From the last paragraph of the proof for Claim V.11.8, [1], we know that
Lemma. There is a unique regualr positive Borel measure $m$ on a compact group $G$ such that $$m(g\Delta)=m(\Delta)=m(\Delta g)\forall g\in G \mbox{ and } \Delta\subset G.$$
Exercise V.11.4, [1]. Let $G$ be a locally compact group. If $m$ is a regular Borel measure on $G$, show that any two of the following properties imply the third:
- $m (\Delta g)= m(\Delta)$ for every Borel set $\Delta$ and every $g$ in $G$;
- $m(g\Delta) = m(\Delta)$ for every Borel set $\Delta$ and every $g$ in $G$;
- $m(\Delta) = m(\Delta^{-1})$ for every Borel set $\Delta$.
My Proof. 1 and 3 imply 2. For every Borel set $\Delta$ and every $g$ in $G$, \begin{align*} m(g\Delta)=m(\Delta^{-1}g^{-1})=m(\Delta^{-1})=m(\Delta) \end{align*}
1 and 2 imply 3. This holds for all compact subsets of $G$ by the above lemma. This also holds for all open subset of $G$ because \begin{align*} m(U)= & \sup\{m(K)|K\subset U \mbox{ is compact.}\}\\ = &\sup\{m(K^{-1})|K\subset U \mbox{ is compact.}\}\\ = &\sup\{m(K^{-1})|K^{-1}\subset U^{-1} \mbox{ is compact.}\}\\ =& \sup\{m(K)|K\subset U^{-1} \mbox{ is compact.}\} \end{align*} Since $m(V)=\inf\{m(U)|U\supset V \mbox{ is open.}\}$ for any Borel set $V$, we can prove the same holds for all Borel subset of $G$ similarily.
Question. Can we prove "1 & 2 $\Rightarrow 3$" directly, without using the uniqueness of $m$ satisfying 1 & 2?
[1] Conway, John B., A course in functional analysis., Graduate Texts in Mathematics, 96. New York etc.: Springer-Verlag. xvi, 399 p. DM 148.00 (1990). ZBL0706.46003.