I'm trying to prove that Haar measures are localizable. we know that Haar measures are decomposable ( see Haar measures are decomposable) in the sense that:
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 $E\in\frak{M}$.
if relation (ii) holds for every $E\in \frak{M}$, then we called $(X,\frak{M},\mu)$ localizable.
$\color{blue}{\textbf{can someone tell me that is my solution correct or not?}}$
$\color{green}{\textbf{my solution:}}$
$\color{red}{\textbf{the idea of proof is to prove that:}}$
$\color{red}{\bigg[}$there exists a measure $\lambda$ on ${\frak{B}}_G$ that is radon and translation invariant and hence is a Haar measure on $G$ and by uniqueness of haar measure, we prove that the first Haar measure $\mu$ is equal with this new measure $\lambda$ and finally, the measure $\lambda$ have a property that localizability is proved.$\color{red}{\bigg]}$
if $G$ be a locally compact group and $(G,{\frak{B}}_G,\mu)$ be Haar measure space where ${\frak{B}}_G$ is the borel $\sigma$-algebra, then if we define a measure $\lambda$ on ${\frak{B}}_G$ such that:
\begin{equation} \lambda(E)= \begin{cases} \underset{i\in I}{\sum}\mu(E\cap X_i) & \;\;\;\;\;\;\;\;\;\;\;\underset{i\in I}{\sum}\mu(E\cap X_i)<\infty \\ \infty & \;\;\;\;\;\;\;\;\;\;\;\underset{i\in I}{\sum}\mu(E\cap X_i)=\infty \end{cases} \end{equation}
$\color{red}{\textbf{proof that $\lambda$ is a translation invariant measure:}}$
if $\{E_j\}_{j\in\mathbb{N}}$ be a family of disjoint measurable sets, then if $\underset{i\in I}{\sum}\mu\bigg[\bigg(\underset{j\in\mathbb{N}}{\bigcup}E_j\bigg)\bigcap X_i\bigg]<\infty$, then $\underset{i\in I}{\sum}\mu(E_j\cap X_i)<\infty$ for every $j\in\mathbb{N}$ and we have
$\lambda\bigg(\underset{j\in\mathbb{N}}{\bigcup}E_j\bigg)=\underset{i\in I}{\sum}\mu\bigg[\bigg(\underset{j\in\mathbb{N}}{\bigcup}E_j\bigg)\bigcap X_i\bigg]=\underset{i\in I}{\sum}\mu\bigg[\underset{j\in \mathbb{N}}{\bigcup}\bigg(E_j\bigcap X_i\bigg)\bigg]=\underset{i \in I}{\sum}\underset{j\in \mathbb{N}}{\sum}\mu(E_j\cap X_i)\color{red}=\underset{j\in \mathbb{N}}{\sum}\underset{i\in I}{\sum}\mu(E_j\cap X_i)=\underset{j\in\mathbb{N}}{\sum}\lambda(E_j)$
equality in red follows from following relation in measure theory. if we consider $\mu$ as counting measure and $X$ as arbitrary set.
$$\int_X\underset{j\in\mathbb{N}}{\sum}f_j(x)d\mu=\underset{j\in \mathbb{N}}{\sum}\int_Xf_j(x)d\mu$$
now if $\underset{i\in I}{\sum}\mu\bigg[\bigg(\underset{j\in\mathbb{N}}{\bigcup}E_j\bigg)\bigcap X_i\bigg]=\infty$, then we have $\underset{j\in\mathbb{N}}{\sum}\lambda(E_j)=\infty$ and $\lambda\bigg(\underset{j\in\mathbb{N}}{\bigcup}E_j\bigg)=\infty$. also it is trivial that $\mu(\phi)=0$ and is proved that $\lambda$ is really a measure on ${\frak{B}}_G$.
it is easy to see that $\lambda$ is translation invariant.
we have to prove that the measure $\lambda$ is radon.
