Def: A collection \mathcal M of a set X is said to be a \sigma-algebra in X if \mathcal X satisfies the following properties:
Then, we call X a measurable space and elements of \mathcal M measurable sets.
Def: Take two measurable spaces X and Y with \sigma-algebras \mathcal M_X, \mathcal M_Y. Then, f: X \to Y is said to be measurable if f^{-1}(E) \in \mathcal M_X for all E \in \mathcal M_Y.
Def: Take a \sigma-algebra \mathcal M on X. A positive measure (or simply a measure) is a function \mu: \mathcal M \to [0, \infty] which is countably additive, i.e. satisfies \mu \left( \bigcup_{n=1}^\infty E_n \right) = \sum_{n=1}^\infty \mu(E_n) for disjoint sets \{E_n\}_{n=1}^\infty. Then, we say that a measure space is a measurable space with a corresponding positive measure, and we write it often as a triple (X, \mathcal M, \mu). Moreover, if the codomain of \mu is \mathbb{R} or \mathbb{C}, we say that it is a signed measure or a complex measure respectively. Finally, if \mu(X) = 1, then we say that \mu is a probability measure.
Def: A simple function is a function whose image has only finitely many points.
Def: For a measurable function f: X \to Y between measurable spaces (X, \mathcal M_X) and (Y, \mathcal M_Y), we say that the \sigma-algebra generated by f is \sigma(f) = \{f^{-1}(S) \mid S \in \mathcal M_Y\}.
Def: For a measure space (X, \mathcal M, \mu), define L^p(\mu) to be all measurable functions f on X for which \left(\int_X |f|^p d\mu\right)^{1/p} < \infty.
Prop (Continuity of Measure): Let \{E_n\}_{n=1}^\infty be an ascending sequence of sets, i.e. E_1 \subset E_2 \subset \dots. Then, we have that \mu \left( \bigcup_{n=1}^\infty E_n \right) = \lim_{n \to \infty} \mu(E_n). Similarly, if \{E_n\}_{n=1}^\infty is a descending sequence of sets, i.e. E_1 \supset E_2 \supset \dots. Then, \mu \left( \bigcap_{n=1}^\infty E_n \right) = \lim_{n \to \infty} \mu(E_n) as well.
Proof: For the ascending sequence, put A_1 = E_1 and A_n = E_n \setminus E_{n-1} for n \geq 2; then we have that \mu(E_n) = \sum_{i=1}^n \mu(A_i) \text{ and } \mu(E) = \sum_{i=1}^\infty \mu(A_n) which immediately gives us what we want.
Corollary (Subadditivity of Measure): For \{E_n\}_{n=1}^\infty, we have that \mu \left( \bigcup_{n=1}^\infty E_n \right) \leq \sum_{n=1}^\infty \mu(E_n).
In the other case, simply put A_n = E_1 \setminus E_n, so that the previous part gives us \mu(E_1) - \mu \left( \bigcap_{n=1}^\infty E_n \right) = \mu \left( E_1 - \bigcap_{n=1}^\infty E_n \right) = \lim_{n \to \infty} \mu(A_n) = \mu(E_1) - \lim_{n \to \infty} \mu(E_n) which then gives us what we want as well. \square
Prop: Let (X, \mathcal M) be a measurable space. If f: X \to [-\infty, \infty] satisfies that f^{-1}((\alpha, \infty]) \in \mathcal M for every real \alpha, then f is measurable.
Proof: TODO
Lemma: If f_n: X \to [-\infty, \infty] is measurable for n \geq 1, and g = \sup_n f_n \text{ and } h = \limsup_{n \to \infty} f_n, then g, h are measurable.
Proof: We have that g^{-1}((\alpha, \infty]) = \bigcup_{n=1}^\infty f_n^{-1}((\alpha, \infty]) so g is measurable, and h = \inf_{m \geq 1} \left\{ \sup_{n \geq m} f_n \right\} as therefore also measurable.
