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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.

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A Borel set in $G$ has the form: for each $y \in \mathbb R_{\rm d}$, choose a Borel set $W_y$ in the line, where the line is considered to be $\mathbb R \times \{y\}$. The disjoint union of all these $W_y$ is your Borel set $W$. In fact, you may as well do your construction (restriction then completion) independently in each $\mathbb R \times \{y\}$. So whatever properties this operation has in the usual line, it still has in $G$.

As I understand it, the strange thing in $G$ is with sets $W$ (even Borel sets) such that every open set $G \supseteq W$ has infinite measure, but every compact set $F \subseteq W$ has measure zero. So-called "locally null" sets. These sets will retain this strange property even after your "restriction then completion" operation.

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    Thanks for answer. But I have still some doubts. For example, a Borel set $W$ is disjoint sum of $W_y \times \{y\}$ but generally uncountable many sets. Is it not important?2011-09-28