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

References:

[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.