Measure (mathematics)

If ${\displaystyle E_{1}}$ and ${\displaystyle E_{2}}$ are measurable sets with ${\displaystyle E_{1}\subseteq E_{2}}$ then

$${\displaystyle \mu (E_{1})\leq \mu (E_{2}).}$$

For any countable sequence ${\displaystyle E_{1},E_{2},E_{3},\ldots }$ of (not necessarily disjoint) measurable sets ${\displaystyle E_{n}}$ in ${\displaystyle \Sigma$ :}

$${\displaystyle \mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)\leq \sum _{i=1}^{\infty }\mu (E_{i}).}$$

If ${\displaystyle E_{1},E_{2},E_{3},\ldots }$ are measurable sets that are increasing (meaning that ${\displaystyle E_{1}\subseteq E_{2}\subseteq E_{3}\subseteq \ldots }$) then the union of the sets ${\displaystyle E_{n}}$ is measurable and

$${\displaystyle \mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)~=~\lim _{i\to \infty }\mu (E_{i})=\sup _{i\geq 1}\mu (E_{i}).}$$

If ${\displaystyle E_{1},E_{2},E_{3},\ldots }$ are measurable sets that are decreasing (meaning that ${\displaystyle E_{1}\supseteq E_{2}\supseteq E_{3}\supseteq \ldots }$) then the intersection of the sets ${\displaystyle E_{n}}$ is measurable; furthermore, if at least one of the ${\displaystyle E_{n}}$ has finite measure then

$${\displaystyle \mu \left(\bigcap _{i=1}^{\infty }E_{i}\right)=\lim _{i\to \infty }\mu (E_{i})=\inf _{i\geq 1}\mu (E_{i}).}$$

This property is false without the assumption that at least one of the ${\displaystyle E_{n}}$ has finite measure. For instance, for each ${\displaystyle n\in \mathbb {N} ,}$ let ${\displaystyle E_{n}=[n,\infty )\subseteq \mathbb {R} ,}$ which all have infinite Lebesgue measure, but the intersection is empty.

A measurable set ${\displaystyle X}$ is called a null set if ${\displaystyle \mu (X)=0.}$ 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 ${\displaystyle Y}$ which differ by a negligible set from a measurable set ${\displaystyle X,}$ that is, such that the symmetric difference of ${\displaystyle X}$ and ${\displaystyle Y}$ is contained in a null set. One defines ${\displaystyle \mu (Y)}$ to equal ${\displaystyle \mu (X).}$

If ${\displaystyle f:X\to [0,+\infty ]}$ is ${\displaystyle (\Sigma ,{\cal {B}}([0,+\infty ]))}$-measurable, then

$${\displaystyle \mu \{x\in X:f(x)\geq t\}=\mu \{x\in X:f(x)>t\}}$$

for almost all ${\displaystyle t\in X.}$[1] This property is used in connection with Lebesgue integral.

Proof

Both ${\displaystyle F(t):=\mu \{x\in X:f(x)>t\}}$ and ${\displaystyle G(t):=\mu \{x\in X:f(x)\geq t\}}$ are monotonically non-increasing functions of ${\displaystyle t,}$ so both of them have at most countably many discontinuities and thus they are continuous almost everywhere, relative to the Lebesgue measure. If ${\displaystyle t<0}$ then ${\displaystyle \{x\in X:f(x)\geq t\}=X=\{x\in X:f(x)>t\},}$ so that ${\displaystyle F(t)=G(t),}$ as desired.

If ${\displaystyle t}$ is such that ${\displaystyle \mu \{x\in X:f(x)>t\}=+\infty }$ then monotonicity implies

$${\displaystyle \mu \{x\in X:f(x)\geq t\}=+\infty ,}$$

so that ${\displaystyle F(t)=G(t),}$ as required. If ${\displaystyle \mu \{x\in X:f(x)>t\}=+\infty }$ for all ${\displaystyle t}$ then we are done, so assume otherwise. Then there is a unique ${\displaystyle t_{0}\in \{-\infty \}\cup [0,+\infty )}$ such that ${\displaystyle F}$ is infinite to the left of ${\displaystyle t}$ (which can only happen when ${\displaystyle t_{0}\geq 0}$) and finite to the right. Arguing as above, ${\displaystyle \mu \{x\in X:f(x)\geq t\}=+\infty }$ when ${\displaystyle t<t_{0}.}$ Similarly, if ${\displaystyle t_{0}\geq 0}$ and ${\displaystyle F\left(t_{0}\right)=+\infty }$ then ${\displaystyle F\left(t_{0}\right)=G\left(t_{0}\right).}$

