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It is well-known that CDFs (Cumulative Distribution Functions) of one dimensional random variables are Borel measurable. But does the same apply to CDFs of multi-dimensional random variables (rvecs)? It suffices, for my purposes, to consider finite dimensional rvecs.


Relevant definitions

Let $n$ be an integer $>= 2$.

Call a probability measure over the Borel field on $\mathbb R^n$ an $n$-dimensional probability measure.

To every $n$-dimensional probability measure, $m$, define the following function $F:\mathbb R^n\rightarrow\mathbb R$, called $m$'s CDF: $F(x)=m\left((-\infty, x]\right)$ where $(-\infty, x]$ is the set of all $y$ in $\mathbb R^n$ such that $y\leq x$ component-wise.

An $n$-dimensional CDF is a function $F:\mathbb R^n\rightarrow\mathbb R$ that can be obtained as some $n$-dimensional probability measure's CDF.

$n$-dimensional CDFs are known to be characterized by certain properties.

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For any $x \in \mathbb R^n$, consider the measurable box-like function $M_x(y) = m((-\infty,x]) \cdot \mathbb1_{[x,\infty)}(y)$. Here $\mathbb1_{[x,\infty)}(y)$ is the indicator function of "$y \ge x$ component-wise".

It should be easy to show that $F(y) = \sup_{x \in \mathbb Q^n} M_x(y)$ for any $y$ (take an increasing sequence of $x$ converging to $y$), and this makes $F$ the countable supremum of measurable functions.

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Thank you. What a great idea, though i suspect your $F(y)$ fails to work, since it's left-continuous, whereas a CDF should be right-continuous. However, a slight modification will fix it: define $M_x(y) = m((-\infty,x])\cdot\mathbb1_{(-\infty,x]}(y) + (1+\epsilon)\cdot\mathbb1_{\mathbb R^n \setminus (-\infty,x]}(y)$ and $F(y) = \inf_{x\in\mathbb Q^n}M_x(y)$. –  Evan Aad Aug 3 '12 at 15:01
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@EvanAad Oops, of course you are correct! Thanks for the fix. How silly of me to try to write $(-\infty,x]$ as a countable union rather than a countable intersection. –  Erick Wong Aug 3 '12 at 22:21
    
Actually the $\epsilon$ is redundant. –  Evan Aad Aug 4 '12 at 5:38

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