Measure, Outer Measure, Premeasure

Definitions#

Let \(X\) be a set equipped with a \(\sigma\)-algebra \(\mathcal{M}\).

• A measure on \(\mathcal{M}\) (or simply on \(X\) if \(\mathcal{M}\) is understood is a function \(\mu:\mathcal{M} \to [0,\infty]\) s.t.
  1. \(\mu(\emptyset)=0\);
  2. if \(\{E_j\}_1^\infty\) is a sequence of disjoint sets in \(\mathcal{M}\), then \(\mu(\cup_1^\infty E_j) = \sum_1^\infty \mu(E_j)\).
• An outer measure on a nonempty set \(X\) is a function \(\mu^\ast:\mathcal{P}(X) \to [0,\infty]\) that satisfies
  1. \(\mu^\ast(\emptyset)=0\);
  2. \(\mu^\ast(A)\leq \mu^\ast(B)\) if \(A\subset B\);
  3. \(\mu^\ast(\cup_1^\infty A_j) \leq \sum_1^\infty \mu^\ast(A_j)\).
• Let \(\mathcal{A}\subset\mathcal{P}(X)\) be an algebra. A premeasure on \(\mathcal{A}\) is a function \(\mu_0:\mathcal{A}\to[0,\infty]\) that satisfies
  1. \(\mu_0(\emptyset)=0\);
  2. if \(\{A_j\}_1^\infty\) is a sequence of disjoint sets in \(\mathcal{A}\) such that \(\cup_1^\infty A_j\in\mathcal{A}\), then \(\mu_0(\cup_1^\infty A_j) = \sum_1^\infty \mu_0(A_j)\).

Note that 

• an outer measure only needs to be countably subadditive, and needs to be defined on the whole power set, instead of a \(\sigma\)-algebra.
• a premeasure needs to be define on an algebra; these are used to extend via Caratheodory's theorem to a \(\sigma\)-algebra and hence to an actual measure.

A measure whose domain inclues all subsets of null sets (measure 0 sets) is called complete. A measure can always be extended to a complete measure:

Theorem. Suppose that \((X,\mathcal{M},\mu)\) is a measure space. Let \(\mathcal{N} = \{N\in\mathcal{M}:\mu(N)=0\}\) and \(\overline{\mathcal{M}} = \{ E\cup F: E\in\mathcal{M}\text{ and }F\subset N\text{ for some }N\in\mathcal{N}\}.\) Then \(\overline{\mathcal{M}}\) is a \(\sigma\)-algebra, and there is a unique extension \(\overline{\mu}\) of \(\mu\) to a complete measure on \(\overline{\mathcal{M}}\) (\(\overline{\mu}(E\cup F):= \mu(E)\)).

The measure \(\overline{\mu}\) is called the completion of \(\mu\) and \(\overline{\mathcal{M}}\) is called the completion of \(\mathcal{M}\) w.r.t. \(\mu\).

Extending pre- and outer measure#

Inducing outer measure from premeasure#

If \(\mu_0\) is a premeasure on \(\mathcal{A}\subset\mathcal{P}(X)\), it induces an outer measure on \(X\), namely

\[\mu^\ast(E) := \inf\{\ \sum_1^\infty \mu_0(A_j): A_j\in\mathcal{A}, E\subset \cup_1^\infty A_j\}.\]

It can be proved that for any \(\mathcal{E}\subset\mathcal{P}(X)\) and \(\rho\to[0,\infty]\) s.t. \(\emptyset,X\in\mathcal{E}\) and \(\rho(\emptyset)=0\), then if we define \(\mu^\ast(E)\) for any \(E\subset X\) as the above, then \(\mu^\ast\) is an outer measure. The definition illustrates the process of approximating arbitrary sets “from the outside” by countable unions of a family \(\mathcal{E}\) of “elementary sets” on which a notion of measure (e.g. premeasure) is defined.

From outer measure to measure#

Let \(\mu^\ast\) be an outer measure on \(X\). A set \(A\subset X\) is called \(\mu^\ast\)-measurable if for all \(E\subset X\),

\[\mu^\ast(E) = \mu^\ast(E\cap A) + \mu^\ast(E\cup A^c).\]

Theorem (Caratheodory). If \(\mu^\ast\) is an outer measure on \(X\), the collection \(\mathcal{M}\) of \(\mu^\ast\)-measurable sets is a \(\sigma\)-algebra, and the restrction of \(\mu^\ast\) to \(\mathcal{M}\) is a complete measure.

Now if \(\mu\) on \(\mathcal{M}\) extends a premeasure \(\mu_0\) by the above two constructions, then if \(\nu\) is another measure on \(\mathcal{M}\) that extends \(\mu_0\), then \(\nu(E)\leq \mu(E)\) for all \(E\in\mathcal{M}\), with equality when \(\mu(E)\) finite; this follows from the meaning of infimum as the greatest lower bound. If \(\mu_0\) is \(\sigma\)-finite (it is the countable union of finite sets), then \(\mu\) is the unique extension of \(\mu_0\) to a measure on \(\mathcal{M}\); this follows from supposing that \(X=\cup_1^\infty A_j\) with \(\mu_0(A_j)\) finite.