Lebesgue extension versus arbitrary sigma additive extension Can anybody help me with the following problem? Let m be a $\sigma $ additive measure on a semiring $F_{m}$ with a unit E, let $\mu $ be the Lebesgue extension of m, and let $\tilde{\mu}$ be an arbitrary $\sigma$ additive extension of m. Prove that $\tilde{\mu}\left ( A \right )=\mu \left ( A \right )$ for every measurable set on which $\tilde{\mu}$ is defined.
Apparently I need to first show that $\mu _{*}\left ( A \right )\leqslant \tilde{\mu}\left ( A \right )\leqslant \mu^{*}\left ( A \right )$
I have no idea how to do this. Can anybody help me?
 A: "Apparently I need to first show that $ \mu_*(A)\leq \overset{\sim}{\mu}(A)\leq \mu^*(A)$" This is sufficient as we need only show the equality for any measurable set $A$ so the outer and inner measures will coincide.
Lets start with showing that $\overset{\sim}{\mu}\leq \mu^*$: let $A$ be some measurable set, using theorem 2' we have $\overset{\sim}{\mu}(A)\leq \overset{\infty}{\underset{i=1}{\sum}}\overset{\sim}{\mu}(P_i)$ when $A\subseteq \overset{\infty}{\underset{i=1}{\bigcup}}P_i, P_i\in \mathscr{S}_m$. This means we also have $\overset{\sim}{\mu}(A)\leq \overset{\infty}{\underset{i=1}{\sum}}\overset{\sim}{\mu}(P_i)=\overset{\infty}{\underset{i=1}{\sum}}\mu(P_i)$ as $\mu$ and $\overset{\sim}{\mu}$ coincide on $\mathscr{S}_m$, this ofcourse means that $\overset{\sim}{\mu}(A)\leq \underset{A\subseteq \cup_i^\infty P_i}{\text{inf}} \overset{\infty}{\underset{i=1}{\sum}}\mu(P_i)=\mu^*(A)$ so our first inequality is done.
For the other inequality we use our earlier result that $\mu^*(A)\geq\overset{\sim}{\mu}(A) \iff -\mu^*(A)\leq -\overset{\sim}{\mu}(A)$.
The definion of inner measure is $\mu_*(A)=m(E)-\mu^*(E-A)$ this combined with our earlier inequality gets us $\mu_*(A)=m(E)-\mu^*(A)\leq m(E)-\overset{\sim}{\mu}(A)$. As $\overset{\sim}{\mu}$ is the extension of $m$ we have that $m(E)-\overset{\sim}{\mu}(A)=\overset{\sim}{\mu}(E)-\overset{\sim}{\mu}(E-A)$ then we may just use theorem 2' again and aquire $\overset{\sim}{\mu}(E)\leq\overset{\sim}{\mu}(E-A)+\overset{\sim}{\mu}(A)$ as $E\subseteq(E-A)\cup A$, so finally $\mu_*(A)\leq\overset{\sim}{\mu}(E)-\overset{\sim}{\mu}(E-A)\leq \overset{\sim}{\mu}(E-A)+\overset{\sim}{\mu}(A)-\overset{\sim}{\mu}(E-A)=\overset{\sim}{\mu}(A)$
And again as we let $A$ be a measurable set the outer and inner measure will conicide and thereof we have proved $\overset{\sim}{\mu}=\mu$
