# Restriction of Haar measure to Borel $\sigma$ -algebra

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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I have TeXified the post to make it more readable. However the part where you define $\mu_1$ makes little sense, I can guess its meaning, but I don't want to assume too much on the possible contents of your question. – Asaf Karagila Dec 7 '11 at 23:04
Thanks, I have corrected. – arc Dec 7 '11 at 23:14
$M_1$ and $M$ are always the same since Haar mesure is defined on Borel sets. – Hsueh-Yung Lin Dec 7 '11 at 23:21
Perhaps anyone answering should provide a reference for the particular construction they are using for Haar measure. Even the comments so far seem to be using different ones. – Gerald Edgar Dec 8 '11 at 0:54
Here's a related thread on math.SE: math.stackexchange.com/q/61878 – Theo Buehler Dec 8 '11 at 2:01

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.

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Very thanks for answer. – arc Dec 9 '11 at 11:19