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  1. Are almost surely constant random variables trivial sigma-algebra-measurable?

These links suggest no:

http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2004&task=show_msg&msg=1121.0001

Is a random variable constant iff it is trivial sigma-algebra-measurable?

This link suggests yes:

http://www.math.duke.edu/~jonm/Courses/Math219/sigmaAlgebra.pdf

Preview:

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I really don't think a random variable $X(\omega) = 2 \ \forall \omega \in \Omega$ except for a set $A \in \mathscr F$ can ever be trivial sigma-algebra-measurable even if A has a probability of zero. Is Dr Mattingly (same link as earlier) wrong?

  1. What about the converse? Are trivial sigma-algebra-measurable random variables necessarily constant (that is never almost surely constant)?
BCLC
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  • All three links say that this is true. I don't understand your question. –  May 19 '15 at 19:31
  • @avid19 Sorry. Going to edit question. The third link says "almost surely constant" rather than "constant". Are almost surely constant but not constant random variables trivial sigma-algebra-measurable? – BCLC May 19 '15 at 19:32
  • So you question is about almost surely constant vs surely constant? Surely constant are trivial sigma algebra measurable but you don't know if almost surely is as well? Thanks for clarifying. –  May 19 '15 at 19:36
  • @avid19 Exactly. – BCLC May 19 '15 at 19:37

2 Answers2

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If a r.v. is constant, you have $[X\in B]$ is empty if the constant is not in $B$ and it is $\Omega$ otherwise. This is measurable in the trivial $\sigma$-algebra. It if is a.s. constant, then $[X\in B]$ is either a set of zero measure or measure 1. The completion of the trivial $\sigma$-algebra is the $\sigma$ algebra consisting of sets of full or zero measure.

ncmathsadist
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Let $\Omega = \{a,b\}$, $\mathcal F=2^{\Omega}$, $\mathbb P(\{a\})=0$, $\mathbb P(\{b\})=1$. Define $X(a)=0$, $X(b)=1$. Then $\mathbb P(X=1)=1$ so $X$ is almost surely constant, but $$X^{-1}(\{1\})=\{b\}\notin\{\varnothing,\Omega\},$$ so $X$ is not trivial $\sigma$-algebra measurable.

The converse is true. Suppose $\sigma(X)=\{\varnothing,\Omega\}$ and $\mathbb P(X=x)=0$ for some $x\in\mathbb R$. Then $X^{-1}(\{x\})=\varnothing$, so there is no $\omega\in\Omega$ such that $X(\omega)=x$.

Math1000
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  • Thanks Math1000. For the converse part, how do we know that X is almost surely constant but not constant? I'll edit question to clarify I mean almost surely constant but not constant. – BCLC May 19 '15 at 20:04
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    If $X$ were almost surely constant but not constant, there would exist $x\in\mathbb R$ with $\mathbb P(X=x)$ but $X(\omega)=x$ for some $\omega\in\Omega$. If $\sigma(X)$ is the trivial $\sigma$-algebra, this cannot be the case. – Math1000 May 19 '15 at 20:20