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 $\mu_1=\mu$ ?

  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Dec 7 '11 at 23:04
  • $\begingroup$ $M_1$ and $M$ are always the same since Haar mesure is defined on Borel sets. $\endgroup$
    – HYL
    Dec 7 '11 at 23:21
  • $\begingroup$ Not necessary. Haar measure is complete, because it is introduced by using Caratheodory theorem, but Borel measure generally is not complete. $\endgroup$
    – arc
    Dec 7 '11 at 23:28
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 8 '11 at 0:54
  • 1
    $\begingroup$ Here's a related thread on math.SE: math.stackexchange.com/q/61878 $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.