Measure of countable unions and intersections

 

Basic properties[edit]

Let ${\displaystyle �}$ be a measure.

Monotonicity[edit]

If ${\displaystyle {�}_1}$ and ${\displaystyle {�}_2}$ are measurable sets with ${\displaystyle {�}_1\subseteq {�}_2}$ then

$${\displaystyle �({�}_1)\le �({�}_2).}$$

Measure of countable unions and intersections[edit]

Subadditivity[edit]

For any countable sequence ${\displaystyle {�}_1,{�}_2,{�}_3,\dots }$ of (not necessarily disjoint) measurable sets ${\displaystyle {�}_{�}}$ in ${\displaystyle \mathrm{\Sigma }:}$

$${\displaystyle �\left(\bigcup _{�=1}^{\mathrm{\infty }}{�}_{�}\right)\le \sum _{�=1}^{\mathrm{\infty }}�({�}_{�}).}$$

Continuity from below[edit]

If ${\displaystyle {�}_1,{�}_2,{�}_3,\dots }$ are measurable sets that are increasing (meaning that ${\displaystyle {�}_1\subseteq {�}_2\subseteq {�}_3\subseteq \dots }$) then the union of the sets ${\displaystyle {�}_{�}}$ is measurable and

$${\displaystyle �\left(\bigcup _{�=1}^{\mathrm{\infty }}{�}_{�}\right)=\underset{�\to \mathrm{\infty }}{lim}�({�}_{�})=\underset{�\ge 1}{sup}�({�}_{�}).}$$

Continuity from above[edit]

If ${\displaystyle {�}_1,{�}_2,{�}_3,\dots }$ are measurable sets that are decreasing (meaning that ${\displaystyle {�}_1\supseteq {�}_2\supseteq {�}_3\supseteq \dots }$) then the intersection of the sets ${\displaystyle {�}_{�}}$ is measurable; furthermore, if at least one of the ${\displaystyle {�}_{�}}$ has finite measure then

$${\displaystyle �\left(\bigcap _{�=1}^{\mathrm{\infty }}{�}_{�}\right)=\underset{�\to \mathrm{\infty }}{lim}�({�}_{�})=\underset{�\ge 1}{inf}�({�}_{�}).}$$

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

Other properties[edit]

Completeness[edit]

Main article: Complete measure

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

μ{x : f(x) ≥ t} = μ{x : f(x) > t} (a.e.)[edit]

If ${\displaystyle �:�\to [0,+\mathrm{\infty }]}$ is ${\displaystyle (\mathrm{\Sigma },\mathcal{�}([0,+\mathrm{\infty }]))}$-measurable, then

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

Proof

Both  and ${\displaystyle �(�):=�\{�\in �:�(�)\ge �\}}$ are monotonically non-increasing functions of ${\displaystyle �,}$ so both of them have at most countably many discontinuities and thus they are continuous almost everywhere, relative to the Lebesgue measure. If  then  so that ${\displaystyle �(�)=�(�),}$ as desired.

If ${\displaystyle �}$ is such that  then monotonicity implies

$${\displaystyle �\{�\in �:�(�)\ge �\}=+\mathrm{\infty },}$$

so that ${\displaystyle �(�)=�(�),}$ as required. If  for all ${\displaystyle �}$ then we are done, so assume otherwise. Then there is a unique ${\displaystyle {�}_0\in \{-\mathrm{\infty }\}\cup [0,+\mathrm{\infty })}$ such that ${\displaystyle �}$ is infinite to the left of ${\displaystyle �}$ (which can only happen when ${\displaystyle {�}_0\ge 0}$) and finite to the right. Arguing as above, ${\displaystyle �\{�\in �:�(�)\ge �\}=+\mathrm{\infty }}$ when  Similarly, if ${\displaystyle {�}_0\ge 0}$ and ${\displaystyle �\left({�}_0\right)=+\mathrm{\infty }}$ then ${\displaystyle �\left({�}_0\right)=�\left({�}_0\right).}$

For  let ${\displaystyle {�}_{�}}$ be a monotonically non-decreasing sequence converging to ${\displaystyle �.}$ The monotonically non-increasing sequence  of members of ${\displaystyle \mathrm{\Sigma }}$ has at least one finitely ${\displaystyle �}$-measurable component, and

Continuity from above guarantees that

The right-hand side ${\displaystyle \underset{{�}_{�}\uparrow �}{lim}�\left({�}_{�}\right)}$ then equals  if ${\displaystyle �}$ is a point of continuity of ${\displaystyle �.}$ Since ${\displaystyle �}$ is continuous almost everywhere, this completes the proof.

Additivity[edit]

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

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

A measure ${\displaystyle �}$ on ${\displaystyle \mathrm{\Sigma }}$ is ${\displaystyle �}$-additive if for any  and any family of disjoint sets  the following hold:

$${\displaystyle \bigcup _{�\in �}{�}_{�}\in \mathrm{\Sigma }}$$

$${\displaystyle �\left(\bigcup _{�\in �}{�}_{�}\right)=\sum _{�\in �}�\left({�}_{�}\right).}$$

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

Sigma-finite measures[edit]

Main article: Sigma-finite measure

A measure space ${\displaystyle (�,\mathrm{\Sigma },�)}$ is called finite if ${\displaystyle �(�)}$ is a finite real number (rather than ${\displaystyle \mathrm{\infty }}$). Nonzero finite measures are analogous to probability measures in the sense that any finite measure ${\displaystyle �}$ is proportional to the probability measure ${\displaystyle \frac{1}{�(�)}�.}$ A measure ${\displaystyle �}$ is called σ-finite if ${\displaystyle �}$ 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 ${\displaystyle [�,�+1]}$ for all integers ${\displaystyle �;}$ 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.^{[original research?]} They can be also thought of as a vague generalization of the idea that a measure space may have 'uncountable measure'.