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I come across this exercise:

Given any probability measure $\mu$ on $(\mathbb{R},\mathcal{B})$, define a random variable such that the probability measure it induces is $\mu$. Can this be done in an arbitrary probability space?

The first construction is simple, just define $X:(\mathbb{R},\mathcal{B})\mapsto (\mathbb{R},\mathcal{B})$ to be the indentity map. Then for every measurable set in $(\mathbb{R},\mathcal{B})$, its preimage doesn't change, and has the same measure under $\mu$. However, for an arbitrary case, I feel it's true but don't how to construct.

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  • $\begingroup$ If the probability space is rich enough to carry a uniform distribution, then it's possible. See this question. I think "arbitrary probability space" is probably asking too much. Consider a "degenerate" probability space... $\endgroup$ Commented Oct 6, 2019 at 19:05

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