let $(X,\Sigma, v)$ be a $\sigma$-finite measure space. i.e., any set in $\Sigma$ can be represented by a countable union of sets whose measures are finite.
now, define $\mu^*(A):=\inf\{v(B)|A\subseteq B, \forall B\in\Sigma\}$. I want to show this set function is an outer measure.
that is, I want to show it has the properties listed below:
i) null empty set: $\mu^*(\emptyset)=0$,
ii) monotonicity: $A_1\subseteq A_2\implies \mu^*(A_1)\leq\mu^*(A_2)$,
iii) countable subadditivity: $\mu^*(\cup A_i)\leq\sum \mu^*(A_i)$.
the first property follows immediately the fact that $v$ is a measure and $\emptyset\in\Sigma$.
but for the second and third, I can't treat the $\inf$ rigorously. like, for ii), I know that $A_2\subseteq B\implies A_1\subseteq B$, but can't deduce $\inf\{v(B)|A_1\subseteq B\}\leq \inf\{v(B)|A_2\subseteq B\}$ from it. intuitively it looks obvious, but every time I write the proof I can't stop thinking that I skipped something and it's not rigorous(in analysis, in general).
so, can you prove them as rigorous as possible? I would really appreciate.
For (ii), if $ U_2\subset U_1\subset [0,\infty]$ then $\inf U_1\leq \inf U_2.$
If $A_1\subset A_2 \subset X,$ then for $j\in \{1,2\}$ let $V_j=\{B:A_i\subset B\in \Sigma\}$ and let $U_j=\{v(B):B\in V_j\}.$ We have $U_2\subset U_1$ because $V_2\subset V_1.$
For (iii): Let $A$ be a countable family of subsets of $X.$ If $A$ is infinite let $A=\{a_n:n\in \mathbb N\}$ where $m\ne n \implies a_m\ne a_n.$ If $A$ is finite, with $k-1$ members (with $k\in \mathbb N$),let $A=\{a_n:n<k\}$, and let $a_n=\phi$ for $n\geq k.$ This is to treat all cases (finite or infinite $A$) at once.
Observe that since $\mu^*(\phi)=0,$ we have $\sum_{n=1}^{\infty} \mu^*(a_n)=\sum_{a\in A}\mu^*(a).$
For $r>0,$ for each $n\in \mathbb N$ let $a_n\subset b_{n,r}\in \Sigma$ such that $\nu(b_{n,r})\leq r\cdot 2^{-n}+\mu^*(a_n).$ We have $\cup A=\cup_na_n\subset \cup_nb_{n,r} \in \Sigma.$ Therefore $$\mu^*(\cup A)\leq v(\cup_nb_{n,r})\leq \sum_{n=1}^{\infty}v(b_{n,r})\leq \sum_{n=1}^{\infty}(r\cdot 2^{-n}+\mu^*(a_n)\;)=r+\sum_{a\in A}\mu^*(a).$$ $$\text {Therefore } \quad \forall r>0 \;\left( \mu^*(\cup A)\leq r+\sum_{a\in A}\mu^*(a)\right).$$