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Prove that ##\mu^*## is an outer measure, given a measure space ##(X, \mathcal A, \mu)## and define

##\mu^*(A) = \inf \{\mu(B) \mid A \subset B, B \in \mathcal A\}##

for all subsets ##A## of ##X##.

Here are what I have gone so far:

**(1)**The first condition is the easiest one:

##\begin{align}

\mu^*(\emptyset) &= \inf \{\mu(B) \mid \emptyset \subset B, B \in \mathcal A\}\\

&= \mu (\emptyset) \\

&= 0

\end{align}##

\mu^*(\emptyset) &= \inf \{\mu(B) \mid \emptyset \subset B, B \in \mathcal A\}\\

&= \mu (\emptyset) \\

&= 0

\end{align}##

**(2)**Now the second condition. Let ##D, E \in X## and ##D \subset E##,

##\begin{align}

\mu^*(D) &= \inf \{\mu(D') \mid D \subset D', D' \in \mathcal A\}\\

\mu^*(E) &= \inf \{\mu(E') \mid E \subset E', E' \in \mathcal A\}\\

\end{align}##

\mu^*(D) &= \inf \{\mu(D') \mid D \subset D', D' \in \mathcal A\}\\

\mu^*(E) &= \inf \{\mu(E') \mid E \subset E', E' \in \mathcal A\}\\

\end{align}##

Here I need to prove ##\mu^* (D) \leq \mu^*(E)##. It looks to me so intuitive especially if I draw Venn diagrams of ##D, E, D'## and ##E'##, but I don't know how to say it in math-speak. I would appreciate helps on this 2nd. condition.

**(3)**And this 3rd. condition is my major stumbling block: Given ##(A_i)_{i \in \mathbb N} \subset X##, I need to arrive at

##\mu^* (\bigcup _{i=1}^{\infty} A_i)\leq \sum_{i=1}^{\infty} \mu^* (A_i).##

Here, I know for sure I need to state this first: ##\forall A_i, \exists B_i ## such that ## A_i \subset B_i, B_i \in \mathcal A##, but I don't think the next step is right:

##\begin{align}

\mu^*(\bigcup_{i=1}^{\infty}A_i) &= \inf \{\bigcup_{i=1}^{\infty}\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}\\

&= \ldots\\

\end{align}##

\mu^*(\bigcup_{i=1}^{\infty}A_i) &= \inf \{\bigcup_{i=1}^{\infty}\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}\\

&= \ldots\\

\end{align}##

I would appreciate any help on this 3rd. condition in addition to the 2nd. above. Thank you for your time and effort.