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Question: Given a borel measure $\mu$ on $\mathbb R^n$ such that $\mu(\mathbb R^n) = 1$, is it induced by a random vector $X$ in a way that $\mu(E) = \mathbb P(X \in E)$ for every borel set $E$?

This is the multi-dimensional version of this question. If $n=1$, we can explicitly construct a random variable using the technique of 'generalized inverse' as is in the link. However, things become much harder if $n>1$. I guess to deal with the multi-dimensional case, some measure-theoretic stuff must be involved.

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    $\begingroup$ yes, the identity function $\mathrm{id}:\mathbb{R}^n\to \mathbb{R}^n,\, (x_1,\ldots ,x_n)\mapsto (x_1,\ldots ,x_n)$ defines this probability measure. The proof is trivial, as $\mathrm{id}^{-1}(A)=A$ for any measurable set $A$ $\endgroup$
    – Masacroso
    Commented Dec 18, 2020 at 1:00
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    $\begingroup$ Both $\mathbb{R}$ and $\mathbb{R}^n$ are Polish spaces (en.wikipedia.org/wiki/Polish_space) and so there is a measurable bijective function (with measurable inverse) from one to the other. Horrible as this map is, composing with it deduces a "yes" answer to your question from the "yes" answer to the 1D version. $\endgroup$
    – Max
    Commented Dec 18, 2020 at 1:00
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    $\begingroup$ @Masacroso Thank you, it is indeed trivial. I can't believe I missed the identity function. $\endgroup$ Commented Dec 18, 2020 at 1:09
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    $\begingroup$ @Zhang dont worry, we generally overlook many trivial things $\endgroup$
    – Masacroso
    Commented Dec 18, 2020 at 1:20

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