The Utilization of Universal Quantifier Statement

Question

This is a measure theory statement, but the thing that I have question on is not really measure theory related but purely a confusion in the quantifiers. Consider a Radon (inner regular on open sets + outer regular on all Borel sets) measure space $(X, \mathcal{B}_X, \mu)$. Let $E \in \mathcal{F}$ be such that $\mu(E) < +\infty$. Now let $\epsilon > 0$ be given. We know by outer regularity of measure, there exists $U_\epsilon \supseteq E$ such that $$ \mu(E) + \epsilon > \mu(U_\epsilon). $$ Equivalently, we get $\mu(E) > \mu(U_\epsilon) - \epsilon$. Now can we by inner regularity on open sets get the existence of $K_\epsilon \subseteq U_\epsilon$ with $\mu(K_\epsilon) > \mu(U_\epsilon) - \epsilon$ even if $U_\epsilon$ is related to $\epsilon$ already? If the answer is yes, can someone break down the definition of quantifiers here so I could see more clearly why?

Answer 1

Yes, such a $K_\epsilon$ is ensured to exist since once we specify $U_\epsilon$ it is simply some open set. If we wanted to relabel everything to make it quite clear we can say

Let $E$ be given as above and we know for all $\epsilon_1>0$ that there is some $U_{\epsilon_1}$ with $$\mu(E)+\epsilon_1>\mu(U_{\epsilon_1})$$ as you have. Now $U_\epsilon$ is just some open set, we can label it just $U$. Via inner regularity as you have pointed out for any $\epsilon_2>0$ we can find $K_{\epsilon_2}\subseteq U$ with

$$\mu(K_{\epsilon_2}) >\mu(U)-\epsilon_2$$

To summarize, we are applying two results back to back, the first is to pick $\epsilon_1>0$ and find an open set $U_{\epsilon_1}$, then once we have this new set, which just so happens to be an open set, we may pick some other $\epsilon_2>0$ which may or may not equal $\epsilon_1$ and then get another new set $K_{\epsilon_2}$.