Measure

Keith A. Lewis

January 26, 2025

Abstract

Do not count twice

A finitely additive measure on a set $S$ is a set function from subsets of $S$ to real numbers $\mu\colon\mathcal{P}(S)\to\boldsymbol{R}$ , where $\mathcal{P}(S) = \{E\subseteq S\}$ is the set of all subsets of $S$ , that satisfies $\mu(E\cup F) = \mu(E) + \mu(F) - \mu(E\cap F)$ for $E,F\subseteq S$ and $\mu(\emptyset) = 0$

Measures don’t count things twice and the measure of nothing is zero.

Exercise. Show if $\nu(E\cup F) = \nu(E) + \nu(F) - \nu(E\cap F)$ for $E,F\subseteq S$ then $\mu = \nu - \nu(\emptyset)$ is measure.

Solution

By $\mu = \nu - \nu(\emptyset)$ we mean $\mu(E) = \nu(E) - \nu(\emptyset)$ for any subset $E\subseteq S$ . Clearly $\mu(E\cup F) = \mu(E) + \mu(F) - \mu(E\cap F)$ for any $E,F\subseteq S$ . Since $\mu(\emptyset) = \nu(\emptyset) - \nu(\emptyset) = 0$ , $\mu$ is a measure.

Exercise. Show if $\mu$ is a measure then $\mu(E\cup F) = \mu(E) + \mu(F)$ for any subsets $E$ and $F$ with empty intersection $E\cap F = \emptyset$ .

Solution

Since $\mu(\emptyset) = 0$ , $\mu(E\cup F) = \mu(E) + \mu(F) - \mu(E\cap F) = \mu(E) + \mu(F) - \mu(\emptyset) = \mu(E) + \mu(F)$ .

Exercise. Show if $\mu$ is a set function with $\mu(E\cup F) = \mu(E) + \mu(F)$ for any subsets $E$ and $F$ having empty intersection then $\mu$ is a measure.

Solution

$\mu(E\cup F) = \mu(E\setminus F) + \mu(E\cap F) + \mu(F\setminus E)$ , $\mu(E) = \mu(E\setminus F) + \mu(E\cap F)$ , and $\mu(F) = \mu(F\setminus E) + \mu(F\cap E)$ so $\mu(E\cup F) - \mu(E) - \mu(F) = -\mu(E\cap F)$ . Also $\mu(\emptyset) = \mu(\emptyset\cup\emptyset) = \mu(\emptyset) + \mu(\emptyset)$ so $\mu(\emptyset) = 0$ .

Exercise. Show if $\mu$ is a measure then $\mu(E) = \mu(E\cap F) + \mu(E\cap F')$ for any subsets $E$ and $F$ where $F' = S\setminus F = \{x\in S\mid x\not\in F\}$ is the complement of $F$ in $S$ .

Solution

Note $(E\cap F)\cup(E\cap F') = E\cap(F\cup F') = E\cap S = E$ and $(E\cap F)\cap(E\cap F') = E\cap(F\cap F') = E\cap\emptyset = \emptyset$ so $\mu(E\cap F) + \mu(E\cap F') = \mu((E\cap F)\cup(E\cap F') = \mu(E)$ .

The space of all (finitely additive) measures on $S$ is denoted $ba(S)$ . Note $ba(S)$ is a vector space with $(a\mu)(E) = a\mu(E)$ and $(\mu + \nu)(E) = \mu(E) + \nu(E)$ , $a\in\boldsymbol{R}$ , $E\subseteq S$ . The norm of a measure $\mu\in ba(S)$ is $\|\mu\| = \sup_{\{E_j\}} \sum_j |\mu(E_j)|$ where $\{E_j\}$ is any finite collection of pairwise disjoint subsets of $S$ .

Exercise. Show $\|\mu\| = 0$ implies $\mu = 0$ , $\|a\mu\| = |a|\|\mu\|$ , and $\|\mu + \nu\| \le \|\mu\| + \|\nu\|$ .

Let $B(S)$ be the vector space of all bounded functions on $S$ . It has norm $\|f\| = \sup_{x\in S}|f(x)|$ . Vector space norms define a metric $d\colon B(S)\times B(S)\to\boldsymbol{R}$ by $d(f,g) = \|f - g\|$ .

Exercise. Show $d$ is a metric.

Hint: Show $d(f,g) = 0$ implies $f = g$ , $d(f,g) = d(g,f)$ , and $d(f,h) \le d(f,g) + d(g, h)$ for $f,g,h\in B(\Omega)$ . Note $d(f,f) \le 2d(f,f)$ so $0\le d(f,f)$ for all $f$ .

The norm is complete so $B(S)$ is a Banach space.

Exercise. Show if $(f_n)$ is Cauchy then there exists $f\in B(\Omega)$ with $\lim_n \|f_n - f\| = 0$ .

Solution

Since $(f_n(x))$ is also Cauchy it has a limit $f(x)$ . Since $(\|f_n\|)$ is Cauchy it has an upper bound so $f$ is bounded.

The (vector space) dual of $B(S)$ is $ba(S)$ . Given $M\in B(S)^*$ define $\mu(E) = M1_E$ and given $\mu\in ba(S)$ define $M(\sum_j a_j 1_{E_j}) = \sum_j a_j\mu(E_j)$ .

Exercise. Show $\mu\mapsto M$ is well-defined.

Solution

Every finite sum of the form $\sum_j a_j 1_{E_j}$ is equal to a function $\sum_k b_k 1_{F_k}$ where the $(F_k)$ are pairwise disjoint. Such functions are uniformly dense in $B(S)$ .

The dual pairing defines the integral $\int_S f\,d\mu = \langle f,\mu\rangle$ , $f\in B(\Omega)$ , $\mu\in ba(\Omega)$ . Finitely additive measures make it challenging to prove theorems of the form ‘If $\lim_n f_n = f$ then $\lim_n \int_S f_n\,d\mu = \int f\,d\mu$ .’ However, Fatou’s Lemma does hold for positive finitely additive measures.

The product of a function and a measure is a measure defined by $\langle f,g\mu\rangle = \langle fg,\mu\rangle$ . Given measures $\mu,\nu\in ba(S)$ we say $g$ is the Radon-Nikodym derivative of $\nu$ with respect to $\mu$ if $g\mu = \nu$ .