A problem on measure restriction Definition of measurable space:
An ordered pair $(\Omega, \mathcal{F})$ is a measurable space if $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$.
Definition of measure:
Let $(\Omega, F)$ be a measurable space, $μ$ is an non-negative function defined on $\mathcal{F}$ (that is $\mu: \mathcal{F} \to [0, +\infty]$). If $\mu(\emptyset) = 0$ and $\mu$ is countably additive (that is $A_n \in \mathcal{F}$, $n \geqslant 1$, $A_n \cap A_m = \emptyset$, $n \neq m \Rightarrow \mu(\cup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty} \mu(A_n)$) then $\mu$ is a measure on $(\Omega, \mathcal{F})$.
Definition of measure space:
Let μ is a measure on $(\Omega, \mathcal{F})$ then $(\Omega, \mathcal{F}, \mu)$ is a measure space.
My problem is 
suppose $(\Omega, \mathcal{F}, \mu)$ is a finite measure space($\mu(\Omega) < + \infty$), $\Omega_0 \subset \Omega$ and $\mu^*(\Omega_0) = \mu(\Omega)$.
Then show $\mu^*(A \cap \Omega_0) = \mu(A)$ for $\forall A \in \mathcal{F}$.
Does anyone have any idea how to prove it? 
Update:
I'm so sorry. I put a word "let" in front of $\mu^*(\Omega_0) = \mu(Ω)$. The original problem have no word "let" there. I've deleted it.
I've searched the $μ^∗$ before this problem and the only place at which $μ^∗$ first came out is $μ^∗ (A) = \inf \{ \sum _{n=1}^ {\infty}\mu(A_n) | A_n \in \mathcal{F}, A \subset \cup _{n=1}^ {\infty} A_n \}$, $A ⊂ \Omega$. The solutions give me a hint that show $\Omega_0 ∩ F$ is a $\sigma$-algebra. I'm not sure what's the relationship between $\sigma$-algebra $\Omega_0 ∩ F$ and the problem I need to prove.
 A: Assuming a correct version of the standard definition of an outer measure $\mu^*$ given a measure $\mu$, and assuming that we're supposed to simply assuume that $\Omega_0\subset\Omega$ satisfies $\mu^*(\Omega_0)=\mu(\Omega)$ (that word "let" threw me off):
Suppose $A\in F$. If we let $A_1=A$ and $A_n=\emptyset$ for $n>1$ then $A_n\in F$ and $A\cap\Omega_0\subset\bigcup A_n$, so $$\mu^*(A\cap\Omega_0)\le\sum\mu(A_n)=\mu(A).$$Now, writing $A^c$ for $\Omega\setminus A$, the same argument shows that $$\mu^*(A^c\cap\Omega_0)\le\mu(A^c),$$hence $$\mu^*(A\cap\Omega_0)+\mu^*(A^c\cap\Omega_0)\le\mu(A)+\mu(A^c)=\mu(\Omega).$$ But $\mu^*$ is subadditive, hence $$\mu(\Omega)=\mu^*(\Omega_0)\le\mu^*(A\cap\Omega_0)+\mu^*(A^c\cap\Omega_0)=\mu(\Omega).$$So we must have equality throughout.
Exercise: Determine exactly where we used the fact that $\mu(\Omega)<\infty$.
Exercise: Show that the result is false without the assumption that $\mu(\Omega)<\infty$.
The two exercises are not the same, btw.
(Exercise: If $x'\le x$, $y'\le y$ and $x'+y'\ge x+y$ then $x=x'$ and $y=y'$.)