For ${\displaystyle t>t_{0},}$ let ${\displaystyle t_{n}}$ be a monotonically non-decreasing sequence converging to ${\displaystyle t.}$ The monotonically non-increasing sequence ${\displaystyle \{x\in X:f(x)>t_{n}\}}$ of members of ${\displaystyle \Sigma }$ has at least one finitely ${\displaystyle \mu }$-measurable component, and

$${\displaystyle \{x\in X:f(x)\geq t\}=\bigcap _{n}\{x\in X:f(x)>t_{n}\}.}$$

Continuity from above guarantees that

$${\displaystyle \mu \{x\in X:f(x)\geq t\}=\lim _{t_{n}\uparrow t}\mu \{x\in X:f(x)>t_{n}\}.}$$

The right-hand side ${\displaystyle \lim _{t_{n}\uparrow t}F\left(t_{n}\right)}$ then equals ${\displaystyle F(t)=\mu \{x\in X:f(x)>t\}}$ if ${\displaystyle t}$ is a point of continuity of ${\displaystyle F.}$ Since ${\displaystyle F}$ is continuous almost everywhere, this completes the proof.

Measures are required to be countably additive. However, the condition can be strengthened as follows. For any set ${\displaystyle I}$ and any set of nonnegative ${\displaystyle r_{i},i\in I}$ define:

$${\displaystyle \sum _{i\in I}r_{i}=\sup \left\lbrace \sum _{i\in J}r_{i}:|J|<\aleph _{0},J\subseteq I\right\rbrace .}$$

That is, we define the sum of the ${\displaystyle r_{i}}$ to be the supremum of all the sums of finitely many of them.

A measure ${\displaystyle \mu }$ on ${\displaystyle \Sigma }$ is ${\displaystyle \kappa }$-additive if for any ${\displaystyle \lambda <\kappa }$ and any family of disjoint sets ${\displaystyle X_{\alpha },\alpha <\lambda }$ the following hold:

$${\displaystyle \bigcup _{\alpha \in \lambda }X_{\alpha }\in \Sigma }$$

$${\displaystyle \mu \left(\bigcup _{\alpha \in \lambda }X_{\alpha }\right)=\sum _{\alpha \in \lambda }\mu \left(X_{\alpha }\right).}$$

Note that the second condition is equivalent to the statement that the ideal of null sets is ${\displaystyle \kappa }$-complete.

A measure space ${\displaystyle (X,\Sigma ,\mu )}$ is called finite if ${\displaystyle \mu (X)}$ is a finite real number (rather than ${\displaystyle \infty }$). Nonzero finite measures are analogous to probability measures in the sense that any finite measure ${\displaystyle \mu }$ is proportional to the probability measure ${\displaystyle {\frac {1}{\mu (X)}}\mu .}$ A measure ${\displaystyle \mu }$ is called σ-finite if ${\displaystyle X}$ 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 numbers with the standard Lebesgue measure are σ-finite but not finite. Consider the closed intervals ${\displaystyle [k,k+1]}$ for all integers ${\displaystyle k;}$ there are countably many such intervals, each has measure 1, and their union is the entire real line. Alternatively, consider the real numbers 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 property of topological spaces. They can be also thought of as a vague generalization of the idea that a measure space may have 'uncountable measure'.

Let ${\displaystyle X}$ be a set, let ${\displaystyle {\cal {A}}}$ be a sigma-algebra on ${\displaystyle X,}$ and let ${\displaystyle \mu }$ be a measure on ${\displaystyle {\cal {A}}.}$ We say ${\displaystyle \mu }$ is semifinite to mean that for all ${\displaystyle A\in \mu ^{\text{pre}}\{+\infty \},}$ ${\displaystyle {\cal {P}}(A)\cap \mu ^{\text{pre}}(\mathbb {R} _{>0})\neq \emptyset .}$[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.)

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

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

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

• ${\displaystyle \mu _{\text{sf}}=(\sup\{\mu (B):B\in {\cal {P}}(A)\cap \mu ^{\text{pre}}(\mathbb {R} _{\geq 0})\})_{A\in {\cal {A}}}.}$[6]
• ${\displaystyle \mu _{\text{sf}}=(\sup\{\mu (A\cap B):B\in \mu ^{\text{pre}}(\mathbb {R} _{\geq 0})\})_{A\in {\cal {A}}}\}.}$[7]
• ${\displaystyle \mu _{\text{sf}}=\mu |_{\mu ^{\text{pre}}(\mathbb {R} _{>0})}\cup \{A\in {\cal {A}}:\sup\{\mu (B):B\in {\cal {P}}(A)\}=+\infty \}\times \{+\infty \}\cup \{A\in {\cal {A}}:\sup\{\mu (B):B\in {\cal {P}}(A)\}<+\infty \}\times \{0\}.}$[8]

