In the real analysis book by Folland, section $11.1$ exercise $9$ have been come that:
if $G$ is a locally compact topological group with Haar measure $\mu$, then $\mu$ is decomposable.
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) $\mu(A)=\sum_{i\in I}\mu(A\cap X_i)$ for every measurable set $A$ of finite measure.
(iii) if $A\subseteq X$ and $A\cap X_i\in \frak{M}$ for all $i$, then $A\in\frak{M}$.
$\textbf{EDITED:}$
$\color{red}{\textbf{I succeeded that prove (i) and (ii). }}$
$\color{red}{\textbf{can someone help me to prove (iii)? }}$
$\color{blue}{\textbf{proof of (i):}}$
if $U\subseteq G$ be a compact simetric ($U^{-1}=U$) neighbourhood of $e$ , then if $U_n=\underbrace{U...U}_{\text{n times}}$, the sets $U_n$ for every $n$, are compact ( because action of group is countinuous) and the set $H=\underset{n\in \mathbb{N}}{\bigcup}U_n$ is an open subgroup generated by $U$ ($H$ is open because, any set $U_n$, have a neighbourhood $U_{n+1}\subseteq H$) and we have $G=\underset{g\in C}{\bigcup}gH=\underset{\underset{n\in \mathbb{N}}{g\in C}}{\bigcup}gU_n$, where $C$ by axiom of choice is a set that contains one point from every coset $gH$. sets $gU_n$ are compact and hence $\mu(gU_n)<\infty$.
now (i) is proved if we consider sets of the form $gU'_n=gU_n\setminus\underset{i=1}{\overset{n-1}{\bigcup}}gU_i$ instead of $gU_n$.
It causes the sets $\{gU'_n| {n\in \mathbb{N}}\}$ for every $g$, be disjoint and since $gH=\underset{n\in C}{\bigcup}gU'_n$ and the cosets $\{gH\}_{g\in C}$ are disjoint, then we have $G=\underset{\underset{n\in \mathbb{N}}{g\in C}}{\bigcup}gU'_n$ where $\mu(gU'_n)<\infty$ and the sets $\{gU'_n|n\in \mathbb{N}, g\in C\}$ are disjoint.
$\color{blue}{\textbf{proof of (ii):}}$
if we substitute $\{gU'_n|g\in C, n\in\mathbb{N}\}$ with $\{X_i\}_{i\in I}$, since
$\underset{i\in F}{\sum}\mu(A\cap X_i)=\mu\Bigg(A\bigcap\bigg[\underset{i\in F}{\bigcup}X_i\bigg]\Bigg)\leq\mu(A)$
for finite set $F\subseteq I$, then
$\underset{i\in I}{\sum}\mu(A\cap X_i )=\underset{\underset{\text{F is finite}}{F\subset I}}{\sup}\left(\underset{i\in F}{\sum}\mu(A\cap X_i)\right)\leq \mu(A)$
and hence
$$\underset{i\in I}{\sum}\mu(A\cap X_i)\leq\mu(A) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1) $$
now if $K\subseteq G$ be compact, then since $K\subseteq \underset{g\in C}{\bigcup}gH$ and the sets $gH$ are open, then $K\subseteq \underset{i=1}{\overset{m}{\bigcup}}g_iH$ where $g_i\in C$.
now if $\mu(A)<\infty$ and $\epsilon>0$, then by inner regularity of haar measure, there exist a compact set $K\subseteq A$ such that $\mu(A)-\epsilon<\mu(K)$. also we have
$\mu(K)=\mu\Big[K\bigcap\Bigg(\underset{i=1}{\overset{m}{\bigcup}}g_iH\Bigg)\Big]=\underset{i=1}{\overset{m}{\sum}}\mu(K\cap g_iH)=\underset{i=1}{\overset{m}{\sum}}\underset{n\in\mathbb{N}}{\sum}\mu(K\cap g_iU'_n)\leq\underset{i=1}{\overset{m}{\sum}}\underset{n\in\mathbb{N}}{\sum}\mu(A\cap g_iU'_n)\leq\underset{i\in I}{\sum}\mu(A\cap X_i)$
then we have $\mu(A)-\epsilon<\mu(K)\leq\underset{i\in I}{\sum}\mu(A\cap X_i)$
and since $\epsilon$ is arbitrary, it follows
$$\mu(A)\leq\underset{i\in I}{\sum}\mu(A\cap X_i) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$$
and by $(1)$ and $(2)$, (ii) also is proved.