Prop: Let f: X \to [0, \infty] be measurable. Then there are simple measurable functions s_n on X such that
Proof: Take \delta_n = 2^{-n}; for each positive integer n and real number t, we can find integer k_{n,t} such that k_{n,t}\delta_n < (k+1)\delta_n. Then, define \varphi_n(t) = \begin{cases} k_{n,t}\delta_n & 0\leq t < n \\ n & n \leq t \leq \infty \end{cases} so that we have
Then, letting s_n = \varphi_n \circ f clearly does the job. \square
Def: If s: X \to [0, \infty) is a measurable simple function, given by s = \sum_{i=1}^n \alpha_i 1_{A_i} and if E \in \mathcal M, we set \int_E s d\mu = \sum_{i=1}^n \alpha_i \mu(A_i \cap E). Then, for any measurable f: X \to [0, \infty], we have \int_E f d\mu = \sup_s \int_E s d\mu where the supremum is overall simple measurable functions less than f. Finally, for f: X \to [-\infty, \infty], we let f^+(x) = \max\{f(x), 0\} \text{ and } f^-(x) = \max\{-f(x), 0\} and define \int_E f d\mu = \int_E f^+ d\mu - \int_E f^- d\mu. Integration of complex valued functions is simply done by integrating over the real and the imaginary parts separately.
Def: We say that f is integrable if \int |f| d\mu < \infty, and we also put f \in L^1(\mu).
Of course, we are referring to monotone convergence, Fatou’s lemma, and dominated convergence.
Theorem (Monotone Convergence): Let \{f_n\}_{n=1}^\infty be a sequence of measurable functions on X, and suppose that
Then f is measurable and \int_X f_n d\mu \to \int_X f d\mu \text{ as } n \to \infty.
Proof: Let \alpha = \lim_{n \to \infty} \int_X f_n d\mu; clearly by construction \alpha \leq \int_X f d\mu. Then, for any 0 < c < 1 and simple function 0 \leq s \leq f, we define E_n = \{x \mid f_n(x) \geq c s(x) \} which is ascending and has \bigcup_{n=1}^\infty E_n = X. Thus, we get \int_X f_n d\mu \geq \int_{E_n} f_n d\mu \geq c \int_{E_n} s d\mu and sending n \to \infty and c \to 1 gives \alpha \geq \sup_s \int_X s d\mu = \int_X f d\mu which gives us what we want. \square
Theorem (Fatou’s Lemma): If f_n: X \to [0, \infty] is measurable for n \geq 1, then \int_X \left(\liminf_{n \to \infty} f_n\right) d\mu \leq \liminf_{n \to \infty} \int_X f_n d\mu.
Proof: Put g_n(x) = \inf_{m \geq n} f_m(x), such that \liminf_{n \to \infty} f_n = \lim_{n \to \infty} g_n. Then, we clearly have that by monotone convergence \int_X \left(\liminf_{n \to \infty} f_n\right) d\mu = \lim_{n \to \infty} \int_X g_n d\mu \leq \lim_{n \to \infty} \int_X f_n d\mu = \liminf_{n \to \infty} \int_X f_n d\mu. \ \square
Theorem (Dominated Convergence): Suppose \{f_n\}_{n=1}^\infty is a sequence of measurable functions on X, and f satisfies that f_n \to f pointwise on X. Then, if there is g \in L^1(\mu) with |f_n(x)| \leq g(x) for all n, then f \in L^1(\mu), \int_X |f_n - f| d\mu = 0, and \int_X f_n d\mu \to \int_X f d\mu.
Proof: Clearly f \in L^1(\mu) since |f| \leq g. Moreover, we have that |f_n - f| \leq 2g, so Fatou applied to 2g - |f_n - f| gives \int_X 2g d\mu \leq \liminf_{n \to \infty} \int_X (2g - |f_n - f|)d\mu = \int_X 2g d\mu - \limsup_{n \to \infty} \int_X |f_n - f|d\mu which gives us that \int_X |f_n - f| d\mu = 0. Jensen’s inequality then gives \lim_{n \to \infty}\left|\int_X f_n - \inf_X f\right| \leq \lim_{n \to \infty} \int_X |f_n - f|d\mu = 0. \ \square