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

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

• Let ${\displaystyle X}$ be nonempty, let ${\displaystyle {\cal {A}}}$ be a ${\displaystyle \sigma }$-algebra on ${\displaystyle X,}$ let ${\displaystyle f:X\to \{0,+\infty \}}$ be not the zero function, and let ${\displaystyle \mu =(\sum _{x\in A}f(x))_{A\in {\cal {A}}}.}$ It can be shown that ${\displaystyle \mu }$ is a measure.
  • ${\displaystyle \mu =\{(\emptyset ,0)\}\cup ({\cal {A}}\setminus \{\emptyset \})\times \{+\infty \}.}$[9]
    • ${\displaystyle X=\{0\},}$ ${\displaystyle {\cal {A}}=\{\emptyset ,X\},}$ ${\displaystyle \mu =\{(\emptyset ,0),(X,+\infty )\}.}$[10]
• Let ${\displaystyle X}$ be uncountable, let ${\displaystyle {\cal {A}}}$ be a ${\displaystyle \sigma }$-algebra on ${\displaystyle X,}$ let ${\displaystyle {\cal {C}}=\{A\in {\cal {A}}:A{\text{ is countable}}\}}$ be the countable elements of ${\displaystyle {\cal {A}},}$ and let ${\displaystyle \mu ={\cal {C}}\times \{0\}\cup ({\cal {A}}\setminus {\cal {C}})\times \{+\infty \}.}$ It can be shown that ${\displaystyle \mu }$ is a measure.[2]

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

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

We say the ${\displaystyle \mathbf {0-\infty } }$ part of ${\displaystyle \mu }$ to mean the measure ${\displaystyle \mu _{0-\infty }}$ defined in the above theorem. Here is an explicit formula for ${\displaystyle \mu _{0-\infty }}$: ${\displaystyle \mu _{0-\infty }=(\sup\{\mu (B)-\mu _{\text{sf}}(B):B\in {\cal {P}}(A)\cap \mu _{\text{sf}}^{\text{pre}}(\mathbb {R} _{\geq 0})\})_{A\in {\cal {A}}}.}$

• Let ${\displaystyle \mathbb {F} }$ be ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} ,}$ and let ${\displaystyle T:L_{\mathbb {F} }^{\infty }(\mu )\to \left(L_{\mathbb {F} }^{1}(\mu )\right)^{*}:g\mapsto T_{g}=\left(\int fgd\mu \right)_{f\in L_{\mathbb {F} }^{1}(\mu )}.}$ Then ${\displaystyle \mu }$ is semifinite if and only if ${\displaystyle T}$ is injective.[13][14] (This result has import in the study of the dual space of ${\displaystyle L^{1}=L_{\mathbb {F} }^{1}(\mu )}$.)
• Let ${\displaystyle \mathbb {F} }$ be ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} ,}$ and let ${\displaystyle {\cal {T}}}$ be the topology of convergence in measure on ${\displaystyle L_{\mathbb {F} }^{0}(\mu ).}$ Then ${\displaystyle \mu }$ is semifinite if and only if ${\displaystyle {\cal {T}}}$ is Hausdorff.[15][16]
• (Johnson) Let ${\displaystyle X}$ be a set, let ${\displaystyle {\cal {A}}}$ be a sigma-algebra on ${\displaystyle X,}$ let ${\displaystyle \mu }$ be a measure on ${\displaystyle {\cal {A}},}$ let ${\displaystyle Y}$ be a set, let ${\displaystyle {\cal {B}}}$ be a sigma-algebra on ${\displaystyle Y,}$ and let ${\displaystyle \nu }$ be a measure on ${\displaystyle {\cal {B}}.}$ If ${\displaystyle \mu ,\nu }$ are both not a ${\displaystyle 0-\infty }$ measure, then both ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are semifinite if and only if ${\displaystyle (\mu \times _{\text{cld}}\nu )}$${\displaystyle (A\times B)=\mu (A)\nu (B)}$ for all ${\displaystyle A\in {\cal {A}}}$ and ${\displaystyle B\in {\cal {B}}.}$ (Here, ${\displaystyle \mu \times _{\text{cld}}\nu }$ is the measure defined in Theorem 39.1 in Berberian '65.[17])

Localizable measures are a special case of semifinite measures and a generalization of sigma-finite measures.

Let ${\displaystyle X}$ be a set, let ${\displaystyle {\cal {A}}}$ be a sigma-algebra on ${\displaystyle X,}$ and let ${\displaystyle \mu }$ be a measure on ${\displaystyle {\cal {A}}.}$

• Let ${\displaystyle \mathbb {F} }$ be ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} ,}$ and let ${\displaystyle T:L_{\mathbb {F} }^{\infty }(\mu )\to \left(L_{\mathbb {F} }^{1}(\mu )\right)^{*}:g\mapsto T_{g}=\left(\int fgd\mu \right)_{f\in L_{\mathbb {F} }^{1}(\mu )}.}$ Then ${\displaystyle \mu }$ is localizable if and only if ${\displaystyle T}$ is bijective (if and only if ${\displaystyle L_{\mathbb {F} }^{\infty }(\mu )}$ "is" ${\displaystyle L_{\mathbb {F} }^{1}(\mu )^{*}}$).[18][14]

A measure is said to be s-finite if it is a countable sum of bounded measures. S-finite measures are more general than sigma-finite ones and have applications in the theory of stochastic processes.