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Measurable Functions
TLDR
Measure functions map elements of a set to a number. Measurable functions are a mapping between elements of a $\sigma$-algebra to another, at least.
Many functions are measurable. Measurability is preserved through various operations, and also $\min, \max$ operations.
Measurable Functions
Now that we have discussed what are measure functions, we move to integration. But the integration function $I(f)$ takes a function $f$ and produces an area under the curve in the Riemann sense. Understanding $\sigma$-algebras and measures helped build the mechanics behind the measurement process, i.e. integrating with respect to a measure. However, the other part the integral is $f$.
$f$ should be a measurable function. Recall a function is a mapping $f: X \to Y$ that induces another function $f^{-1}: Y\to X$. Namely, $f^{-1}(E) := {x \in X : f(x) \in E}$. This notation preserves under unions, intersections, and complements.
Exercise: Given $f: X\to Y$, Prove that $\cup_i f^{-1}(Y_i) = f^{-1}(\cup_i Y_i)$, and ${f^{-1}(Y)}^C = f^{-1}(Y^C)$. Hint: Use the set representation of $f^{-1}$.
Given two measurable spaces $(X,\mathcal{X}), (Y, \mathcal{Y})$, the function $f: X \to Y$ is measurable if $f^{-1}(E) \subseteq \mathcal{X}$ for $E \in \mathcal{Y}$. We call this function $f$ to be $\mathcal{X}/\mathcal{Y}$-measurable. For example, the constant function $f(x)=c$ is continuous and thus measurable, but also we can construct the trivial $\sigma$-algebra in the image, and any measurable set in the domain and show the mapping, to demonstrate measurability.
Exercise: If $f: A \to B$ is $\mathcal{A}/\mathcal{B}$-measurable and $g: B \to C$ is $\mathcal{B}/\mathcal{C}$-measurable, show that $f \circ g$ is $\mathcal{A}/\mathcal{C}$-measurable.
Exercise: Let $X, Y$ be any topological space. Prove that any continuous function $f:X\to Y$ is $\mathcal{B}_X/\mathcal{B}_Y$-measurable. Hint: Continuity condition of $f$: the set ${x : f(x) \in O}$ is open for any open set $O$ in the image. Then recall Borel sets on a topological space is the $\sigma$-algebra generated by its respectful open sets.
Measurable Functions on Subsets
For a given function $f: X \to \mathbf{R}$ that is $\mathcal{X}/\mathcal{B}$-measurable, it is sufficient to check that $f^{-1}(\text{generating set}) \in \mathcal{X}$ to determine measurability.
Exercise: Prove the above statement. That is, let $\mathcal{B} = \sigma(A)$. If the preimage condition $f^{-1}(A) \in \mathcal{X}$ holds, then for any $B \in \mathcal{B}$, the preimage also holds. Hint: Prove that that ${a \in A : f^{-1}(a) \in \mathcal{X}}$ is a $\sigma$-algebra and it contains $A$, hence contains $\mathcal{B}$.
Suppose we have a measurable space $(\Omega, \mathcal{M})$ and $f: \mathcal{M} \to \mathbf{R}$ and is measurable. Now consider $E \in \mathcal{M}$. How do you prove that the restriction of $f$ onto $E$ is also measurable?
Exercise: Prove the above, that $f|_E$ is measurable. Hint: Show that ${F \cap E: F \in \mathcal{M}}$ is a $\sigma$-algebra that contains $E$ or show that $f^{-1}(B) \cap E \in \mathcal{M}$ for any Borel set $B$.
Induced Sigma Algebras
Up to now, the measurable functions we deal with involve first defining a $\sigma$-algebra for the domain and range of some function $f$. However, we can see that for a function $f$, we can also induce a measurable domain, for which $f$ is measurable. So a function $f : X \to Y$ where the image is a measurable space $(Y, \mathcal{Y})$, can induce a $\sigma$-algebra on the domain:
\[\sigma(\{f^{-1}(Y) \subset 2^X : Y \in \mathcal{Y} \})\]In shorter notation, we write this as $\sigma(f)$, or the smallest $\sigma$-algebra on the domain $X$, such that $f$ is measurable.
There are several properties of measurable functions that are pretty straight forward to prove at this point.
Exercise: Show that if $f,g: X-> Y$ are both $\mathcal{X}/\mathcal{Y}$-measurable, then $f+g$ and $fg$ are measurable. Hint: Use the fact that continuous functions are measurable.
Exercise: Suppose that $f_i : X \to \bar{\mathbf{R}}$ is a sequence of $\mathcal{X}/\mathcal{Y}$-measurable functions. Show that $\sup_i f_i, \inf_i f_i$ are both measurable. Using those, show that $\limsup_i f_i$ and $\liminf_i f_i$ are both measurable. Then show that if $\lim_i f_i$ exists, then it is also measurable. (By measurable, it means $\mathcal{X}/\mathcal{Y}$-measurable.) Hint: Define the Borel $\sigma$-algebra using a half open interval. Use the preimage definition of measurability and leverage the set operations that work well with preimages. Depending on how the Borel $\sigma$-algebra is generated, think of what $\sup, \inf, \lim$ all mean in terms of set operations.
Exercise: For measurable $f,g$ prove $\max(f,g)$ and $\min(f,g)$ are measurable. Hint: Construct a Borel $\sigma$-algebra with half intervals. Then consider what it means for $f,g$ when $\max(f,g) < a$. Argue analogously for $\min(f,g)$.
This previous exercise will demonstrate that decomposition of a function into its negative and positive parts is still measurable. This will become critical for building up integration theory.
Define $f^- = \max(-f(x), 0)$ as the negative part of $f$, and $f^+ = \max(f(x), 0)$ as the positive part of $x$. See that $f = f^+ - f^-$. If $f$ is measurable, then so is $f^-$ and $f^+$, by the previous exercises.
Now, measurable functions have been introduced. So in the integral $\int_A f d\mu$, we have discussed what $\mu$ means (a measure function), and what $A$ means (a measurable set), and we have imposed some restrictions on $f$ (measurability of the function). Now we have enough to learn how what the entire expression $\int_A f d\mu$ means.