Are there other important measure spaces which are not obtained directly from an outer measure or from the Caratheodory extension theorem

measure-theoryouter-measurereal-analysis

I have just started to study measure theory and I have a question. But before presenting it, I will provide the context from which it comes.

Given a nonempty set $\Omega$, we say that a set function $\mu$ defined on a algebra $\mathcal{F}\subseteq\mathcal{P}(\Omega)$ is a measure if

  • $\mu(A)\geq 0$ for all $A\in\mathcal{F}$,

  • $\mu(\varnothing) = 0$,

  • $\mu$ satisfies the countable additivity property.

We say the set function $\mu^{*}:\mathcal{P}(\Omega)\to[0,+\infty)$ is an outer measure if $\mu^{*}(\varnothing) = 0$, it satisfies the monotonicity property and the countable subadditivity property. We also say that $A\subseteq\Omega$ is $\mu^{*}$-measurable if, for any set $E\subseteq\Omega$, one has that
\begin{align*}
\mu^{*}(E) = \mu^{*}(E\cap A) + \mu^{*}(E\cap A^{c})
\end{align*}

Then we have the following theorem:

Let $\mu^{*}$ be an outer measure on $\mathcal{P}(\Omega)$. Let $\mathcal{M} := \mathcal{M}_{\mu^{*}} := \{A:A\,\text{is}\,\mu^{*}\text{-measurable}\}$. Then

  • $\mathcal{M}$ is a $\sigma$-algebra
  • $\mu^{*}$ restriced to $\mathcal{M}$ is a measure, and
  • $\mu^{*}(A) = 0$ implies that $\mathcal{P}(A)\subset\mathcal{M}$.

This result makes $(\Omega,\mathcal{M}_{\mu^{*}},\mu^{*})$ a complete measure space.

Moreover, it gives an inexhaustible source method to construct measure spaces (as far as I have understood).

We may now state the Caratheodory's extension theorem, which says:

Let $\mu$ be a measure on a semi-algebra $\mathcal{C}$ and let $\mu^{*}$ be the set function induced by $\mu$ defined on $\mathcal{P}(\Omega)$ s.t.
\begin{align*}
\mu^{*}(A) = \inf\left\{\sum_{i=1}^{\infty}\mu(A_{i}):\{A_{n}\}_{n\geq1}\subset\mathcal{C},\,A\subset\bigcup_{i=1}^{\infty}A_{n}\right\}
\end{align*}

Then we have that

  • $\mu^{*}$ is an outer measure,

  • $\mathcal{C}\subset\mathcal{M}_{\mu^{*}}$, and

  • $\mu^{*} = \mu$ on $\mathcal{C}$

Now let us consider the semialgebra
\begin{align*}
\mathcal{C} = \{(a,b]:-\infty\leq a\leq b<\infty\}\cup\{(a,\infty):-\infty\leq a < \infty\}
\end{align*}
as well as the nondecreasing functions $F:\textbf{R}\to\textbf{R}$ which induces the following measure on $\mathcal{C}$:
\begin{align*}
\begin{cases}
\mu_{F}((a,b]) = F(b+) – F(a+)\\\\
\mu_{F}((a,\infty)) = F(\infty) – F(a+)
\end{cases}
\end{align*}

Let $(\textbf{R},\mathcal{M}_{\mu^{*}_{F}},\mu^{*}_{F})$ be the Caratheodory extesion of $\mu_{F}$. Then the book defines such measure space as the Lebesgue-Stieltjes measure space and $\mu^{*}_{F}$ is the Lebesgue-Stieltjes measure generated by $F$.

My question is: are there other important measure spaces which are not obtained directly from an outer measure or from the Caratheodory extension theorem?

I am new to this so any contribution is appreciated.

Best Answer

Another extension result provides a different construction of measures. It starts with a $\sigma$-finite “measure” $\nu$ that is only defined on an algebra $\mathcal{A}$ of sets (meaning, it satisfies countable additivity $\nu(\bigcup_n Q_n) = \sum_n \nu(Q_n)$ only in case the countable union belongs to $\mathcal{A}$). Then there is a unique measure $\nu'$ defined on $\sigma(A)$ such that extends $\nu$ (Theorem A in Sect. 13 of Halmos Measure Theory).

This construction is used, for instance, to the define the product measure. Given measure spaces $\langle X_i, \mathcal{F}_i, \mu_i\rangle$ ($i=1,2$), we can define $\nu$ on measurable rectangles of $X_1\times X_2$ by $$ \nu(Q_1\times Q_2) = \mu_1(Q_1)\cdot\mu_2(Q_2) $$ Since every element of the algebra $\mathcal{A}$ generated by rectangles is a finite disjoint union of rectangles, this works as a definition of $\mathcal{A}$. Hence you cant construct the product of the two measure spaces, $$ \langle X_1\times X_2, \mathcal{F}_1\otimes \mathcal{F}_2, \mu_1\times\mu_2\rangle, $$ where $\mathcal{F}_1\otimes \mathcal{F}_2$ is $\sigma(\mathcal{A})$ and $\mu_1\times\mu_2$ is the extension of $\nu$.

In general, the product measure space is not complete, even if $\langle X_i, \mathcal{F}_i, \mu_i\rangle$ are; for instance, this happens when the factors are taken to be $\mathbb{R}$ with the Lebesgue measurable sets and Lebesgue measure. So it is not obtained by Carathéodory.

A final word is that the measurable space $\langle X_1\times X_2, \mathcal{F}_1\otimes \mathcal{F}_2\rangle$ is not an ad-hoc construction, it is indeed the categorical product of the measurable spaces $\langle X_i, \mathcal{F}_i\rangle$.

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