Tag Archive: measures

Definition. Let X be a set equipped with a \sigma-algebra \mathcal M. A measure on \mathcal M (or on (X, \mathcal M), or simply on X if \mathcal M\to[0, \infty] such that

1). \mu(\emptyset)=0.

2). (Countable additivity) if \{E_j\}_1^\infty is a sequence of disjoint sets in \mathcal M, then \mu(\cup_1^\infty E_j)=\sum_1^\infty\mu(E_j).

2)’. if E_1, ..., E_n are disjoint sets in \mathcal M, then \mu(cup_1^n E_j)=\sum_1^n\mu(E_j),

because one can take E_j=\emptyset for j>n. A function \mu that satisfies 1) and 2)’ but not necessarily 2) is called a finitely additive measure.


If X is a set and \mathcal M\subset\mathcal P(X) is a \sigma-algebra, (X, \mathcal M) is called a measurable space and the sets in \mathcal M are called measurable sets. If \mu is a measure on (X, \mathcal M), then (X, \mathcal M, \mu) is called a measure space.


Let (X, \mathcal M, \mu) be a measure space. If \mu(X)<\infty (which implies that \mu(E)<\infty for all E\in\mathcal M since \mu(X)=\mu(E)+\mu(E^c)), \mu is called finite. If X=\cup_1^\infty E_j where E_j\in\mathcal M and \mu(E_j)<\infty for all j, the set E is said to be \sigma-finite for \mu. More generally, if E=\cup_1^\infty E_j where E_j\in\mathcal M and \mu(E_j)<\infty for all j, the set E is said to be \sigma-finite for \mu. If for each E\in\mathcal M with \mu(E)=\infty there exists F\in\mathcal M with F\subset E and 0<\mu(F)<\infty, \mu is called semifinite.


Remark. Every \sigma-finite measure is semifinite, but not conversely.


Examples. 1) Let X be any nonempty set, \mathcal M=\mathcal P(X), and f any function from X to [0, \infty]. Then f determines a measure \mu on \mathcal M by the formula \mu(E)=\sum_{x\in E}f(x). \mu is semifinite iff f(x)<\infty for every x\in X, and \mu is \sigma-finite iff \mu is semifinite and \{x: f(x)>0\} is countable.

a. If f(x)=1 for all x, \mu is called counting measure;

b. If for some x_0\in X, f is defined by f(x_0)=1 and f(x)=0 for x\not=x_0, \mu is called the point mass or Dirac measure at x_0.


2) Let X be an uncountable set, and let \mathcal M be the \sigma-algebra of countable or co-countable sets. The function \mu on \mathcal M defined by \mu(E)=0 if E is countable and \mu(E)=1 if E is co-countable is easily seen to be a measure.


3) Let X be an infinite set and M=\mathcal P(X). Define \mu(E)=0 if E is finite, \mu(E)=\infty if E is infinite. Then \mu is a finitely additive measure but not a measure.


Theorem. Let (X, \mathcal M, \mu) be a measure space.

a). (Monotonicity) If E, F\in\mathcal M and E\subset F, then $\mu(E)\le\mu(F)$.

b). (Subadditivity) If \{E_j\}_1^\infty\subset \mathcal M, then \mu(\cup_1^\infty E_j)\le\sum_1^\infty \mu(E_j).

c). (Continuity from below) If \{E_j\}_1^\infty\subset \mathcal M and E_1\subset E_2\subset..., then \mu(\cup_1^\infty E_j)=\lim_{j\to\infty}\mu(E_j).

d). (Continuity from above) If \{E_j\}_1^\infty\subset \mathcal M, E_1\supset E_2\supset..., and \mu(E_1)<\infty, then \mu(\cap_1^\infty E_j)=\lim_{j\to\infty}\mu(E_j).

