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Let $F:mathbb Rto [0,1]$ a function s.t. $F$ is right continuous, $lim_xto -infty F(x)=0$, $lim_xto infty F(x)=1$ and non decreasing.

I have a theorem in my lecture that says :

A function that the above properties is the distribution of some random variable.

I'm a bit confused by this theorem. Does it mean that

A) There exist a probability space $(Omega ,mathcal F,mathbb P)$ and a random variable $X:Omega to mathbb R$ s.t. $$F(x)=mathbb P(Xleq x),$$

B) Or, in the previous theorem a probability space is already fixed ? I.e the theorem is :

Let $(Omega ,mathcal F,mathbb P)$ a probability space. If a function $F$ has the properties above, then there is a r.v. $X:Omega to mathbb R$ s.t. $F(x)=mathbb P(Xleq x)$.

Why such a question ? My previous question could look a bit weird, but when in an exercise we say : Let $(Omega ,mathcal P,mathbb P)$ a probability space and let $X,Y$ two random variable s.t. $X$ is normally distributed and $Y$ is exponentially distributed.

The question arise is : why two such r.v. exist on the same space ? If B) is not true, even if $X$ is normally distributed on $(Omega ,mathcal F,mathbb P)$ (take $F(x)=int_-infty ^x e^-x^2/2/sqrt2pidx$) and $Y$ is normally distributed on $(tilde Omega ,tilde mathcal F,tilde mathcal P)$, ((take $F(x)=int0^infty lambda e^-lambda xdx$) there is no reason that such two r.v. exist on the same space.

Given $F$ there exists some probability space on which there is a random variable with distribution $F$. You cannot always do this on an arbitrary space. For example if the probability space consists of a finite set we cannot construct $X$ with normal distribution on it. [However, it can be shown that we can always find $X$ on $(0,1)$]. To construct $X$ and $Y$ with given distributions on the same space we can use product spaces. Kolomogorov's Existence Theorem (also known as Kolomogorov's Consistency Theorem) guarantees existence any number of random variables with given distributions on a single space.