Let's consider the measure space $(G, \mathfrak{M}, \mu)$, where $\mu$ is the Haar measure on topological group $G:=\mathbf{R} \times \mathbf{R_d}$, ($\mathbf{R}$ is the group of reals with the natural topology whereas $\mathbf{R_d}$ is the group of reals with the discrete topology) and $\mathfrak{M}$ is the $\sigma$-algebra of all Haar measurable subsets of $G$.
What happens when we first restrict $\mu$ to the measure $\mu_0:=\mu|_{\cal{B}}$, where $\cal{B}$ is the $\sigma$-algebra of all Borel subsets in $G$, and next we extend $\mu_0$ to the smallest completion $(G,\mathfrak{M_1}, \mu_1)$ of the measure space $(G, \cal{B}, \mu_o)$ ?
( $E\in \mathfrak{M_1}$ iff $E$ is of the form $E=A \cup B$, where $A \in \mathfrak{B}$, $B \subset Z \in \mathfrak{B}$, $\mu_0(Z)=0$; $\mu_1(E)=\mu_0(A)=\mu(A)$).
Does we obtain that $\mathfrak{M_1}=\mathfrak{M}$ and consequently $\mu_1=\mu$ ?
Thanks.