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Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space satisfying $\mathbb{P}(F) \in \{0,1\}$ for all $F \in \mathcal{F}$ and let $X$ be a random variable on $(\Omega, \mathcal{F}, \mathbb{P})$.

I want to prove that there exists a $c \in \mathbb{R}$ such that $\mathbb{P}(X=c)=1$.

Therefore, I am considering \begin{align} \mathbb{P}(X\leq x). \end{align} However I do not see directly how this should work. Any suggestions?

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here is one way to prove this: suppose $X$ can take two different values $c_1 \neq c_2$ with positive probability: $$ \mathbb{P}(X=c_1) \in (0,1), \quad \mathbb{P}(X=c_2) \in (0,1). $$ Then from the measurability of $X$ we get that $$ F_1:=X^{-1}(\{c_1\}) \in \mathcal{F}. $$ But $$ \mathbb{P}(F_1)= \mathbb{P}(X=c_1) \notin \{0,1\}, $$ which is a contradiction.


Edit: new part(thanks to iJup for pointing out that there is an argument missing)

This implies that $$ \mathbb{P}(X=c) \in \{0,1\} \quad \text{for all} \ c \in \mathbb R. $$ From here there are two possible scenarios: either there exists $c \in \mathbb R$ such that $\mathbb{P}(X=c) =1$, or the law of $X$ is absolutely continuous with respect to the Lebesgue measure and has a probability density $f$.

I will argue that second scenario also yields a contradiction. If the law of $X$ were continuous there would exist a Borel set $A$ such that $\mathbb{P}(X\in A) =\frac 12$ (for example $A=(-\infty,q)$, where $q$ is the $50\%$ quantile). Now we can use a similar argument as before: $$ F:=X^{-1}(A) \in \mathcal{F}. $$ But $$ \mathbb{P}(F)= \mathbb{P}(X\in A) \notin \{0,1\}, $$ which is a contradiction.

Therefore there exist $c \in \mathbb R$ such that $\mathbb{P}(X=c) =1$

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  • $\begingroup$ Thanks Cettt for your answer. So you say that $ 0 < \mathbb{P}(X=c_1)<1$ and since $\mathbb{P}(F) \in \{0,1\}$ for all $F \in \mathcal{F}$ there is a contradiction. However, for my it is not directly clear how does this prove that $\exists c \in \mathbb{R} : \mathbb{P}(X=c)=1$. $\endgroup$ – iJup Dec 23 '16 at 13:13
  • $\begingroup$ well my argument shows that whenever $X$ takes one value with probability $p \notin \{0,1\}$ there is a contradiction. This is equivalent of saying that we have to have $\mathbb P(X=c) \in \{0,1\}$ for all $c \in \mathbb R$. And this implies the existence of $c \in \mathbb R$ such that $\mathbb P (X=c)=1$. $\endgroup$ – Cettt Dec 23 '16 at 13:23
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    $\begingroup$ Okido. If I understand you correctly, you introduce $c_1$ and $c_2$ since $\mathbb{P}(X=c_1)$ could be equal to $0$? If this is the case, why does it follow then that $\mathbb{P}(X=c_2)$ equals $1$? $\endgroup$ – iJup Dec 23 '16 at 13:29
  • $\begingroup$ yeah you are right about that, I will edit the answer $\endgroup$ – Cettt Dec 23 '16 at 13:31
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Since every probability measure is continuous from below, set $$C=\{c \in \mathbb R \mid P(X \leq c)=1 \}$$ must have at least one element; otherwise, from $$\mathbb R = \bigcup_{n=1}^\infty (-\infty,n]$$ we get $\mathbb P(\Omega)=P(X \in \mathbb R)=\lim_{n \to \infty} P(X \leq n)=0$. But then, $P(X = \inf C)=1$.

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Need a bit more clarity to precisely answer your question is $\mathcal{F}$ constructed from $\mathbb{R}$? I am not sure your statement is true in general, as while you could instead say that it has to be true in the limit as c goes to infinity, your condition would for instance by violated by the normal distribution?

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    $\begingroup$ It doesn't matter where the probability space comes from. $\endgroup$ – Ian Dec 23 '16 at 12:40

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