Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Measure (mathematics)
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Other properties== ===Completeness=== {{Main|Complete measure}} A measurable set <math>X</math> is called a ''[[null set]]'' if <math>\mu(X) = 0.</math> A subset of a null set is called a ''negligible set''. A negligible set need not be measurable, but every measurable negligible set is automatically a null set. A measure is called ''complete'' if every negligible set is measurable. A measure can be extended to a complete one by considering the σ-algebra of subsets <math>Y</math> which differ by a negligible set from a measurable set <math>X,</math> that is, such that the [[symmetric difference]] of <math>X</math> and <math>Y</math> is contained in a null set. One defines <math>\mu(Y)</math> to equal <math>\mu(X).</math> ==="Dropping the Edge"=== If <math>f:X\to[0,+\infty]</math> is <math>(\Sigma,{\cal B}([0,+\infty]))</math>-measurable, then <math display=block>\mu\{x\in X: f(x) \geq t\} = \mu\{x\in X: f(x) > t\}</math> for [[almost everywhere|almost all]] <math>t \in [-\infty,\infty].</math><ref>{{citation | last = Fremlin | first = D. H. | title = Measure Theory | volume = 2 | year = 2010 | edition = Second | page = 221}}</ref> This property is used in connection with [[Lebesgue integral]]. {{math proof| proof= Both <math>F(t) := \mu\{x\in X : f(x) > t\}</math> and <math>G(t) := \mu\{x\in X : f(x) \geq t\}</math> are monotonically non-increasing functions of <math>t,</math> so both of them have [[Discontinuities of monotone functions|at most countably many discontinuities]] and thus they are continuous almost everywhere, relative to the Lebesgue measure. If <math>t < 0</math> then <math>\{x\in X : f(x) \geq t\} = X = \{x\in X : f(x) > t\},</math> so that <math>F(t) = G(t),</math> as desired. If <math>t</math> is such that <math>\mu\{x\in X : f(x) > t\} = +\infty</math> then [[#Monotonicity|monotonicity]] implies <math display=block>\mu\{x\in X : f(x) \geq t\} = +\infty,</math> so that <math>F(t) = G(t),</math> as required. If <math>\mu\{x\in X : f(x) > t\} = +\infty</math> for all <math>t</math> then we are done, so assume otherwise. Then there is a unique <math>t_0 \in \{-\infty\} \cup [0,+\infty) </math> such that <math>F</math> is infinite to the left of <math>t</math> (which can only happen when <math>t_0 \geq 0</math>) and finite to the right. Arguing as above, <math>\mu\{x\in X : f(x) \geq t\} = +\infty </math> when <math>t < t_0.</math> Similarly, if <math>t_0 \geq 0</math> and <math>F\left(t_0\right) = +\infty</math> then <math>F\left(t_0\right) = G\left(t_0\right).</math> For <math>t > t_0,</math> let <math>t_n</math> be a monotonically non-decreasing sequence converging to <math>t.</math> The monotonically non-increasing sequences <math>\{x\in X : f(x) > t_n\}</math> of members of <math>\Sigma</math> has at least one finitely <math>\mu</math>-measurable component, and <math display=block>\{x\in X : f(x) \geq t\} = \bigcap_n \{x\in X : f(x) > t_n\}.</math> Continuity from above guarantees that <math display=block>\mu\{x\in X : f(x) \geq t\} = \lim_{t_n \uparrow t} \mu\{x\in X : f(x) > t_n\}.</math> The right-hand side <math>\lim_{t_n \uparrow t} F\left(t_n\right)</math> then equals <math>F(t) = \mu\{x\in X : f(x) > t\}</math> if <math>t</math> is a point of continuity of <math>F.</math> Since <math>F</math> is continuous almost everywhere, this completes the proof. }} ===Additivity=== Measures are required to be countably additive. However, the condition can be strengthened as follows. For any set <math>I</math> and any set of nonnegative <math>r_i,i\in I</math> define: <math display=block>\sum_{i\in I} r_i=\sup\left\lbrace\sum_{i\in J} r_i : |J|<\infty, J\subseteq I\right\rbrace.</math> That is, we define the sum of the <math>r_i</math> to be the supremum of all the sums of finitely many of them. A measure <math>\mu</math> on <math>\Sigma</math> is <math>\kappa</math>-additive if for any <math>\lambda<\kappa</math> and any family of disjoint sets <math>X_\alpha,\alpha<\lambda</math> the following hold: <math display=block>\bigcup_{\alpha\in\lambda} X_\alpha \in \Sigma</math> <math display=block>\mu\left(\bigcup_{\alpha\in\lambda} X_\alpha\right) = \sum_{\alpha\in\lambda}\mu\left(X_\alpha\right).</math> The second condition is equivalent to the statement that the [[Ideal (set theory)|ideal]] of null sets is <math>\kappa</math>-complete. ===Sigma-finite measures=== {{Main|Sigma-finite measure}} A measure space <math>(X, \Sigma, \mu)</math> is called finite if <math>\mu(X)</math> is a finite real number (rather than <math>\infty</math>). Nonzero finite measures are analogous to [[probability measure]]s in the sense that any finite measure <math>\mu</math> is proportional to the probability measure <math>\frac{1}{\mu(X)}\mu.</math> A measure <math>\mu</math> is called ''σ-finite'' if <math>X</math> can be decomposed into a countable union of measurable sets of finite measure. Analogously, a set in a measure space is said to have a ''σ-finite measure'' if it is a countable union of sets with finite measure. For example, the [[real number]]s with the standard [[Lebesgue measure]] are σ-finite but not finite. Consider the [[closed interval]]s <math>[k, k+1]</math> for all [[integer]]s <math>k;</math> there are countably many such intervals, each has measure 1, and their union is the entire real line. Alternatively, consider the [[real number]]s with the [[counting measure]], which assigns to each finite set of reals the number of points in the set. This measure space is not σ-finite, because every set with finite measure contains only finitely many points, and it would take uncountably many such sets to cover the entire real line. The σ-finite measure spaces have some very convenient properties; σ-finiteness can be compared in this respect to the [[Lindelöf space|Lindelöf property]] of topological spaces.{{OR inline|date=May 2022}} They can be also thought of as a vague generalization of the idea that a measure space may have 'uncountable measure'. ===Strictly localizable measures=== {{Main|Decomposable measure}} ===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}}--> ===Localizable measures=== Localizable measures are a special case of semifinite measures and a generalization of sigma-finite 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> * 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 localizable if and only if <math>T</math> is bijective (if and only if <math>L_\mathbb{F}^\infty(\mu)</math> "is" <math>L_\mathbb{F}^1(\mu)^*</math>).{{sfn|Fremlin|2016|loc=part (b) of Theorem 243G, p. 159}}{{sfn|Fremlin|2016|loc=Section 243K, p. 162}} ===s-finite measures=== {{Main|s-finite measure}} A measure is said to be s-finite if it is a countable sum of finite measures. S-finite measures are more general than sigma-finite ones and have applications in the theory of [[stochastic processes]].
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)