probability theory - Why a normally distributed random variable and an exponential distributed r.v. exist on the same space?

Let $F:\mathbb R\to [0,1]$ a function s.t. $F$ is right continuous, $\lim_{x\to -\infty }F(x)=0$, $\lim_{x\to \infty }F(x)=1$ 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.

$$F(x)=\mathbb P(X\leq x),$$

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. $F(x)=\mathbb P(X\leq x)$.

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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 $F(x)=\int_{-\infty }^x e^{-x^2/2}/\sqrt{2\pi}dx$) and $Y$ is normally distributed on $(\tilde \Omega ,\tilde {\mathcal F},\tilde {\mathcal P})$, ((take $F(x)=\int0^\infty \lambda e^{-\lambda x}dx$) there is no reason that such two r.v. exist on the same space.