Proof. a) If E\subset F, then \mu(F)=\mu(E)+\mu(F\setminus E)\ge\mu(E).

b) Let F_1=E_1 and F_k=E_k\setminus(\cup_1^{k-1} E_j) for k>1. Then the F_k's are disjoint and \cup_1^n F_j=\cup_1^n E_j for all n. Therefore, by a)

\mu(cup_1^\infty E_j)=\mu(\cup_1^\infty F_j)=\sum_1^\infty\mu(F_j)\le\sum_1^\infty\mu(E_j).

c) Setting E_0=\emptyset, we have

\mu(\cup_1^\infty E_j)=\sum_1^\infty\mu(E_j\setminus E_{j-1})=\lim_{n\to\infty}\sum_1^n\mu(E_j\setminus E_{j-1})=\lim_{n\to\infty}\mu(E_n).

d) Let F_j=E_1\setminus E_j; then F_1\subset F_2\subset ..., \mu(E_1)=\mu(F_j)+\mu(E_j), and \cup_1^\infty F_j=E_1\setminus(\cap_1^\infty E_j). By c) then

\mu(\cap_1^\infty E_j)+\lim_{j\to\infty}\mu(F_j)=\mu(\cap_1^\infty E_j)+\lim_{j\to\infty}[\mu(E_1)-\mu(E_j)].

Since \mu(E_1)<\infty, we may subtract it from both sides to yield the desired result. \Box

Remark.  The condition \mu(E_1)<\infty in d) could be replaced by \mu(E_n)<\infty for some n>1, as the first n-1 E_j‘s can be discarded from the sequence without affecting the intersection. However, some finiteness assumption is necessary, as it can happen that \mu(E_j)=\infty for all j but \mu(cap_1^\infty E_j)<\infty.

Definition. If (X, \mathcal M, \mu) is a measure space, a set E\in\mathcal M such that \mu(E)=0 is called a null set. By subadditivity, any countable union of null sets is a null set, a fact which we shall use frequently. If a statement about points x\in X is true except for x in some every x. (If more precision is needed, we shall speak of a \mu-null set, or \mu-almost everywhere).

If \mu(E)=0 and F\subset E, then \mu(F)=0 by monotonicity provided that F\in\mathcal M, but in general it need not be true that F\in\mathcal M. A measure whose domain includes all subsets of null sets is called complete. Completeness can sometimes obviate annoying technical points, and it can always be achieved by enlarging the domain of \mu, as follows.

Theorem. Suppose that (X, \mathcal M, \mu) is a measure space. Let \mathcal N=\{N\in\mathcal: \mu(N)=0\} and \bar{\mathcal M}=\{E\cup F: E\in\mathcal M\} and F\subset N for some N\in\mathcal N\}. Then \bar{\mathcal M} is a \sigma-algebra, and there is a unique extension \bar{\mu} of \mu to a complete measure on \bar{\mathcal M}.

Proof. Since \mathcal M and \mathcal N are closed under countable unions, so is \bar{\mathcal M}. If E\cup F\in\bar{\mathcal M} where E\in\mathcal M and F\subset N\in\mathcal N, we can assume that E\cap F=\emptyset (otherwise, replace F and N by F\setminus E and N\setminus E). Then E\cup F=(E\cup N)\cap(N^c\cup F), so (E\cup F)^c=(E\cup N)^c\cup(N\setminus F). But (E\cup N)^c\in\mathcal M and N\setminus F\subset N, so that (E\cup F)^c\in\bar{\mathcal M}. Thus \bar{M} is a \sigma-algebra.

If E\cup F\in\bar{\mathcal M} as above, we set \bar\mu(E\cup F)=\mu(E). This is well defined, since if E_1\cup F_1=E_2\cup F_2 where $F_j\subset N_j\in\mathcal N$, then $E_1\subset E_2\cup N_2$ and so \mu(E_1)\le\mu(E_2)+\mu(N_2)=\mu(E_2), and likewise \mu(E_2)\le\mu(E_1). It is easily verified that \bar\mu is a complete measure on \mu\mathcal M, and that \bar\mu is the only measure on \mathcal M that extends \mu.

