cooloicons62

Answered

2022-07-09

Let $X(\omega )$ be a continuous random variable defined on a probability space $(\mathrm{\Omega},\mathcal{F},\mathbb{P})$.

How does the measurability condition of this random variable (i.e. $\mathrm{\forall}r\in \mathbb{R},\{\omega :X(\omega )\le r\}\in \mathcal{F}$) cohere with the $\sigma $-additivity of the $\sigma $-algebra $\mathcal{F}$?

I understand that a $\sigma $-algebra requires that the union of countable many elements of $\mathcal{F}$ must also be contained in $\mathcal{F}$. But if we consider a continuous random variable, then the events that the $\sigma $-algebra $\mathcal{F}$ must contain are uncountable, since $r$ can be any value in $\mathbb{R}$ and $X(\omega )$ is a continuous r.v.What am I missing?

How does the measurability condition of this random variable (i.e. $\mathrm{\forall}r\in \mathbb{R},\{\omega :X(\omega )\le r\}\in \mathcal{F}$) cohere with the $\sigma $-additivity of the $\sigma $-algebra $\mathcal{F}$?

I understand that a $\sigma $-algebra requires that the union of countable many elements of $\mathcal{F}$ must also be contained in $\mathcal{F}$. But if we consider a continuous random variable, then the events that the $\sigma $-algebra $\mathcal{F}$ must contain are uncountable, since $r$ can be any value in $\mathbb{R}$ and $X(\omega )$ is a continuous r.v.What am I missing?

Answer & Explanation

diamondogsaz

Expert

2022-07-10Added 12 answers

A $\sigma $-algebra is closed under countable unions. This does not mean that its elements cannot be uncountable, or that it's impossible that some (or even all) uncountable unions of elements in the $\sigma $-algebra are in the $\sigma $-algebra.

For example, if $\mathrm{\Omega}$ is an arbitrary nonempty set (possibly an uncountable one, like $\mathbf{R}$ or ${\mathbf{R}}^{\mathbf{R}}$), both $\{\mathrm{\varnothing},\mathrm{\Omega}\}$ and $\mathcal{P}(\mathrm{\Omega})$ (the whole power set) are perfectly good $\sigma $-algebras of subsets of $\mathrm{\Omega}$, and both of them are closed under arbitrary (even uncountable) unions and have uncountable elements, if $\mathrm{\Omega}$ is not countable.

In fact, unless $\mathrm{\Omega}$ is countable, there is always an uncountable elements in any $\sigma $-algebra: namely, $\mathrm{\Omega}$ itself.

Further, if $\mathcal{F}$ is a $\sigma $-algebra of subsets of an uncountable set $\mathrm{\Omega}$, and $\mathcal{F}$ contains singletons, then there is always at least one uncountable disjoint union of elements of $\mathcal{F}$ which is in $\mathcal{F}$, namely, $\mathrm{\Omega}=\bigcup _{\omega \in \mathrm{\Omega}}\{\omega \}$.

For example, if $\mathrm{\Omega}$ is an arbitrary nonempty set (possibly an uncountable one, like $\mathbf{R}$ or ${\mathbf{R}}^{\mathbf{R}}$), both $\{\mathrm{\varnothing},\mathrm{\Omega}\}$ and $\mathcal{P}(\mathrm{\Omega})$ (the whole power set) are perfectly good $\sigma $-algebras of subsets of $\mathrm{\Omega}$, and both of them are closed under arbitrary (even uncountable) unions and have uncountable elements, if $\mathrm{\Omega}$ is not countable.

In fact, unless $\mathrm{\Omega}$ is countable, there is always an uncountable elements in any $\sigma $-algebra: namely, $\mathrm{\Omega}$ itself.

Further, if $\mathcal{F}$ is a $\sigma $-algebra of subsets of an uncountable set $\mathrm{\Omega}$, and $\mathcal{F}$ contains singletons, then there is always at least one uncountable disjoint union of elements of $\mathcal{F}$ which is in $\mathcal{F}$, namely, $\mathrm{\Omega}=\bigcup _{\omega \in \mathrm{\Omega}}\{\omega \}$.

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