Measure

Keith A. Lewis

April 25, 2024

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.