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 $\mu\_1=\mu$ ?



Restriction of Haar measure to Borel $\sigma$ -algebra
**One construction.** 


$G$ is a locally compact Hausdorff topological group.  

Let $C\_c(G)$ denote the collection of continouous, real-valued functions on $G$ with compact support. 


(1) We begin with a *Haar integral*, a linear
functional $\Lambda : C\_c(G) \to \mathbb R$. The Haar
integral is unique up to a constant factor.


(2) Then we construct
a set-function $\mu\_1$ on the open subsets of $G$: If $U \subseteq G$
is open, let
$$
 \mu\_1(U) = \sup\{\Lambda(f) : f \in C\_c(G), 0 \le f \le 1\_U\} .
$$
Here, $1\_U$ is the indicator function (characteristic function) of the set $U$. 


(3) Now we construct a set function $\mu\_2$ on all sets.
If $E \subseteq G$, let
$$
 \mu\_2(E) = \inf\{\mu\_1(U) : U \text{ open, } U \supseteq E\} .
$$
This $\mu\_2$ is a Carathéodory outer measure.


(4) This version of the *Haar measure* is the restriction $\mu\_3$
of $\mu\_2$ to the collection $\mathcal M$ of $\mu\_2$-measurable sets.


So, I understand the question to be: If $E \in \mathcal M$, then is there
a Borel set $B$ and a $\mu\_3$-null set $N$ so that $E = B \Delta N$?


**This Example**


Now consider the case $G = \mathbb R \times \mathbb R\_d$.
The Haar integral we will use is:
$$
 \Lambda(f) = \sum\_{y \in \mathbb R\_d} \int\_{\mathbb R} f(x,y)\,dx .
$$
If $f \in C\_c(G)$, then for all but finitely many $y$
the integral is identically zero (so it is a finite sum),
and for the remaining $y$, the integrand has compact
support in $\mathbb R$. 


Next: an open set $U \subseteq G$ is obtained by arbitrarily choosing
open sets $U\_y \subseteq \mathbb R$, one for each $y$, and taking
$$
 U = \bigcup\_y \big(U\_y \times \{y\}\big) .
$$
For such $U$ we get
$$
 \mu\_1(U) = \sum\_y \lambda(U\_y) .
$$
Here, $\lambda$ is Lebesgue measure in $\mathbb R$. Note that
an uncountable sum has value $\infty$ unless all but countably
many terms are zero. And the only open set in $\mathbb R$
with measure zero is the empty set. So $\mu\_1(U) < \infty$
implies that $U\_y = \varnothing$ except for countably many $y$.


Now compute $\mu\_2$. An arbitrary set $E \subseteq G$ is of
course obtained by taking arbitrary sets $E\_y \subseteq \mathbb R$,
one for each $y$, and then
$$
 E = \bigcup\_y \big(E\_y \times \{y\}\big) .
$$
Let $U \supseteq E$ open, so that
$U\_y \supseteq E\_y$ open for all $y$.
If $E\_y = \varnothing$ for all but countably many $y$, we get
$$
 \mu\_2(E) = \sum\_y \lambda(E\_y)
$$
by taking the $U\_y$ close to the corresponding $E\_y$.
But if $E\_y \ne \varnothing$ for uncountably many $y$, we get
$\mu\_2(E) = \infty$, even if the series $\sum\_y \lambda(E\_y)$
has finite value. So, for example, if $E$ is an uncountable
subset of the $y$-axis in $G$, then $\mu\_2(E) = \infty$
even though $\lambda(E\_y)=0$ for all $y$. (This is used
for a standard counterexample to show a limitation in
Fubini's theorem.)


What about $\mathcal M$? Write $\mathcal L$ for the collection
of Lebesgue measurable sets in $\mathbb R$.
Let $E$ be an arbitrary set in $G$ as before.
We have $E \in \mathcal M$ if and only if $E\_y \in \mathcal L$
for all $y$. **\* proof omitted \***


Write $\mathcal B$
for the collection of Borel sets in $\mathbb R$. If $E$ is Borel in $G$
then $E\_y \in \mathcal B$ for all $y$. (The converse is false, but
we won't need it.)


Now, consider a set $Q \subseteq [0,1]$ in
$\mathcal L \setminus \mathcal B$. The example to consider is
$E = Q \times \mathbb R\_d$. So that $E\_y = Q$ for all $y$.
Can it be that $E = B \Delta N$ with $B$ Borel and $N$ null?
Since $Q \not\in \mathcal B$, we would need
$N\_y \ne \varnothing$ for all $y$. But then
$\mu\_2(N) = \infty$ and it is not a null set after all.


**Conclusion** ... for this particular construction of Haar measure,
the question has a negative answer.


