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Let $\mu$ be a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ denotes the Borel sets.

Then, is it true that there exists a probability space $(\Omega,\Sigma,\mathbb{P})$ and a random variable $X$ defined on this probability space such that

$$ P(X \in B) = \mu(B)$$ for every borel set $B$?

I know the "converse" of the claim is true: given a random variable $X$ on some probability space, there exists a probability measure on $\mathbb{R}$ such that $ P(X \in B) = \mu(B)$.

user74261
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1 Answers1

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You can always take the probability space to be the interval $(0,1)$ with the Borel sigma-algebra and the Lebesgue measure on it. Define the random variable $X$ to be $X(\omega)=G(\omega)$ for $\omega\in(0,1)$, where $G$ is the quasi-inverse of the cumulative distribution function $F$ of $\mu$, meaning

\begin{equation} F(x)=\mu((-\infty,x]) \quad \textrm{and} \quad G(\omega)=\sup F^{-1} ([0,\omega])=\sup\{s\in\mathbb{R}\ |\ F(s)\leq \omega \}. \end{equation}

Here $F^{-1}$ denotes the preimage under the function $F$. This is a very nice and very important lemma, you should prove it for yourself.

Alexisz
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    As @David commented, you can also take the identity, if you have $\mu$. In the above construction, you don't need $\mu$, just $F$ which has to be monoton increasing, right-continuous, with limits 1 and 0 at infinity and minus infinity. Then the constructed $X$ will have the distribution function F. In almost every situation, you can think about an abstract probability space to be the interval $(0,1)$ with the uniform distribution. – Alexisz Aug 18 '15 at 19:51
  • I know I am late to the party, but this answer is awesome, it allows us to assume the existence of random variables describing our "unknown information" – QED Mar 10 '23 at 23:39