Outer measure defined as infimum of a measure Let $\mathcal F$ be a $\sigma$-algebra on $\Omega$
 and let $\mu$: $\mathcal F \rightarrow [0,\infty]$ be a measure. 
For each $X \subseteq \Omega$ define $\mu^*(X) := \inf\{\mu(A) : X \subseteq A,A\in\mathcal F\}$. 
I'm trying to show that $\mu^*$ is an outer measure on $\Omega$, and that $\mathcal F$ is contained in $\mathcal F_{\mu^*}$ (the set of $\mu^*$ measurable sets). 
So far for showing $\mu^*$ is an outer measure: 1) clearly $\mu^*(\emptyset)=0$. 2) If $A \subseteq B \subseteq \Omega$, then I want to say $\mu^*(A) \leq \mu(B) \leq\mu^*(B)$ but am unsure if this works. 
Any help would be appreciated, thanks. 
 A: Regarding 2): You cannot claim that if $A \subseteq B$ then $\mu^*(A) \le \mu(B) \le \mu^*(B)$, because you need to have $B \in \mathcal F$ for the expression $\mu(B)$ to be meaningful.
However, you can still claim that $\mu^*(A) \le \mu^*(B)$. Here's a sketch of the proof: Consider all sets $C \in \mathcal F$ such that $B \subseteq C$. The outer measure of $B$ is exactly the infimum of the measures of those sets. But all those sets participate (maybe together with even more sets) in the definition of the outer measure of $A$...
(If you have a set of numbers, adding more numbers to it cannot increase its infimum.)
You also need to show countable subadditivity. Hint: Notice that if $X_n \subseteq A_n$ for every $n$ then also $\bigcup X_n \subseteq \bigcup A_n$. Select "covers" $A_n \supseteq  X_n$ which become tighter and tighter exponentially, and use the countable subadditivity of $\mu$.
Regarding $\mathcal F \subseteq \mathcal F_{\mu^*}$: This is easy, it follows from the definition of $\mu^*$. Edit: I forgot the definition of $\mathcal F_{\mu^*}$ when I wrote this, and it is less immediate than I imagined. You need to prove that given $M \in \mathcal F$, for all $X \subseteq \Omega$, we have $\mu^*(X) = \mu^*(X \cap M) + \mu^*(X \cap M^C)$. Hint: One way to prove equality is to prove inequality in both directions. One direction is easy and follows directly from the subadditivity of $\mu^*$. For the other direction, replace $X$ with a slightly bigger set which is in $\mathcal F$ (think about the definition of $\mu^*(X)$) and see that you get almost the inequality you need. Then, let "slightly" tend to 0.
