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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?

StefanH
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    You just have to show that a non-decreasing function $F$ that can only attain the values 0 and 1 has this form. – Hans Engler Jul 07 '15 at 14:25
  • 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? – StefanH Jul 07 '15 at 14:48
  • It means that $P(a - \varepsilon < X < a + \varepsilon) = 1$ for all $\varepsilon > 0$. – Hans Engler Jul 07 '15 at 14:51
  • 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? – StefanH Jul 07 '15 at 15:06
  • 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 $\epsilon > 0$. – Hans Engler Jul 07 '15 at 15:42

2 Answers2

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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.

shalin
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Just complementing the answer by shalop. In fact, a and b may be infinite values. Suppose, for example: $\Omega = \mathbb{N}$, $\mathcal{G} = \mathcal{P}(\mathbb{N})$ with probability measure \begin{equation*} P(A) = \begin{cases} 0, & \text{ if A is a finite set},\\ 1, & \text{ if A is an infinite set}. \end{cases} \end{equation*} Indeed, $P(A) \in \{0,1\}$, $\forall A \in \mathcal{G}$. Now, take $X(n) = n$, $\forall n \in \mathbb{N}$. Then X is a random variable (i.e. $\mathcal{G}$-measurable), but $P(X = x) = 0$, and $P(X > x) = 1$, $\forall x \in \mathbb{R}$. In this case $c = \infty$. So I think it would be necessary to include that $c \in \overline{\mathbb{R}}$.

nhilario
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  • This isn't a probability measure. Let $i ∈ \mathbb{N}$ and consider $1 = P(\mathbb{N}) = P \left ( \bigcup_i {i} \right ) = \sum_i P({i}) = \sum_i 0 = 0$. It isn't countably additive. – Mister Feb 07 '25 at 17:42