Strictly localizable measures

 

Strictly localizable measures[edit]

Main article: Decomposable measure

Semifinite measures[edit]

Let ${\displaystyle �}$ be a set, let ${\displaystyle \mathcal{�}}$ be a sigma-algebra on ${\displaystyle �,}$ and let ${\displaystyle �}$ be a measure on ${\displaystyle \mathcal{�}.}$ We say ${\displaystyle �}$ is semifinite to mean that for all ${\displaystyle �\in {�}^{\text{pre}}\{+\mathrm{\infty }\},}$ ^[2]

Semifinite measures generalize sigma-finite measures, in such a way that some big theorems of measure theory that hold for sigma-finite but not arbitrary measures can be extended with little modification to hold for semifinite measures. (To-do: add examples of such theorems; cf. the talk page.)

Basic examples[edit]

• Every sigma-finite measure is semifinite.
• Assume ${\displaystyle \mathcal{�}=\mathcal{�}(�),}$ let ${\displaystyle �:�\to [0,+\mathrm{\infty }],}$ and assume ${\displaystyle �(�)=\sum _{�\in �}�(�)}$ for all ${\displaystyle �\subseteq �.}$
  • We have that ${\displaystyle �}$ is sigma-finite if and only if  for all ${\displaystyle �\in �}$ and  is countable. We have that ${\displaystyle �}$ is semifinite if and only if  for all ${\displaystyle �\in �.}$^[3]
  • Taking ${\displaystyle �=�\times \{1\}}$ above (so that ${\displaystyle �}$ is counting measure on ${\displaystyle \mathcal{�}(�)}$), we see that counting measure on ${\displaystyle \mathcal{�}(�)}$ is
    • sigma-finite if and only if ${\displaystyle �}$ is countable; and
    • semifinite (without regard to whether ${\displaystyle �}$ is countable). (Thus, counting measure, on the power set ${\displaystyle \mathcal{�}(�)}$ of an arbitrary uncountable set ${\displaystyle �,}$ gives an example of a semifinite measure that is not sigma-finite.)
• Let ${\displaystyle �}$ be a complete, separable metric on ${\displaystyle �,}$ let ${\displaystyle \mathcal{�}}$ be the Borel sigma-algebra induced by ${\displaystyle �,}$ and let  Then the Hausdorff measure ${\displaystyle {\mathcal{�}}^{�}|\mathcal{�}}$ is semifinite.^[4]
• Let ${\displaystyle �}$ be a complete, separable metric on ${\displaystyle �,}$ let ${\displaystyle \mathcal{�}}$ be the Borel sigma-algebra induced by ${\displaystyle �,}$ and let  Then the packing measure ${\displaystyle {\mathcal{�}}^{�}|\mathcal{�}}$ is semifinite.^[5]

Involved example[edit]

The zero measure is sigma-finite and thus semifinite. In addition, the zero measure is clearly less than or equal to ${\displaystyle �.}$ It can be shown there is a greatest measure with these two properties:

Theorem (semifinite part)^[6] — For any measure ${\displaystyle �}$ on ${\displaystyle \mathcal{�},}$ there exists, among semifinite measures on ${\displaystyle \mathcal{�}}$ that are less than or equal to ${\displaystyle �,}$ a greatest element ${\displaystyle {�}_{\text{sf}}.}$

We say the semifinite part of ${\displaystyle �}$ to mean the semifinite measure ${\displaystyle {�}_{\text{sf}}}$ defined in the above theorem. We give some nice, explicit formulas, which some authors may take as definition, for the semifinite part:

• ${\displaystyle {�}_{\text{sf}}=(sup\{�(�):�\in \mathcal{�}(�)\cap {�}^{\text{pre}}({\mathbb{�}}_{\ge 0})\}{)}_{�\in \mathcal{�}}.}$^[6]
• ${\displaystyle {�}_{\text{sf}}=(sup\{�(�\cap �):�\in {�}^{\text{pre}}({\mathbb{�}}_{\ge 0})\}{)}_{�\in \mathcal{�}}\}.}$^[7]
• ^[8]

Since ${\displaystyle {�}_{\text{sf}}}$ is semifinite, it follows that if ${\displaystyle �={�}_{\text{sf}}}$ then ${\displaystyle �}$ is semifinite. It is also evident that if ${\displaystyle �}$ is semifinite then ${\displaystyle �={�}_{\text{sf}}.}$

Non-examples[edit]

Every ${\displaystyle 0-\mathrm{\infty }}$ measure that is not the zero measure is not semifinite. (Here, we say ${\displaystyle 0-\mathrm{\infty }}$ measure to mean a measure whose range lies in ${\displaystyle \{0,+\mathrm{\infty }\}}$: ${\displaystyle (\mathrm{\forall }�\in \mathcal{�})(�(�)\in \{0,+\mathrm{\infty }\}).}$) Below we give examples of ${\displaystyle 0-\mathrm{\infty }}$ measures that are not zero measures.

