April 25, 2024
A function and a measure determine an integral. Integrals are linear in both the function and the measure. Measures are functions on subsets of the domain of the function that do not count things twice. They can be identified with the dual of the vector space of functions.
If A and B are sets then A^B = \{f\colon B\to A\} is the set of all functions from B to A.
Exercise. If \#A is the number of elements of the set A show \#A^B = \#A^{\#B} when A and B are finite.
The set \mathbf{R}^\Omega is a vector space of real-valued functions on \Omega where scalar multiplication and addition are defined pointwise: \begin{aligned} (af)(\omega) &= af(\omega) \\ (f + g)(\omega) &= f(\omega) + g(\omega), \\ \end{aligned} for a\in\mathbf{R}, f,g\in\mathbf{R}^\Omega, \omega\in \Omega. We can replace \mathbf{R} by any field \mathbf{F} and get a vector space over \mathbf{F}.
The set of all subsets of a set \Omega is \mathcal{P}(\Omega) = \{A\mid A\subseteq\Omega\}. A measure on a set \Omega is a function \mu\colon \mathcal{P}(\Omega)\to\mathbf{R} satisfying \mu(A\cup B) = \mu(A) + \mu(B) - \mu(A\cap B), A,B\subseteq\Omega and \mu(\emptyset) = 0. Measures do not count things twice and the measure of nothing is zero.
Exercise If \nu\colon \mathcal{P}(\Omega)\to\mathbf{R} and \nu(A\cup B) = \nu(A) + \nu(B) - \nu(A\cap B), A,B\subseteq\Omega, then \mu(A) = \nu(A) - \nu(\emptyset) is a measure.
Hint: c = c + c - c.
Exercise. If \Omega is finite show \mu(A) = \#A is a measure.
Exercise. If \Omega is finite show \mu(A) = \sum_{\omega\in A} \mu(\{\omega\}).
This shows measures on finite sets are determined by the measure of each singleton \{\omega\}, \omega\in\Omega. In this case a measure can be identified with an element of \mathbf{R}^\Omega.
The set ba(\Omega) of all measures on \Omega is a vector space where scalar multiplication and addition are defined setwise: \begin{aligned} (a\mu)(A) &= a\mu(A) \\ (\mu + \nu)(A) &= \mu(A) + \nu(A), \\ \end{aligned} for a\in\mathbf{R}, \mu,\nu\in ba(\Omega), A\subseteq\Omega.
Integration is a linear functional M\colon\mathbf{R}^\Omega\to\mathbf{R}: M(af) = aMf and M(f + g) = Mf + Mg, a\in\mathbf{R}, f,g\in\mathbf{R}^\Omega. Each linear functional M determines a measure \mu by \mu(A) = M1_A, A\subseteq\Omega, where 1_A\in\mathbf{R}^\Omega is the indicator function defined by 1_A(\omega) = 1 if \omega\in A and 1_A(\omega) = 0 if \omega\not\in A.
Exercise. Show \mu is a measure.
Hint: 1_{A\cup B} = 1_A + 1_B - 1_{A\cap B} and 1_\emptyset = 0.
Exercise. Show \mathcal{P}(\Omega) can be identified with \{0,1\}^\Omega.
Hint: A\subset\Omega corresponds to 1_A\in \{0,1\}^\Omega. We also write 2^\Omega for this where 2 = \{0,1\}.
Each measure \mu determines a linear functional on a subspace of \mathbf{R}^\Omega. Functions that are finite linear combinations of indicator functions are called elementary. If f = \sum_j a_j 1_{A_j}, a_j\in\mathbf{R}, A_j\subseteq\Omega is elementary then define Mf = \sum_j a_j \mu(A_j). For this to be well-defined we must show if \sum_j a_j 1_{A_j} = \sum_k b_k1_{B_k} are elementary functons then \sum_j a_j \mu(A_j) = \sum_k b_k\mu(B_k). Since M is linear this is equivalent to showing if \sum_j a_j 1_{A_j} = 0 then \sum_j a_j \mu(A_j) = 0.
Exercise. If T\colon V\to W is a linear transformation between vector spaces then Tf = Tg if and only if T(f - g) = 0, f,g\in V.
Hint: T(f - g) = Tf - Tg.
Every elementary function can be written as \sum_j a_j 1_{A_j} where (A_j) are pairwise disjoint.
Exercise. If f = a_1 1_{A_1} + a_2 1_{A_2} show f = a_1 1_{A_1\setminus A_2} + a_2 1_{A_2\setminus A_1} + (a_1 + a_2) 1_{A_1\cap A_2}.
