$\mu^{*}(G)=1$ implies that $\mu^{*}(F\cap G)=\mu(F)$, $\forall F \in \mathcal{F}$...

Question

where $\mu^{*}(G):=\inf\{\mu(F):F\in\mathcal{F}, G \subset F\}$, $(\Omega,\mathcal{F},\mu)$ a probability space.

How would you prove it? Thank you.

Answer 1

By definition, $\mu^{*}(F \cap G):=\inf\{\mu(H):H\in\mathcal{F}, F \cap G \subset H\}$.

Since $F \cap G \subset F$ and $F \in \mathcal{F}$, the definition tells use $\mu^{*}(F \cap G) \leq \mu(F)$.

Now we will show that, for any $H \in \mathcal{F}$ with $F \cap G \subset H$, $\mu(F) \leq \mu(H)$. This would imply $\mu^{*}(F \cap G) \geq \mu(F)$, and so complete the proof.

So let $H \in \mathcal{F}$ with $F \cap G \subset H$. Then $G \subset H \cup (\Omega \backslash F)$, and since $\mu^{*}(G) = 1$ we have $\mu(H \cup (\Omega \backslash F)) \geq 1$. We also have $\mu(H \cup (\Omega \backslash F)) \leq \mu(H) + \mu(\Omega \backslash F) = \mu(H) + 1 - \mu(F)$ (since it is a probability measure). Putting it all together:

$$1 \leq \mu(H) + 1 - \mu(F)$$ $$\Rightarrow \mu(F) \leq \mu(H)$$