TITLE: Constructing an independent uniform random variable from two independent ones
QUESTION [10 upvotes]: Does there exist a continuous (differentiable) function $h:[0,1]\times [0,1] \to [0,1]$ such that if $\alpha,\beta\in [0,1]$ are independent and uniformly distributed on $[0,1]$, the random variable $h(\alpha,\beta)$ is uniformly distributed on $[0,1]$ independent of $\alpha,\beta$?
Clarification: By independent I mean pairwise independent, i.e.
$\mathbb{P}[h(\alpha,\beta)\leq x\mid \alpha]=x$ for all $x,\alpha\in[0,1]$ 
and
$\mathbb{P}[h(\alpha,\beta)\leq x\mid \beta]=x$ for all $x,\beta\in[0,1]$..
Thanks a lot!

REPLY [3 votes]: Inspired by Nate's answer, here is why it cannot be differentiable. First for each $y \in [0,1]$, there must be some $x \in [0,1]$ such that $h(x,y) = 0$, otherwise by compactness of the unit interval and continuity of $h$, we would have $P(h(\alpha, y) < \epsilon) = 0$ for some small $\epsilon$.
Next I claim that such $x$ cannot be in the interior of $[0,1]$. If it is, then by linear approximation near $x$ and the limit definition of derivative, $P(h(\alpha, y) < \epsilon) > \epsilon$ for sufficiently small $\epsilon$, because the derivative $\partial_1 h(x,y)$ must be zero. This is clearly enough to get a contradiction: it means that $h(0,y) = 0$ or $h(1,y)=0$ for all $y \in [0,1]$. Reversing the role of $x$ and $y$, we see that $P(h(0, \beta) < \epsilon) \geq 1/2$ or $P(h(1, \beta) < \epsilon) \geq 1/2$. Since $h$ is continuous, this argument also works in the almost sure category.