If a random variable $X : \Omega \to \mathbb R$ is $\{ \emptyset, \Omega \}$-measurable, then it is constant. I want to generalise this result:

Now if $\mathcal G$ is a $\sigma$-algebra such that for each $A \in \mathcal G$ we have $P(A) = 0$ or $P(A) = 1$, then if a random variable $X : \Omega \to \mathbb R$ is $\mathcal G$-measurable, then it is almost surely constant, meaning that $P(X = c) = 1$ for some $c \in \mathbb R$.

I am looking for a proof of this fact. I conjecture that $$ X \mbox{ is almost surely constant } \Leftrightarrow F(x) = 1_{[a,\infty)} $$ for some $a \in \mathbb R$, where $F(x) := P(X \le x)$ denotes the distribution function, but this I am also unable to prove?

  • 2
    $\begingroup$ You just have to show that a non-decreasing function $F$ that can only attain the values 0 and 1 has this form. $\endgroup$ – Hans Engler Jul 7 '15 at 14:25
  • $\begingroup$ Yes, I see. If $F$ is i) non-decreasing and ii) $\lim_{x\to \infty} F(x) = 1, \lim_{x\to -\infty} F(x) = 0$ then $F = 1_{[a,\infty)}$ for some $a$. But why does this imply that $X$ is almost surely constant? $\endgroup$ – StefanH Jul 7 '15 at 14:48
  • $\begingroup$ It means that $P(a - \varepsilon < X < a + \varepsilon) = 1$ for all $\varepsilon > 0$. $\endgroup$ – Hans Engler Jul 7 '15 at 14:51
  • $\begingroup$ I think there is some continuity argument left, but $P(a-\varepsilon < X < a+\varepsilon)$ is this a continous function, $F$ is just right-continuous? $\endgroup$ – StefanH Jul 7 '15 at 15:06
  • $\begingroup$ By construction $F$ is right continuous. You can also just show that $P(a - \varepsilon < X \le a + \varepsilon) = F(a + \varepsilon) - F(a - \varepsilon) = 1$ for all $\epsilon > 0$. $\endgroup$ – Hans Engler Jul 7 '15 at 15:42

Suppose $X$ is $\mathcal G$-measurable. Then $F(x):=P(X \leq x)\in \{0,1\}$ for all $x \in \Bbb R$.

Let $a = \sup\{x \in \Bbb R: F(x) = 0\}$, and similarly let $b = \inf \{ x\in \Bbb R: F(x)=1\}$.

I claim that $a=b$. Otherwise $a<b$, so we could pick some $c \in (a,b)$, and we would have that $F(c) \in (0,1)$ since $F$ is nondecreasing, which contradicts the fact that $X$ is $\cal G$-measurable.

Since $a=b$, it follows that $P\big(X \leq a+k^{-1}\big)=1$ and $P\big(X\leq a-k^{-1}\big)=0$ for all $k \in \Bbb N$, by definition of $a,b$. Therefore $P\big(|X-a|\leq k^{-1}\big) = 1$ for all $k \in \Bbb N$. Hence, by the basic properties of probability, we see that $$P(|X-a|=0) = \lim_{k\to \infty} P\big(|X-a|\leq k^{-1}\big) = 1$$ Hence $X=a$ almost surely.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.