• Let ${\displaystyle �}$ be nonempty, let ${\displaystyle \mathcal{�}}$ be a ${\displaystyle �}$-algebra on ${\displaystyle �,}$ let ${\displaystyle �:�\to \{0,+\mathrm{\infty }\}}$ be not the zero function, and let ${\displaystyle �=(\sum _{�\in �}�(�){)}_{�\in \mathcal{�}}.}$ It can be shown that ${\displaystyle �}$ is a measure.
  • ${\displaystyle �=\{(\mathrm{\varnothing },0)\}\cup (\mathcal{�}\setminus \{\mathrm{\varnothing }\})\times \{+\mathrm{\infty }\}.}$^[9]
    • ${\displaystyle �=\{0\},}$ ${\displaystyle \mathcal{�}=\{\mathrm{\varnothing },�\},}$ ${\displaystyle �=\{(\mathrm{\varnothing },0),(�,+\mathrm{\infty })\}.}$^[10]
• Let ${\displaystyle �}$ be uncountable, let ${\displaystyle \mathcal{�}}$ be a ${\displaystyle �}$-algebra on ${\displaystyle �,}$ let ${\displaystyle \mathcal{�}=\{�\in \mathcal{�}:�\text{ is countable}\}}$ be the countable elements of ${\displaystyle \mathcal{�},}$ and let ${\displaystyle �=\mathcal{�}\times \{0\}\cup (\mathcal{�}\setminus \mathcal{�})\times \{+\mathrm{\infty }\}.}$ It can be shown that ${\displaystyle �}$ is a measure.^[2]

Involved non-example[edit]

Measures that are not semifinite are very wild when restricted to certain sets.^{[Note 1]} Every measure is, in a sense, semifinite once its ${\displaystyle 0-\mathrm{\infty }}$ part (the wild part) is taken away.

— A. Mukherjea and K. Pothoven, Real and Functional Analysis, Part A: Real Analysis (1985)

Theorem (Luther decomposition)^[11][12] — For any measure ${\displaystyle �}$ on ${\displaystyle \mathcal{�},}$ there exists a ${\displaystyle 0-\mathrm{\infty }}$ measure ${\displaystyle �}$ on ${\displaystyle \mathcal{�}}$ such that ${\displaystyle �=�+�}$ for some semifinite measure ${\displaystyle �}$ on ${\displaystyle \mathcal{�}.}$ In fact, among such measures ${\displaystyle �,}$ there exists a least measure ${\displaystyle {�}_{0-\mathrm{\infty }}.}$ Also, we have ${\displaystyle �={�}_{\text{sf}}+{�}_{0-\mathrm{\infty }}.}$

We say the ${\displaystyle \mathbf{0}\mathbf{-}\mathrm{\infty }}$ part of ${\displaystyle �}$ to mean the measure ${\displaystyle {�}_{0-\mathrm{\infty }}}$ defined in the above theorem. Here is an explicit formula for ${\displaystyle {�}_{0-\mathrm{\infty }}}$: ${\displaystyle {�}_{0-\mathrm{\infty }}=(sup\{�(�)-{�}_{\text{sf}}(�):�\in \mathcal{�}(�)\cap {�}_{\text{sf}}^{\text{pre}}({\mathbb{�}}_{\ge 0})\}{)}_{�\in \mathcal{�}}.}$

Results regarding semifinite measures[edit]

• Let ${\displaystyle \mathbb{�}}$ be ${\displaystyle \mathbb{�}}$ or ${\displaystyle \mathbb{�},}$ and let ${\displaystyle �:{�}_{\mathbb{�}}^{\mathrm{\infty }}(�)\to {\left({�}_{\mathbb{�}}^1(�)\right)}^{\ast }:�\mapsto {�}_{�}={\left(\int ����\right)}_{�\in {�}_{\mathbb{�}}^1(�)}.}$ Then ${\displaystyle �}$ is semifinite if and only if ${\displaystyle �}$ is injective.^[13][14] (This result has import in the study of the dual space of ${\displaystyle {�}^1={�}_{\mathbb{�}}^1(�)}$.)
• Let ${\displaystyle \mathbb{�}}$ be ${\displaystyle \mathbb{�}}$ or ${\displaystyle \mathbb{�},}$ and let ${\displaystyle \mathcal{�}}$ be the topology of convergence in measure on ${\displaystyle {�}_{\mathbb{�}}^0(�).}$ Then ${\displaystyle �}$ is semifinite if and only if ${\displaystyle \mathcal{�}}$ is Hausdorff.^[15][16]
• (Johnson) Let ${\displaystyle �}$ be a set, let ${\displaystyle \mathcal{�}}$ be a sigma-algebra on ${\displaystyle �,}$ let ${\displaystyle �}$ be a measure on ${\displaystyle \mathcal{�},}$ let ${\displaystyle �}$ be a set, let ${\displaystyle \mathcal{�}}$ be a sigma-algebra on ${\displaystyle �,}$ and let ${\displaystyle �}$ be a measure on ${\displaystyle \mathcal{�}.}$ If ${\displaystyle �,�}$ are both not a ${\displaystyle 0-\mathrm{\infty }}$ measure, then both ${\displaystyle �}$ and ${\displaystyle �}$ are semifinite if and only if ${\displaystyle (�{\times }_{\text{cld}}�)}$${\displaystyle (�\times �)=�(�)�(�)}$ for all ${\displaystyle �\in \mathcal{�}}$ and ${\displaystyle �\in \mathcal{�}.}$ (Here, ${\displaystyle �{\times }_{\text{cld}}�}$ is the measure defined in Theorem 39.1 in Berberian '65.^[17])