We know that a saturated measure is a measure space $(X,\mathcal{M},\mu)$ such that $\mu:\mathcal{M}\rightarrow [0,\infty]\;$ and $\;\mathcal{M}=\widetilde{\mathcal{M}}\;$
where
$\;\widetilde{\mathcal{M}}=\{E\subseteq X\;|\;E\cap M\in \mathcal{M}\;,\;\forall M\in\mathcal{M} \;\text{with}\;\mu(M)<\infty\}\;$ is the set of all locally measurable sets.
Are Haar measures defined on borel $\sigma$-algebra ${\frak{B}}_G$ of locally compact group $G$ saturated?
$\textbf{EDITED:}$
I found a solution, But I doubt whether it is true or not! Can anyone tell me if my solution is correct or not?
$\textbf{my solution:}$
if ${\frak{B}}_G$ be borel $\sigma$-algebra on $G$, then we know that ${\frak{B}}_G\subseteq\widetilde{{\frak{B}}_G}$ and it's enough that prove $\widetilde{{\frak{B}}_G}\subseteq {\frak{B}}_G$.
let $E\in \widetilde{{\frak{B}}_G}$, then we have $E\cap M\in {\frak{B}}_G$ for every $M\subseteq G$ with $\mu(M)<\infty$. since Haar measures are $\;$decomposable$^*$ (for definition, see below ($*$) and for proof, see Haar measures are decomposable), then according to (i) in below $(*)$ there exists $\{X_i\}_{i\in I}\subseteq {\frak{B}}_G$ with $\mu(X_i)<\infty$. since $\mu(X_i)<\infty$, $\forall i\in I$ and $E\cap M\in {\frak{B}}_G$ for all $M$ with finite measure, then we have $E\cap X_i\in {\frak{B}}_G$, $\forall i\in I$ and according to (iii) in below $(*)$, we have $E\in {\frak{B}}_G$. hence $\widetilde{{\frak{B}}_G}\subseteq{\frak{B}}_G$ and $\widetilde{{\frak{B}}_G}={\frak{B}}_G$.
$\color{\red}{\bigg[}$actually, I think that is proved that every decomposable measure is saturated.$\color{\red}{\bigg]}$
$\textbf{is my answer correct?}$
$(*)$ A measure space $(X,\mathfrak{M},\mu)$ is decomposable if:
(i) $X$ is a disjoint union of measurable subsets, $X=\bigcup_{i\in I}X_{i}$, with $\mu(X_{i})<\infty$ for all $i$.
(ii) $\mu(E)=\sum_{i\in I}\mu(E\cap X_i)$ for every measurable set $E$ of finite measure.
(iii) if $E\subseteq X$ and $E\cap X_i\in \frak{M}$ for all $i$, then $E\in\frak{M}$.
