Are as constant but not constant random variables trivial sigma-algebra-measurable? Converse?

Question

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:

---

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?

---

2. What about the converse? Are trivial sigma-algebra-measurable random variables necessarily constant (that is never almost surely constant)?

Answer 1

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 $X(\omega)=x$.

Answer 2

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.