# Is every probability measure of $\mathbb R^n$ induced by a random vector?

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.

• 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 $A$ Commented Dec 18, 2020 at 1:00
• 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.
– Max
Commented Dec 18, 2020 at 1:00
• @Masacroso Thank you, it is indeed trivial. I can't believe I missed the identity function. Commented Dec 18, 2020 at 1:09
• @Zhang dont worry, we generally overlook many trivial things Commented Dec 18, 2020 at 1:20