Editing Measure (mathematics) (section)

====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>