Hint: Recall the set difference is defined by A\setminus B = \{a\in A\mid a\not\in B\}.
Exercise. Show a_1\mu(A_1) + a_2\mu(A_2) = a_\mu(A_1\setminus A_2) + a_2\mu(A_2\setminus A_1) + (a_1 + a_2)\mu(A_1\cap A_2).
Hint: A = (A\setminus B)\cup(A\cap B) is a disjoint union so \mu(A) = \mu(A\setminus B) + \mu(A\cap B).
The case for any finite linear combination can be proved by induction.
Exercise. If \sum_j a_j 1_{A_j} = 0 and (A_j) are pairwise disjoint then a_j = 0 for all j.
Hint: Consider when \omega\in A_j.
This shows M is well-defined.
Given a linear functional M on \mathbf{R}^\Omega we always have a measure \mu on \Omega given by \mu(A) = M1_A. A measure only determines a linear functional on the subspace of elementary functions in \mathbf{R}^\Omega.
If \Omega is finite then every function is elementary since f = \sum_{\omega\in\Omega} f(\omega) 1_{\{\omega\}}
We write Mf = \int_\Omega f\,d\mu.
A function f\in\mathbf{R}^\Omega is bounded if there exists N\in\mathbf{R} with |f(\omega)| < N for \omega\in\Omega. The set of all bounded functions is denoted B(\Omega). The norm of a bounded function is the greatest lower bound (infimum) \|f\| = \inf\{N\in\mathbf{R}\mid |f| < N\}, where |f| < N is shorthand for |f(\omega)| < N for all \omega\in\Omega.
Exercise. Show \|f\| = \sup\{|f(\omega)|\mid \omega\in\Omega\}.
Exercise. Show \|f\| = 0 implies f = 0, \|af\| = |a|\|f\| and \|f + g\| \le \|f\| + \|g\|, a\in\mathbf{R}, f,g\in B(\Omega).
A vector space having a norm is a normed linear space. It is also a metric space with distance defined by d(f,g) = \|f - g\|.
Exercsie. Show d(f,g) = 0 implies f = g and d(f,g) \le d(f,h) + d(h,g) for f,g,h\in B(\Omega).
A metric space is complete if every Cauchy sequence converges to a point in the space.
Exercise. If (f_n) is a Cauchy sequence in B(\Omega) show it converges to some f\in B(\Omega).
Hint: (f_n(\omega)) is a Cauchy sequence in \mathbf{R} for each \omega\in\Omega and \mathbf{R} is complete.
Let B(\Omega)^* be the space of all linear functionals on B(\Omega) that are continuous in the metric topology: M\colon B(\Omega)\to\mathbf{R} where f_n\to f implies Mf_n \to Mf. Note B(\Omega)^* is a vector space and we can define a norm for each M\in B(\Omega)^* by \|M\| = \sup\{|Mf|\mid f\in B(\Omega), \|f\| \le 1\}.
Exercise. Show this ia a norm.
Hint: Show \|M\| = 0 implies M = 0, \|aM\| = |a| \|M\|, and \|M + N\| \le \|M\| + \|N\|, a\in\mathbf{R}, M,N\in B(\Omega)^*.
As before, each M\in B(\Omega)^* defines a measure \mu\in ba(\Omega) by \mu(A) = M1_A. Again, each measure \mu\in ba(\Omega) defines a linear functional on the subspace of elementary functions by M(\sum_j a_j 1_{A_j}) = \sum_j a_j \mu(A_j). This can be extended to all of B(\Omega) by continuity.
The set of elementary functions in B(\Omega) is dense: given f\in B(\Omega) and \epsilon > 0 there exists an elementary function e with \|f - e\| < \epsilon. Suppose \|f\| < N and let -N = a_0 < a_1 < \cdots < a_n = N, a_j\in\mathbf{R} be any partition of [-N, N]. Define A_j = \{\omega\in\Omega\mid a_j \le f(\omega) < a_{j+1}\}, 0\le j < n.
Exercise. Show \|f - \sum_j a_j 1_{A_j}\| \le \max_{0\le j < n} a_{j + 1} - a_j.
For any N and any \epsilon > 0 we can find a partition with \max_{0\le j < n} a_{j + 1} - a_j < \epsilon.
This shows B(\Omega)^* is in 1-to-1 correspondence with ba(\Omega). The map M\mapsto\mu is linear and continuous.
