How is $\{\mu (U): A^{-1} \subset U, U \text{ is open}\}=\{ \mu (U^{-1}): A \subset U, U \text{ is open}\}$?

Question

I am reading these notes by Jonathan Gleason on the Existence and Uniqueness of the Haar Measure. On page no. 5 (Proof of Proposition 4.1, STEP: 2), I am not able to understand why $$ \{\mu (U): A^{-1} \subseteq U, U \text{ is open}\}=\{ \mu (U^{-1}): A \subseteq U, U \text{ is open}\}. $$ Here $A$ is a measurable subset of a topological group $G$, $U$ is open in $G$, and $\mu$ is a Haar measure on $G$. I have read the lemmas and definitions before the propositions but I couldn't get it. I'm sure it is something trivial, though; any help would be appreciated!

Answer 1

Just before the mentioned assertion, it is said that "$A^{-1}\subseteq U$ and $U$ is open if and only if $A\subseteq U^{-1}$ and $U^{-1}$ is open". Therefore, we have $$\tag{*} \left\{\mu\left(U\right),A^{-1}\subseteq U, U\mbox{ is open} \right\}= \left\{\mu\left(U\right),A\subseteq U^{-1}, U^{-1}\mbox{ is open} \right\}.$$ Now, as user342207 pointed out, $U$ is open if and only if $U^{-1}$ is open. Accounting the fact that $\left(U^{-1}\right)^{-1}=U$, we thus have \begin{align} \left\{\mu\left(U\right),A\subseteq U^{-1}, U^{-1}\mbox{ is open} \right\}&=\left\{\mu\left(\left(\color{red}{ U^{-1}} \right)^{-1} \right),A\subseteq \color{red}{ U^{-1}},\color{red}{ U^{-1}}\mbox{ is open} \right\}\\ &=\left\{\mu\left(\left(\color{red}{ V} \right)^{-1} \right),A\subseteq \color{red}{ V},\color{red}{V}\mbox{ is open} \right\}. \end{align}