Editing Measure (mathematics) (section)

===Semifinite measures===
Let <math>X</math> be a set, let <math>{\cal A}</math> be a sigma-algebra on <math>X,</math> and let <math>\mu</math> be a measure on <math>{\cal A}.</math> We say <math>\mu</math> is '''semifinite''' to mean that for all <math>A\in\mu^\text{pre}\{+\infty\},</math> <math>{\cal P}(A)\cap\mu^\text{pre}(\R_{>0})\ne\emptyset.</math>{{sfn|Mukherjea|Pothoven|1985|p=90}} 

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====
* Every sigma-finite measure is semifinite.
* Assume <math>{\cal A}={\cal P}(X),</math> let <math>f:X\to[0,+\infty],</math> and assume <math>\mu(A)=\sum_{a\in A}f(a)</math> for all <math>A\subseteq X.</math>
** We have that <math>\mu</math> is sigma-finite if and only if <math>f(x)<+\infty</math> for all <math>x\in X</math> and <math>f^\text{pre}(\R_{>0})</math> is countable. We have that <math>\mu</math> is semifinite if and only if <math>f(x)<+\infty</math> for all <math>x\in X.</math>{{sfn|Folland|1999|p=25}}
** Taking <math>f=X\times\{1\}</math> above (so that <math>\mu</math> is counting measure on <math>{\cal P}(X)</math>), we see that counting measure on <math>{\cal P}(X)</math> is
*** sigma-finite if and only if <math>X</math> is countable; and
*** semifinite (without regard to whether <math>X</math> is countable). (Thus, counting measure, on the power set <math>{\cal P}(X)</math> of an arbitrary uncountable set <math>X,</math> gives an example of a semifinite measure that is not sigma-finite.)
* Let <math>d</math> be a complete, separable metric on <math>X,</math> let <math>{\cal B}</math> be the [[Borel sigma-algebra]] induced by <math>d,</math> and let <math>s\in\R_{>0}.</math> Then the [[Hausdorff measure]] <math>{\cal H}^s|{\cal B}</math> is semifinite.{{sfn|Edgar|1998|loc=Theorem 1.5.2, p. 42}}
* Let <math>d</math> be a complete, separable metric on <math>X,</math> let <math>{\cal B}</math> be the [[Borel sigma-algebra]] induced by <math>d,</math> and let <math>s\in\R_{>0}.</math> Then the [[Packing dimension#Definitions|packing measure]] <math>{\cal H}^s|{\cal B}</math> is semifinite.{{sfn|Edgar|1998|loc=Theorem 1.5.3, p. 42}}

====Involved example====
The zero measure is sigma-finite and thus semifinite. In addition, the zero measure is clearly less than or equal to <math>\mu.</math> It can be shown there is a greatest measure with these two properties:
{{Math theorem|name=Theorem (semifinite part){{sfn|Nielsen|1997|loc=Exercise 11.30, p. 159}}|math_statement=
For any measure <math>\mu</math> on <math>{\cal A},</math> there exists, among semifinite measures on <math>{\cal A}</math> that are less than or equal to <math>\mu,</math> a [[Greatest element and least element|greatest]] element <math>\mu_\text{sf}.</math>
}}
We say the '''semifinite part''' of <math>\mu</math> to mean the semifinite measure <math>\mu_\text{sf}</math> defined in the above theorem. We give some nice, explicit formulas, which some authors may take as definition, for the semifinite part:
* <math>\mu_\text{sf}=(\sup\{\mu(B):B\in{\cal P}(A)\cap\mu^\text{pre}(\R_{\ge0})\})_{A\in{\cal A}}.</math>{{sfn|Nielsen|1997|loc=Exercise 11.30, p. 159}}
* <math>\mu_\text{sf}=(\sup\{\mu(A\cap B):B\in\mu^\text{pre}(\R_{\ge0})\})_{A\in{\cal A}}\}.</math>{{sfn|Fremlin|2016|loc=Section 213X, part (c)}}
* <math>\mu_\text{sf}=\mu|_{\mu^\text{pre}(\R_{>0})}\cup\{A\in{\cal A}:\sup\{\mu(B):B\in{\cal P}(A)\}=+\infty\}\times\{+\infty\}\cup\{A\in{\cal A}:\sup\{\mu(B):B\in{\cal P}(A)\}<+\infty\}\times\{0\}.</math>{{sfn|Royden|Fitzpatrick|2010|loc=Exercise 17.8, p. 342}}

Since <math>\mu_\text{sf}</math> is semifinite, it follows that if <math>\mu=\mu_\text{sf}</math> then <math>\mu</math> is semifinite. It is also evident that if <math>\mu</math> is semifinite then <math>\mu=\mu_\text{sf}.</math>

====Non-examples====
Every ''<math>0-\infty</math> measure'' that is not the zero measure is not semifinite. (Here, we say ''<math>0-\infty</math> measure'' to mean a measure whose range lies in <math>\{0,+\infty\}</math>: <math>(\forall A\in{\cal A})(\mu(A)\in\{0,+\infty\}).</math>) Below we give examples of <math>0-\infty</math> measures that are not zero measures.
* Let <math>X</math> be nonempty, let <math>{\cal A}</math> be a <math>\sigma</math>-algebra on <math>X,</math> let <math>f:X\to\{0,+\infty\}</math> be not the zero function, and let <math>\mu=(\sum_{x\in A}f(x))_{A\in{\cal A}}.</math> It can be shown that <math>\mu</math> is a measure.
** <math>\mu=\{(\emptyset,0)\}\cup({\cal A}\setminus\{\emptyset\})\times\{+\infty\}.</math>{{sfn|Hewitt|Stromberg|1965|loc=part (b) of Example 10.4, p. 127}}
*** <math>X=\{0\},</math> <math>{\cal A}=\{\emptyset,X\},</math> <math>\mu=\{(\emptyset,0),(X,+\infty)\}.</math>{{sfn|Fremlin|2016|loc=Section 211O, p. 15}}
* Let <math>X</math> be uncountable, let <math>{\cal A}</math> be a <math>\sigma</math>-algebra on <math>X,</math> let <math>{\cal C}=\{A\in{\cal A}:A\text{ is countable}\}</math> be the countable elements of <math>{\cal A},</math> and let <math>\mu={\cal C}\times\{0\}\cup({\cal A}\setminus{\cal C})\times\{+\infty\}.</math> It can be shown that <math>\mu</math> is a measure.{{sfn|Mukherjea|Pothoven|1985|p=90}}