$\color{red}{\textbf{proof that $\lambda$ is finite on compact sets:}}$
if $K\in {\frak{B}}_G$ is a compact set, then $\mu(K)<\infty$ and by property (ii) of decomposability, it follows that $\underset{i\in I}{\sum}\mu(K\cap X_i)=\mu(K)<\infty$. hence $\lambda(K)=\underset{i\in I}{\sum}\mu(K\cap X_i)<\infty$ and $\lambda$ is finite on compact sets.
$\color{red}{\textbf{proof that $\lambda$ is inner regular on sets of finite measure:}}$
if $E\in{\frak{B}}_G$ and $\lambda(E)<\infty$ and for arbitrary $\epsilon>0$, then
$\lambda(E)=\underset{i\in I}{\sum}\mu(E\cap X_i)$
and by definition, there exists a finite set $F\subseteq I$ suchthat
$\underset{i\in I}{\sum}\mu(E\cap X_i)-\frac{\epsilon}{2}<\underset{i\in F}{\sum}\mu(E\cap X_i)$
now since $\mu\bigg[\underset{i\in F}{\bigcup}(E\cap X_i)\bigg]<\mu(E)<\infty$, then there exists a compact set $K\subseteq \underset{i\in F}{\bigcup}(E\cap X_i)\subseteq E$ such that
$$\mu\bigg[\underset{i\in F}{\bigcup}(E\cap X_i)\bigg]-\frac{\epsilon}{2}<\mu(K)$$
and it follows that:
$\lambda(E)-\epsilon=\underset{i\in I}{\sum}\mu(E\cap X_i)-\epsilon<\underset{i\in F}{\sum}\mu(E\cap X_i)-\frac{\epsilon}{2}=\mu\bigg[\underset{i\in F}{\bigcup}(E\cap X_i)\bigg]-\frac{\epsilon}{2}<\mu(K)$
hence we have
$$ \lambda(E)-\epsilon<\mu(K) $$
where $K\subseteq E$. hence inner regularity on sets with finite measure is proved.
$\color{red}{\textbf{proof that $\lambda$ is outer regular}}$
if $E\in {\frak{B}}_G$ with $\lambda(E)=\infty$. then since $\lambda$ is a measure and $E\subseteq G$, then we have $\lambda(G)=\infty$ and since $G$ is open, then $\lambda$ is inner regular on sets with infinite measure.
now if $E\in {\frak{B}}_G$ and $\lambda(E)<\infty$ and for arbitrary $\epsilon>0$, since $\underset{i\in I}{\sum}\mu(E\cap X_i)=\lambda(E)<\infty$, then $\mu(E\cap X_i)=0$ except for countable number of $i\in \mathbb{N}\subseteq I$. hence we have
$$\lambda(E)=\underset{i\in I}{\sum}\mu(E\cap X_i)=\underset{i\in \mathbb{N}}{\sum}\mu(E\cap X_i)=\mu\bigg[E\bigcap\bigg(\underset{i\in \mathbb{N}}{\bigcup}X_i\bigg)\bigg]$$
now, there exists an open set $O_1\supseteq \bigg[E\bigcap\bigg(\underset{i\in \mathbb{N}}{\bigcup}X_i\bigg)\bigg]$ such that $\mu(O_1)<\mu\bigg[E\bigcap\bigg(\underset{i\in \mathbb{N}}{\bigcup}X_i\bigg)\bigg]+\epsilon$
and it follows that
$$\mu(O_1)<\lambda(E)+\epsilon\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
now, there exists a compct set $K\subseteq \bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]$ such that
$$\mu\bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]<\frac{\epsilon}{2}+\mu(K)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$$
but in the case of Haar measure, the set $\{X_i\}_{i \in I}$ is equal with $\{gU'_n|g\in C, n\in\mathbb{N}\}$ and $H=\underset{n\in \mathbb{N}}{\bigcup}U'_n$ where $H$ is a open subgroup of $G$ and the set $C$ contains one poin from every coset $gH$ (for details, see Haar measures are decomposable). know if $K\subseteq G$ be a compact set, then since $K\subseteq \underset{g\in C}{\bigcup}gH$ and $\{gH|g\in C\}$ is an open cover of $K$, then since $K$ is compact, there exists a finite set $g_1,...,g_n$ such that $K\subseteq \underset{i=1}{\overset{n}{\bigcup}}g_iH$. hence $K$ is subset of union of countable numbers of $X_i$ $\Bigg(K\subseteq \underset{i\in \mathbb{N'}}{\bigcup}X_i\Bigg)$ where $\mathbb{N'}$ is a countable subset of $I$.
