Sigma-Algebra/Examples/Trivial Sigma-Algebra

Examples of $\sigma$-Algebra

Let $X$ be a set.

The trivial $\sigma$-algebra on $X$ is the $\sigma$-algebra defined as:

$\set {\O, X}$

Proof

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Verify the axioms for a $\sigma$-algebra in turn:

Axiom $(\text {SA} 1)$

From Intersection with Empty Set, there exists $\O \in \set {\O, X}$, such that:

$\O \cap X = \O$

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And from Intersection with Subset is Subset and Set is Subset of Itself, there exists $X \in \set {\O, X}$ such that:

$ X \cap X = X$

So, we have:

$\forall A \in \set {\O, X}: A \cap X = A$

Hence $X \in \Sigma$.

There is believed to be a mistake here, possibly a typo. In particular: What a logic here? $X\in\Sigma$ by definition, if you mean $\Sigma := \set {\O, X}$. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Mistake}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Hence $X$ is the unit of $\set {\O, X}$.

$\Box$

Axiom $(\text {SA} 2)$

By Set Difference with Self is Empty Set again:

$\relcomp X X = \O \in \Sigma$.

Where $\relcomp X X$, and is the relative complement of $X$ in $X$

Now suppose:

$\O \in \Sigma$

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Then by Set Difference with Empty Set is Self:

$\relcomp X \O = X \in \Sigma$.

So $\forall A \in \Sigma$:

$\relcomp X A \in \Sigma$

Hence $\Sigma$ is closed under complement.

$\Box$

Axiom $(\text {SA} 3)$

Let the set of sets $\mathbb S$ be such that $\mathbb S$ is the empty set $\O$.

Then by Union is Empty iff Sets are Empty, the union of $\mathbb S$ is $\O$:

$\mathbb S = \O \implies \bigcup \mathbb S = \O$

By Union with Empty Set we have:

$X \cup \O = X$

By Union of Subsets is Subset and Set is Subset of Itself, we have:

$\ds \bigcup \mathbb S \subseteq X$

So we have:

$\forall A_n \in \Sigma: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma$

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Hence $\sigma$ is closed under countable unions.

$\Box$

Hence the result.

$\blacksquare$