Remark. The measure \bar\mu is called the completion of \mu, and \bar{\mathcal M} is called the completion of {\mathcal M} with respect to \mu.

“Whenever you meet a ghost, don’t run away, because the ghost will capture the substance of your fear and materialize itself out of your own substance… it will take over all your own vitality… So then, whenever confronted with a ghost, walk straight into it and it will disappear.” ~ Allan Watts


[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 24-27.

[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.


For n\in\mathbb N we would like to have a function \mu that assigns to each E\subset\mathbb R^n a number \mu(E)\in[0, \infty], the n-dimensional measure of E, such that \mu(E) is given by the usual integral formulas when the latter apply. The function \mu has the following properties:

1) If E_1, E_2, ... is a finite or infinite sequence of disjoint sets, then

\mu(E_1\cup E_2\cup ...)=\mu(E_1)+\mu(E_2)+....

2) If E is congruent to F (that is, if E can be transformed into F by translations, rotations, and reflections), then \mu(E)=\mu(F).

3) \mu(Q)=1, where Q is the unit cube

Q=\{x\in\mathbb R^n: 0\le x_j\le 1 for j=1, ..., n\}.

However, 1)-3) are mutually inconsistent. To see why, we can define an equivalence relation on [0, 1) by declaring that x~y iff x-y\in Q is rational. Let N be a subset of [0, 1) that contains precisely one member of each equivalence class. Let R=\mathbb Q\cap[0, 1), and for each r\in R let

N_r=\{x+r: x\in N\cap[0, 1-r)\}\cup\{x+r-1: x\in N\cap[1-r, 1)\}

To obtain N_r, shift N to the right by r units and then shift the part that sticks out beyond [0, 1) one unit to the left. Then N_r\subset[0, 1), every x\in[0, 1) belongs to precisely one N_r. Indeed, if y is the element of N that belongs to the equivalence class of x, then x\in N_r where r=x-y if x\ge y or r=x-y+1 if x<y; on the other hand, if x\in N_r\cap N_s, then x-r (or x-r+1) and x-s (or x-s+1) would be distinct elements of N belonging to the same equivalence class, which is impossible.

Suppose that now \mu: \mathcal P(\mathbb R)\to [0, \infty] satisfies (i), (ii), and (iii). By (i) and (ii),

\mu(N)=\mu(N\cap[0, 1-r))+\mu(N\cap[1-r, 1))=\mu(N_r)

for any r\in R. Also, since R is countable and $[0, 1)$ is the disjoint union of the N_r‘s,

\mu([0, 1))=\sum_{r\in R}\mu(N_r)

by (1) again. But \mu([0,1))=1 by 3), since $\mu(N_r)=\mu(N)$, the sum on the right is either 0 (if \mu(N)=0) or \infty (if \mu(N)>0). Hence no such \mu can exist.

But in 1924, Banach and Tarski proved the following amazing result:

Let U and V be arbitrary bounded open sets in \mathbb R^n, n\ge 3. There exist k\in\mathbb N and subsets E_1, ..., E_k, F_1, ..., F_k of \mathbb R^n such that:

1) the E_j‘s are disjoint and their union is U;

2) the F_j‘s are disjoint and their union is V;

3) E_j is congruent to F_j for j=1,..., k.

The construction of sets E_j and F_j depends on the axiom of choice. But their existence precludes the construction of any \mu: \mathcal P(\mathbb R^n)\to[0,\infty] that assigns positive, finite values to bounded open sets and satisfies (1) for finite sequences as well as (2).


“Don’t take anything Personally Nothing others do is because of you. What others say and do is a projection of their own reality, their own dream. When you are immune to the opinions and actions of others, you won’t be the victim of needless suffering” ~ Miguel Ruiz


To be continued tomorrow 🙂



[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 19-21.

[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.