mu(A \cap B)$, for all $A,B \in \mathcal{A}$
4) $(A_i)_{i\geq 0}$ ascending chain in $\mathcal{A} \Rightarrow \mu(\bigcup A_i) = \sup \mu(A_i)$
Define the function $\mu^*:\wp(X) \to [0, \infty]$ by $\mu^*(Y)=\inf\{\mu(A): A \in \mathcal{A}, Y \subseteq A\}$.
I want to prove $\mu^*$ is an outer measure, i.e. the following hold:
1') $\mu^*(\emptyset) = 0$ (clear, since $\mu^*=\mu$ on $\mathcal{A}$)
2') $Y \subseteq Z \Rightarrow \mu^*(Y) \leq \mu^*(Z)$ (clear, since any $A\in\mathcal{A}$ containing $Z $ also contains $Y$)
3') $\mu^*(Y)+\mu^*(Z) \geq \mu^*(Y \cup Z) + \mu^*(Y \cap Z)$
4') $(Y_i)_{i\geq 0}$ ascending chain in $\wp(X) \Rightarrow \mu^*(\bigcup Y_i) = \sup \mu^*(Y_i)$
I know the usual definitions of measure and outer measure are different, but I'm working with measure on $\sigma$-frames. Anyway, I put $\mathcal{A}$ and $P(X)$ for the sake of simplicity. So no worries if you ignore what a frame is. I will adapt any suggestion, if possible.Please use the conditions I wrote. Thanks! Any reference may be useful.
3')
Let $A,B\in\mathcal A$ with $Y\subseteq A$ and $Z\subseteq B$.
Then $Y\cap Z\subseteq A\cap B$ and $Y\cup Z\subseteq A\cup B$ so that $\mu^*(Y\cap Z)\leq\mu(A\cap B)$ and $\mu^*(Y\cup Z)\leq\mu(A\cup B)$.
Then $\mu^*(Y\cap Z)+\mu^*(Y\cup Z)\leq\mu(A\cap B)+\mu(A\cup B)=\mu(A)+\mu(B)$
This for every pair $A,B\in\mathcal A$ with $Y\subseteq A$ and $Z\subseteq B$ so that: $$\mu^*(Y\cap Z)+\mu^*(Y\cup Z)\leq\mu^*(Y)+\mu^*(Z)$$
4')
Let $A\in\mathcal A$ with $\bigcup_iY_i\subseteq A$. Then $Y_i\subseteq A$ and consequently $\mu^*(Y_i)\leq\mu(A)$ for every $i$.
Then $\sup\mu^*(Y_i)\leq\mu(A)$ and this for every $A\in\mathcal A$ with $\bigcup_iY_i\subseteq A$, so that:$$\sup\mu^*(Y_i)\leq\mu^*(\bigcup_iY_i)$$
If $\sup\mu^*(Y_i)=\infty$ then we are ready so assume that $\sup\mu^*(Y_i)<\infty$. Let $\epsilon>0$ and let $A_i\in\mathcal A$ with $\mu(A_i)<\mu^*(Y_i)+\epsilon$.
Now define $B_i=\bigcap_{j\geq i}A_j$ and observe that $B_i\in\mathcal A$ with $Y_i\subseteq B_i\subseteq A_i$ so that:$$\mu^*(Y_i)\leq\mu(B_i)\leq\mu(A_i)\leq\mu^*(Y_i)+\epsilon$$
Also observe that $i<j\implies B_i\subseteq B_j$.
We have $\bigcup_iY_i\subseteq\bigcup_iB_i\in\mathcal A$ so that: $$\mu^*(\bigcup_i Y_i)\leq\mu(\bigcup_i B_i)=\sup\mu(B_i)\leq\sup\mu^*(Y_i)+\epsilon$$
This for every $\epsilon$ so we are allowed to conclude that also:$$\mu^*(\bigcup_i Y_i)\leq\sup\mu^*(Y_i)$$
Proved is now that: $$\mu^*(\bigcup_i Y_i)=\sup\mu^*(Y_i)$$