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Let $(G,M,\mu)$ be a measure space, where $\mu$ is the Haar measure on topological group $G:=\mathbb R\times\mathbb R_d$, ($\mathbb R$ is the group of reals with the natural topology whereas $\mathbb R_d$ is the group of reals with the discrete topology) and $M$ isbe the $\sigma$-algebra of all Haar measurable subsets of $G$.

Let $\mu_0 :=\mu|_B$, where $B$ is the $\sigma$-algebra of all Borel subsets of $G$, and let $\mu_0$ to $(G,M_1,\mu_1)$ be the smallest completion $(G,M_1,\mu_1)$ of the measure space $(G,B,\mu_0)$?.

Is it true that $M_1=M$ and consequently $\mu_1=\mu$ ?

Let $(G,M,\mu)$ be a measure space, where $\mu$ is the Haar measure on topological group $G:=\mathbb R\times\mathbb R_d$, ($\mathbb R$ is the group of reals with the natural topology whereas $\mathbb R_d$ is the group of reals with the discrete topology) and $M$ is the $\sigma$-algebra of all Haar measurable subsets of $G$.

Let $\mu_0 :=\mu|_B$, where $B$ is the $\sigma$-algebra of all Borel subsets of $G$, and let $\mu_0$ to the smallest completion $(G,M_1,\mu_1)$ of the measure space $(G,B,\mu_0)$?

Is it true that $M_1=M$ and consequently $\mu_1=\mu$ ?

Let $(G,M,\mu)$ be a measure space, where $\mu$ is the Haar measure on topological group $G:=\mathbb R\times\mathbb R_d$, ($\mathbb R$ is the group of reals with the natural topology whereas $\mathbb R_d$ is the group of reals with the discrete topology) and $M$ be the $\sigma$-algebra of all Haar measurable subsets of $G$.

Let $\mu_0 :=\mu|_B$, where $B$ is the $\sigma$-algebra of all Borel subsets of $G$, and let $(G,M_1,\mu_1)$ be the smallest completion of the measure space $(G,B,\mu_0)$.

Is it true that $M_1=M$ and consequently $\mu_1=\mu$ ?

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Asaf Karagila
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