----- > [!proposition] Proposition. ([[generator criterion for function measurability]]) > Let $(X,\Sigma)$ and $(Y, T)$ be [[σ-algebra|measurable spaces]], and $f:X \to Y$ a function. Suppose that $\mathscr{A}$ is a collection of subsets of $Y$ such that $\textcolor{Skyblue}{T \subset \sigma(\mathscr{A}) }\ \ \ \ \text{ and } \ \ \ \ \textcolor{Thistle}{f ^{-1}(A) \in \Sigma \text{ for all }A \in \mathscr{A}}. $ > Then $f$ is a [[measurable function]]. > The proof is obvious: $f$ is measurable iff $T$ is a subset of ${S}:=\{ B \subset Y: f ^{-1}(B) \in \Sigma \}.$ The set $S$ is manifestly a $\sigma$-algebra[^1] that (by assumption) contains $\mathscr{A}$, and hence (by [[σ-algebra generated by a set collection|definition]]) contains $\sigma(\mathscr{A})$, which we assumed contains $T$. So $T \subset S$ as required. > The $\textcolor{Skyblue}{\text{blue}}$ and $\textcolor{Thistle}{\text{pink}}$ conditions are usually in tension: if showing one condition is difficult for a given choice of $\mathscr{A}$, showing the other may not be. [^1]: To see this, recall that [[preimages and unions commute]] and [[complements and inverse images commute|preimages and complements commute]]. > [!proposition] Corollary. > If $f:(X, \Sigma) \to (Y,T)$ is a function between [[σ-algebra|measurable spaces]] where $T=\sigma(\mathscr{A})$ is [[σ-algebra generated by a set collection|generated by]] a subcollection $\mathscr{A}$, then $f$ is a [[measurable function]] iff $f ^{-1} (A) \in \Sigma$ for all $A \in \mathscr{A}$. ^proposition > [!basicexample] > Suppose $(X, \Sigma)$ is a [[σ-algebra|measurable space]] and $f:X \to \mathbb{R}$ is a function such that $f ^{-1}\big( (a, \infty)\big) \in \Sigma$ for all $a \in \mathbb{R}$. Then $f$ is a [[measurable function]]. ^basic-example > [!proof] > Recall that, by default, $T$ is the [[Borel set|Borel algebra]] on $\mathbb{R}$ in this situation. We are given that the pink condition holds, and so need to show the blue condition, i.e. every element of $T$ (that is, every [[Borel set|Borel subset]] of $\mathbb{R}$) can be obtained using complements and countable unions of open [[ray|rays]] $(a, \infty)$, $a \in \mathbb{R}$. It is enough to check that every [[open interval]] can be obtained in such a way.[^2] Indeed, any open interval $(a,b)$ equals a countable union $(a,b)=\bigcup_{k=1}^{\infty}\left( a, b- \frac{1}{k} \right]$ > and $\left( a, b-\frac{1}{k} \right]=\underbrace{ (-\infty, a] }_{ =(a, \infty)^{c} } \cap\left( b- \frac{1}{k}, \infty \right)$. > [^2]: Indeed, every open subset of $\mathbb{R}$ can be obtained as a [[countably infinite|countable]] union of open intervals ($\mathbb{R}$ is [[second-countable space|second-countable]]), and every Borel subset can be obtained from the generators via complements and countable unions. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```