Let $G$ be a locally compact abelian group.
Let $f\in L^{2}(G)$, i.e. $f:G\to\mathbb{C}$ is measurable and $\int_{G}|f(s)|^2ds <\infty$ where the integral is taken with respect to Haar measure.
Define the function $\tilde{f}(s) = f(s^{-1})$ a.e. on $G$.
Is $\tilde{f}\in L^{2}(G)$?
If $G$ is a group and $\mu$ is a measure on $G$, then defining, for every measurable subset $A$, $\bar{\mu}(A)=\mu(A^{-1})$ gives a measure on $G$.
It's easy to see that if $\mu$ is right invariant (that is, $\mu(Ag)=\mu(A)$, for every measurable $A$ and every $g\in G$), then $\bar{\mu}$ is left invariant.
In particular, if $G$ is locally compact and $\mu$ is a right invariant Haar measure on $G$, then $\bar{\mu}$ is a left invariant Haar measure on $G$, because $g\mapsto g^{-1}$ is a homeomorphism of $G$ onto itself.
By uniqueness of (left or right) invariant Haar measures on $G$, if $G$ is abelian, then $\bar{\mu}=c\mu$, for some scalar $c\ne0$ (the modulus of $G$).
This answer your question: $\mu$ and $\bar{\mu}$ share the measurable sets and also the integrable functions.