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I am taking an introductory probability theory course and we defined a random variable as a function $X: \Omega \to \mathbb{R}$. We defined $X$ to have uniform distribution on $[0,1]$ if $$\mathbb{P}(\{\omega \in \Omega: X(\omega)=x\})= \mathbb{P}(X=x)= \begin{cases} 1 &\text{if} \ \ 0 \leq x \leq 1 \\ 0 &\text{else}\end{cases}$$ where $\mathbb{P}((a,b))=b-a$. However, we now have $\mathbb{P}(X=0) = \mathbb{P}(X=1)=1$ and so seemingly $(X=1)=[0,1]=(X=0)$. Hence for $\omega \in [0,1]$ we have $X(\omega)=0$ and $X(\omega)=1$ so $X$ is not a function (or we could use any $x\in [0,1]$ not just 0,1 to apparently get uncountably many different values of $X(\omega)$).

What am I missing here? Is this simply because I am not defining things with measure theory?

Austin Shiner
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    The definition does not make sense. If $X$ has (continuous) uniform distribution on a set of positive Lebesgue-measure, then $\mathbb{P}(X=x)=0$ for any $x$. I suspect that it is an instance of some old-school notation abuse. Indeed, in some old textbooks $\mathbb{P}(X=x)$ is used to denote the PDF of $X$ when $X$ has continuous distribution (even we can find one in Durrett's celebrated textbook!). Not to mention, such practice is now highly depreciated for obvious reasons. – Sangchul Lee Sep 14 '19 at 04:27
  • Ah yes, that would make sense. Thanks. – Austin Shiner Sep 14 '19 at 04:33
  • also note that we doesn't know what is $\Omega$, so $\omega\in[0,1]$ doesn't make sense. $[0,1]$ is referring to a subset of $\Bbb R$, not of $\Omega$. In any case, as Lee says, the definition doesn't make sense – Masacroso Sep 14 '19 at 05:19

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