----- > [!proposition] Proposition. ([[σ-algebra generated by a set collection]]) > [[lattice|Meets and joins]] exist for [[σ-algebra|σ- algebras]]: given a set $X$, there is a smallest $\sigma$-algebra on $X$ containing a given collection $\mathscr{A}$ of subsets of $X$, namely, the intersection $\Sigma$ of all $\sigma$-algebras on $X$ that contain $\mathscr{A}$. We call $\Sigma$ the **$\sigma$-algebra generated by $\mathscr{A}$**. We write $\sigma(\mathscr{A})$. > > Thus, the collection of $\sigma$-algebras on a set $X$ forms a [[lattice|complete lattice]]. ^proposition > [!definition] Definition. ($\sigma$-algebra generated by a function collection) > Let $(Y_{i}, \mathcal{T}_{i})_{\in I}$ be a [[σ-algebra|measurable spaces]] and suppose $(f_{i}:X \to Y_{i})_{i \in I}$ is a collection of functions. The **$\sigma$-algebra generated by $(f_{i}:X\to Y_{i})_{i \in I}$** is smallest $\sigma$-algebra on $X$ making all the $f_{i}$ [[measurable function|measurable]]. Namely, $\sigma\big( (f_{i})_{i \in I} \big):= \sigma ( \{ f_{i} ^{-1} (E): i \in I, E \in \mathcal{T}_{i} \}),$ where the RHS is the $\sigma$-algebra generated by the various preimages of measurable sets under the $f_{i}$. > Given the obvious analogy to [[initial topology]], $\sigma\big( (f_{i})_{i \in I} \big)$ is often called **initial $\sigma$-algebra on $X$ wrt $(f_{i})$**. > [!basicproperties] Properties of the $\sigma$-algebra generated by a function collection. > - Since the preimage of a $\sigma$-algebra is already a $\sigma$-algebra, for a single function $f: X \to (Y, \mathcal{T})$ one has[^1] $\sigma(f)=\{ f ^{-1} (E): E \in \mathcal{T} \}.$ > If moreover $\mathcal{T}=\sigma(\mathscr{A})$ for some collection $\mathscr{A} \subset \mathcal{T}$, then also $\sigma(f)= \sigma\{ f ^{-1} (A): A \in \mathscr{A} \},$ > or in nicer notation: $\underbrace{ f ^{-1} \big( \sigma(\mathscr{A}) }_{ =: \sigma(f) } \big)=\sigma\big(f^{-1}(\mathscr{A}) \big).$ > See also [[generator criterion for function measurability]], which basically holds the same discussion. ^properties [^1]: (That is, we don't need to apply $\sigma(\cdot)$ to the set of preimages in the case of one function.) > [!proof] > The first part is obvious, since the preimage of a $\sigma$-algebra is already a $\sigma$-algebra ([[preimages and unions commute]] and [[complements and inverse images commute|preimages and complements commute]]). For the second part, the inclusion $\sigma(f^{-1}(\mathscr{A})) \subset f ^{-1}(\sigma(\mathscr{A}))$ is obvious. As to the reverse inclusion, set $\mathcal{S}:=\{ E \subset Y: f ^{-1} (E) \in \sigma( f ^{-1} (\mathscr{A})) \}$. Check $\mathcal{S}$ is a $\sigma$-algebra containing $\mathscr{A}$. Thus $\mathcal{S} \supset \sigma(\mathscr{A})$, and since $f^{-1}(\mathcal{S}) \subset \sigma(f ^{-1} (\mathscr{A}))$ by construction, so too $f ^{-1} (\sigma(\mathscr{A})) \subset \sigma(f^{-1} (\mathscr{A}))$. ^proof > [!basicexample] > - Suppose $X$ is a set and $\mathscr{A}$ is the collection of subsets of $X$ that consist of exactly one element: $\mathscr{A}=\{ \{ x\} : x \in X\} .$ > Then the smallest $\sigma$-algebra on $X$ that contains $\mathscr{A}$ is the countable-cocountable $\sigma$-algebra: the collection $\Sigma$ of all subsets $E$ of $X$ such that $E$ is [[countably infinite|countable]] or $X-E$ is [[countably infinite|countable]]. Indeed, if $\mathcal{T}$ is a $\sigma$-algebra on $X$ containing $\mathscr{A}$, then $\mathcal{T}$ has to contain every countable subset of $X$ (since every countable subset of $X$ arises as a countable union of elements of $\mathscr{A}$) and every cocountable subset (since such a subset's complement belongs to the $\sigma$-algebra and is countable). Thus, $\Sigma \subset \mathcal{T}$. > > - Suppose $\mathscr{A}=\{ (0,1), (0, \infty) \}$. Then, by taking complements and then taking countable unions, we see that the smallest $\sigma$-algebra on $\mathbb{R}$ containing $\mathscr{A}$ is > $\Sigma=\{ \emptyset, (0,1), (0,\infty), (-\infty, 0] \cup [1, \infty), (-\infty, 1] , \mathbb{R} \}.$ > > > [!proof]- Proof. ([[σ-algebra generated by a set collection]]) > By construction, $\Sigma$ contains $\mathscr{A}$ and $\Sigma \subset \mathcal{T}$ for all $\sigma$-algebras $\mathcal{T}$ on $X$ that contain $\mathscr{A}$. We just have to show $\Sigma$ is indeed a $\sigma$-algebra. Clearly $\emptyset \in \Sigma$. If $E \in \Sigma$, so that $E$ is in every $\sigma$-algebra on $X$ that contains $\mathscr{A}$, then $X-E$ is in every $\sigma$-algebra on $X$ that contains $\mathscr{A}$. If $E_{1},E_{2},\dots$ is a sequence of elements in $\Sigma$, so that $E_{1},E_{2}$ are in every $\sigma$-algebra on $X$ that contains $\mathscr{A}$, then $\bigcup_{k=1}^{\infty}E_{k}$ is in every $\sigma$-algebra on $X$ that contains $\mathscr{A}$, hence in $\Sigma$. ----- #### ---- #### 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 > ```