Measure Algebra

# Measure Algebra

Let $G$ be a locally compact group, that is, $G$ is a topological group whose topology is locally compact and Hausdorff. Let $\mathcal B$ be the set of Borel subsets of $G$ (subsets of $G$ that are obtained as countable unions, countable intersections, or complements of open sets in $G$) and let:

(1)
\begin{align} \quad M(G) = \left \{ \mu : \mathcal B \to \mathbb{C} \biggr | \mu \: \mathrm{is \: a \: measure \: on \:} \mathcal B \right \} \end{align}

That is, $M(G)$ consists of all complex-valued measures of the collection of Borel subsets of $G$. For each $\mu, \nu \in M(G)$ and $a \in \mathbf{F}$, define $\mu + \nu$, $a \mu$, $mu * \nu$ for all Borel subsets $E \in \mathcal B$ by:

(2)
\begin{align} \quad (\mu + \nu)(E) &= \mu(E) + \nu(E) \\ \quad (a \mu)(E) &= a \mu(E) \\ \quad (\mu * \nu)(E) &= \int_G \mu(Es^{-1}) \nu(ds) \end{align}

And define a function $\| \cdot \| : M(G) \to [0, \infty)$ for all $\mu \in M(G)$ by:

(3)
\begin{align} \quad \| \mu \| = \sup_{\mathrm{\{ E_1, E_2, ..., E_n \} \subset \mathcal B, \: \mathrm{disjoint}}} \left \{ \sum_{k=1}^n |\mu(E_k)| \right \} \end{align}

The supremum above runs over all finite disjoint subcollections $\{ E_1, E_2, ..., E_n \}$ of $\mathcal B$.

• 1. Showing that $(\mu * \nu) * \lambda = \mu * (\nu * \lambda)$: Let $\mu, \nu, \lambda \in M(G)$ and let $E \in \mathcal B$. Then:
(4)
\begin{align} \quad [(\mu * \nu) * \lambda](E) &= \int_G (\mu * \nu)(Es^{-1}) \: \lambda(ds) \\ &= \int_G \left [ \int_G \mu (Es^{-1}t^{-1}) \nu(dt) \right ] \: \lambda (ds) \\ &= \int_G \int_G \mu (Es^{-1}t^{-1}) \nu (dt) \lambda (ds) \end{align}
(5)
\begin{align} \quad [\mu * (\nu * \lambda)](E) &= \int_G \mu(Es^{-1}) (\nu * \lambda) (ds) \\ &= \int_G \mu (Es^{-1}) \left [ \int_G \nu(dst^{-1}) \lambda(dt) \right ] \end{align}
• 2. Showing that $\mu * (\nu + \lambda) = \mu * \nu + \mu * \lambda$: Let $\mu, \nu, \lambda \in M(G)$. Then for every $E \in \mathcal B$ we have that:
(6)
\begin{align} \quad [\mu * (\nu + \lambda)](E) = \int_G \mu(Es^{-1}) \: (nu + \lambda)(ds) = \int_G \mu(Es^{-1}) \: d (\nu) + \int_G \mu(Es^{-1}) \: \nu(ds) + \int_G \mu(Es^{-1}) \: \lambda (ds) = (\mu * \nu)(E) + (\mu * \lambda)(E) \end{align}
• So $\mu * (\nu + \lambda) = \mu * \nu + \mu * \lambda$.
• 3. Showing that $(a \mu) * nu = a (\mu * \nu) = \mu * (a \nu)$: Let $\mu, \nu \in M(G)$ and let $a \in \mathbf{F}$. Then for every $E \in \mathcal B$ we have that:
(7)
\begin{align} \quad [(a \mu) * \nu](E) = \int_G a \mu(Es^{-1}) \: \nu (ds) = a \int_G \mu(Es^{-1}) \: \nu (ds) = [a (\mu * \nu)](E) \end{align}
• And also:
(8)
\begin{align} \quad a (\mu * \nu)(E) = a \int_G \mu(Es^{-1}) \: \nu (ds) = \int_G \mu(Es^{-1}) a \nu (ds) = [\mu * (a \nu)](E) \end{align}
• Therefore $(a \mu) * \nu = a (\mu * \nu) = \mu * (a \nu)$.
• Hence $M(G)$ is an algebra. We now show that $M(G)$ is a normed algebra.
• 1. Suppose that $\| \mu \| = 0$. Then $\displaystyle{\sup \left \{ \sum_{k=1}^{n} |\mu(E_k)| = 0 \right \}}$ (where the supremum runs over all finite subcollections $\{E_1, E_2, ..., E_n \}$ of $\mathcal B$. In particular, $|\mu(E)| = 0$ for every $E \in \mathcal B$ and so $\mu = 0$. On the other hand, if $\mu = 0$ then clearly $\| \mu \| = 0$.
• 2. Let $\mu \in M(G)$ and let $a \in \mathbf{F}$. Then:
(9)
\begin{align} \quad \| a \mu \| = \sup \left \{ \sum_{k=1}^{n} |a\mu(E_k)| \right \} = \sup \left \{ |a| \sum_{k=1}^{n} |\mu(E_k)| \right \} = |a| \sup \left \{ \sum_{k=1}^{n} |\mu(E_k)| \right ) = |a| \| mu \| \end{align}
• 3. Let $\mu, \nu \in M(G)$. Then:
(10)
\begin{align} \quad \| \mu + \nu \| = \sup \left \{ \sum_{k=1}^{n} |\mu(E_k) + \nu(E_k) | \right \} \leq \sup \left \{ \sum_{k=1}^{n} [|\mu(E_k)| + |\nu(E_k)|] \right \} = \sup \left \{ \sum_{k=1}^{n} |\mu(E_k)| \right \} + \sup \left \{ \sum_{k=1}^{n} |\nu(E_k)| \right \} = \| \mu \| + \| \nu \| \end{align}
• 4. Lastly, let $\mu, \nu \in M(G)$. Then:
(11)
\begin{align} \quad \| \mu * \nu \| &= \sup \left \{ \sum_{k=1}^{n} |(\mu * \nu)(E_k)| \right \} \\ &= \sup \left \{ \sum_{k=1}^{n} \left | \int_G \mu(E_ks^{-1}) \: \nu(ds) \right | \right \} \\ & \leq \sup \left \{ \sum_{k=1}^{n} \int_G |\mu(E_ks^{-1})| \: \nu(ds) \right \} \end{align}