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 $\geq 2$.

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.


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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    $\begingroup$ 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 $F(y) = \inf_{x\in\mathbb Q^n}M_x(y)$. $\endgroup$
    – Evan Aad
    Aug 3 '12 at 15:01
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    $\begingroup$ @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. $\endgroup$
    – Erick Wong
    Aug 3 '12 at 22:21
  • $\begingroup$ Actually the $\epsilon$ is redundant. $\endgroup$
    – Evan Aad
    Aug 4 '12 at 5:38

Actually the Wikipedia properties are not sufficient. You also need n-monotonicity, which is defined here on pp. 265-266: https://books.google.fi/books?id=GjPUBwAAQBAJ&printsec=frontcover&hl=fi#v=onepage&q&f=false A more readable treatment is here: http://home.uchicago.edu/hickmanbr/uploads/PT_REVIEW_CH3.pdf but the monotonicity condition in Theorem 3 is too weak, use the n-monotonicity instead. However, in Theorem 2 the n-monotonicity is correctly stated for R^2: F(b, d) − F(b, c) − F(a, d) + F(a, c) ≥ 0 for all a < b, c < d.

Beware: the Springer link it seems to use left-continuous CDFs.


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