now we have
$$K\subseteq \bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]\bigcap\bigg[\underset{i\in \mathbb{N'}}{\bigcup}X_i\bigg]=\bigg[E\bigcap\bigg(\underset{i\in\mathbb{N''}}{\bigcup}X_i\bigg)\bigg]$$
where $\mathbb{N''}\subseteq I\setminus\mathbb{N}$ is a countable subset of $I$ and we have
$$\mu(K)\leq\mu\bigg[E\bigcap\bigg(\underset{i\in\mathbb{N''}}{\bigcup}X_i\bigg)\bigg]=\underset{i\in \mathbb{N''}}{\sum}\mu(E\cap X_i)=0$$
because we know that if $i\in I\setminus\mathbb{N}$, then $\mu(E\cap X_i)=0$.
hence $\mu(K)=0$ and from formula $(2)$, we have $\mu\bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]<\frac{\epsilon}{2}+\mu(K)=\frac{\epsilon}{2}$
also there exists an open set $O_2\supseteq \bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]$ such that $\mu(O_2)<\mu\bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]+\frac{\epsilon}{2}$
hence we have
$$\mu(O_2)<\mu\bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3)$$
and by formula $(1)$ and $(3)$, we have
$$\mu(O_1\cup O_2)\leq\mu(O_1)+\mu(O_2)<[\lambda(E)+\epsilon]+\epsilon=\lambda(E)+2\epsilon$$
where $E\subseteq \bigg[E\bigcap\bigg(\underset{i\in \mathbb{N}}{\bigcup}X_i\bigg)\bigg]\bigcup \bigg[E\bigcap\bigg(\underset{i\in I\setminus\mathbb{N}}{\bigcup}X_i\bigg)\bigg]\subseteq O_1\cup O_2$.
but since $\lambda(E)<\infty$, then $\mu(O_1\cup O_2)<\infty$ and by second decomposability property of Haar measure, we have
$$\infty>\mu(O_1\cup O_2)=\underset{i\in I}{\sum}\mu\bigg[\bigg(O_1\cup O_2\bigg)\cap X_i\bigg]$$
hence by definition of $\lambda$, it follows that
$$\lambda(O_1\cup O_2)=\underset{i\in I}{\sum}\mu\bigg[\bigg(O_1\cup O_2\bigg)\cap X_i\bigg]=\mu(O_1\cup O_2)<\lambda(E)+2\epsilon$$
hence we have
$$ \lambda(O_1\cup O_2)<\lambda(E)+2\epsilon $$
where $E\subseteq O_1\cup O_2$. and is proved that $\lambda$ is outer regular.
$\color{red}{\textbf{proof that $\mu$ is localizable}}$
now by uniqueness of Haar measure, we know that $\lambda=\mu$ and we have:
\begin{equation} \mu(E)=\lambda(E)= \begin{cases} \underset{i\in I}{\sum}\mu(E\cap X_i) & \;\;\;\;\;\;\;\;\;\;\;\underset{i\in I}{\sum}\mu(E\cap X_i)<\infty \\ \infty & \;\;\;\;\;\;\;\;\;\;\;\underset{i\in I}{\sum}\mu(E\cap X_i)=\infty \end{cases} \end{equation}
and it follows that in any case we have
$\mu(E)=\underset{i\in I}{\sum}\mu(E\cap X_i)$
and second decomposability property of Haar measure holds for every $E\in {\frak{B}}_G$ and it follows that Haar measure $\mu$ is localizable.