Exercise. Show the map is isometric.
Hint: Show if M\mapsto\mu then \|M\| = \|\mu\|.
This establishes the fact bounded linear functionals on B(\Omega) are isometrically isomorphic to measures on \Omega.
Possible outcomes are modeled by a set \Omega. For example, the outcome of rolling a six-sided die can be modeled by the set \Omega = \{1,2,3,4,5,6\}. Subsets of \Omega are called events. The event “rolling an even number” corresponds to the subset \{2,4,6\}. Partial information is modeled by a partition of \Omega. The partition \{\{1,3,5\},\{2,4,6\}\} represents knowing whether the roll was odd or even. Partial information is knowing only whether \omega\in\{1,3,5\} or \omega\in\{2,4,6\}.
Full information corresponds to the partition of singletons \{\{\omega\}\mid\omega\in\Omega\}. No information corresponds to the singleton partition \{\Omega\}.
Partial information is modeled by a partition of a set \Omega: a collection of subsets of \Omega that are pairwise disjoint and have union \Omega. Every \omega\in\Omega belongs to a unique set, A_\omega, in the partition. We call A_\omega the atom containing \omega.
Exercise. Show either A_\omega = A_{\omega'} or A_\omega\cap A_{\omega'} = \emptyset for \omega,\omega'\in\Omega.
Define a relation on \Omega by \omega\sim\omega' if and only if A_\omega = A_{\omega'}.
Exercise. Show \sim is an equivalence relations.
Hint: Show \omega\sim\omega, \omega\sim\omega' imples \omega'\sim\omega, and \omega\sim\omega', \omega'\sim\omega'' imply \omega\sim\omega''.
Exercie. Show every equivalence relation determines a partition.
Hint: Let A_\omega = \{\omega'\in\Omega\mid \omega'\sim\omega\} and show \{A_\omega\mid\omega\in\Omega\} is a partition.
An algebra of sets on \Omega is a subset of the power set of \Omega that is closed under complement and union. Sets in the algebra are called events. These conditions are necessary in order to talk about an event not occuring or either of two events occuring. If E is an event then “not E” corresponds to \Omega\setminus E. if E and F are events then “E or F” corresponds to E\cup F.
Exercise. Show algebras are closed under intersection.
Hint: De Morgan.
If E and F are events in the algebra then “E and F” corresponds to the event E\cap F.
Exercise. If \mathcal{A} and \mathcal{A}' are algebras then \mathcal{A}\cap\mathcal{A}' is an algebra.
Extra Credit. Find algebras \mathcal{A} and \mathcal{A}' where \mathcal{A}\cup\mathcal{A}' is not an algebra.
Given any collection of subsets \mathcal{B} of \Omega define \mathop{\rm{Alg}}\mathcal{B} to be the smallest algebra containing \mathcal{B}.
Exercise Show \mathop{\rm{Alg}}\mathcal{B} is the intersection of all algebras containing \mathcal{B}.
If \mathcal{A} is an algebra let A_\omega = \cap\{A\in\mathcal{A}\mid\omega\in A\}, \omega\in\Omega.
Exercise. Show either A_\omega = A_{\omega'} or A_\omega\cap A_{\omega'} = \emptyset for \omega,\omega'\in\Omega.
Exercise. If \mathcal{A} is an algebra then \{A_\omega\mid\omega\in\Omega\} is a parition.
Exercise. If \mathcal{A} is an algebra show \mathcal{A}= \mathop{\rm{Alg}}\{A_\omega\mid\omega\in\Omega\}.
This shows algebras are in 1-to-1 correspondence with their partition of atoms. We will use \mathcal{A} to denote either the algebra or partition.
If \mathcal{A} is an algebra on \Omega and f\colon\Omega\to\mathbf{R} then f is \mathcal{A}-measurable if \{\omega\mid f(\omega) \le a\} belongs to \mathcal{A} for every a\in\mathbf{R}.
Exercise. If \mathcal{A} is finite then f\colon\Omega\to\mathbf{R} is measurable if and only if f is constant on the atoms of \mathcal{A}.
This shows f is measurable if and only if f\colon\mathcal{A}\to\mathbf{R} is a function on the atoms of \mathcal{A}.
The atoms of \mathcal{A} are a set so the theory of integration above applies. B(\mathcal{A}) is the set of bounded \mathcal{A} measurable functions. Its dual B(\mathcal{A})^* is isometrically isomorphic to ba(\mathcal{A}), the set of all measures on \mathcal{A}.