We know that semifinite measure space $(X,\mathcal{M},\mu)$ is a measure space that for every measurable set $E\in\mathcal{M}$ with measure $\mu(E)=\infty$, there exist a measurable set $B\subseteq E$ such that $0<\mu(B)<\infty$.
by
it follows that Harar measures are localizable and since by
Why every localizable measure space is semifinite measure space?
every localizable measure is semifinite, it follows that Haar measures are semifinite.
but I'm doubted that every Haar measure is localizable or not!
can someone tell me is my argument true or not?
Haar measures need not be semifinite.
For example, let $G= \mathbb{R}_E\times \mathbb{R}_D$ where $\mathbb{R}_E$ is $\mathbb{R}$ with the usual topology and $\mathbb{R}_D$ is $\mathbb{R}$ with the discrete topology.
Let $\lambda$ be Lebesgue measure on $\mathbb{R}$ and $\sharp$ counting measure on $\mathbb{R}$. Then consider $\mu:= \lambda \times \sharp$ where $\times$ denotes the Radon product of Radon measures (see e.g. section 7.4 in Folland's book "Real analysis: Modern techniques and their applications"). Then $\mu$ is a Haar measure on $G$.
Let $F:= \{0\}\times \mathbb{R}_D\subseteq G$. Then one can show that $\mu(F)=\infty$ and $\mu(K)= 0$ for all compact subsets $K\subseteq F$ (Folland 7.2 exercise 12). Radon measures are inner regular on sets with finite measure, so if $S$ is a Borel set with $\mu(S) <\infty$ and $S\subseteq F$, then necessarily $\mu(S)= 0$. We conclude that $\mu$ is not semi-finite.