====Involved non-example====
{{Blockquote
|text=Measures that are not semifinite are very wild when restricted to certain sets.<ref group=Note>One way to rephrase our definition is that <math>\mu</math> is semifinite if and only if <math>(\forall A\in\mu^\text{pre}\{+\infty\})(\exists B\subseteq A)(0<\mu(B)<+\infty).</math> Negating this rephrasing, we find that <math>\mu</math> is not semifinite if and only if <math>(\exists A\in\mu^\text{pre}\{+\infty\})(\forall B\subseteq A)(\mu(B)\in\{0,+\infty\}).</math> For every such set <math>A,</math> the subspace measure induced by the subspace sigma-algebra induced by <math>A,</math> i.e. the restriction of <math>\mu</math> to said subspace sigma-algebra, is a <math>0-\infty</math> measure that is not the zero measure.</ref> Every measure is, in a sense, semifinite once its <math>0-\infty</math> part (the wild part) is taken away.
|author=A. Mukherjea and K. Pothoven
|source=''Real and Functional Analysis, Part A: Real Analysis'' (1985)
}}
{{Math theorem|name=Theorem (Luther decomposition){{sfn|Luther|1967|loc=Theorem 1}}{{sfn|Mukherjea|Pothoven|1985|loc=part (b) of Proposition 2.3, p. 90}}|math_statement=
For any measure <math>\mu</math> on <math>{\cal A},</math> there exists a <math>0-\infty</math> measure <math>\xi</math> on <math>{\cal A}</math> such that <math>\mu=\nu+\xi</math> for some semifinite measure <math>\nu</math> on <math>{\cal A}.</math> In fact, among such measures <math>\xi,</math> there exists a [[Greatest element and least element|least]] measure <math>\mu_{0-\infty}.</math> Also, we have <math>\mu=\mu_\text{sf}+\mu_{0-\infty}.</math>
}}
We say the '''<math>\mathbf{0-\infty}</math> part''' of <math>\mu</math> to mean the measure <math>\mu_{0-\infty}</math> defined in the above theorem. Here is an explicit formula for <math>\mu_{0-\infty}</math>: <math>\mu_{0-\infty}=(\sup\{\mu(B)-\mu_\text{sf}(B):B\in{\cal P}(A)\cap\mu_\text{sf}^\text{pre}(\R_{\ge0})\})_{A\in{\cal A}}.</math>

====Results regarding semifinite measures====
<!--Help appreciated, please add results / check the c.l.d. product measure thing-->
* Let <math>\mathbb F</math> be <math>\R</math> or <math>\C,</math> and let <math>T:L_\mathbb{F}^\infty(\mu)\to\left(L_\mathbb{F}^1(\mu)\right)^*:g\mapsto T_g=\left(\int fgd\mu\right)_{f\in L_\mathbb{F}^1(\mu)}.</math> Then <math>\mu</math> is semifinite if and only if <math>T</math> is injective.{{sfn|Fremlin|2016|loc=part (a) of Theorem 243G, p. 159}}{{sfn|Fremlin|2016|loc=Section 243K, p. 162}} (This result has import in the study of the [[Lp space#Dual spaces|dual space of <math>L^1=L_\mathbb{F}^1(\mu)</math>]].)
* Let <math>\mathbb F</math> be <math>\R</math> or <math>\C,</math> and let <math>{\cal T}</math> be the topology of convergence in measure on <math>L_\mathbb{F}^0(\mu).</math> Then <math>\mu</math> is semifinite if and only if <math>{\cal T}</math> is Hausdorff.{{sfn|Fremlin|2016|loc=part (a) of the Theorem in Section 245E, p. 182}}{{sfn|Fremlin|2016|loc=Section 245M, p. 188}}
* (Johnson) Let <math>X</math> be a set, let <math>{\cal A}</math> be a sigma-algebra on <math>X,</math> let <math>\mu</math> be a measure on <math>{\cal A},</math> let <math>Y</math> be a set, let <math>{\cal B}</math> be a sigma-algebra on <math>Y,</math> and let <math>\nu</math> be a measure on <math>{\cal B}.</math> If <math>\mu,\nu</math> are both not a <math>0-\infty</math> measure, then both <math>\mu</math> and <math>\nu</math> are semifinite if and only if [[Product measure|<math>(\mu\times_\text{cld}\nu)</math>]]<math>(A\times B)=\mu(A)\nu(B)</math> for all <math>A\in{\cal A}</math> and <math>B\in{\cal B}.</math> (Here, <math>\mu\times_\text{cld}\nu</math> is the measure defined in Theorem 39.1 in Berberian '65.{{sfn|Berberian|1965|loc=Theorem 39.1, p. 129}}) <!--To check: Is this actually the ''c.l.d. product measure''?{{sfn|Fremlin|2016|loc=Definition 251F, p. 206}}-->