Pushforward via $(x,y,z) \mapsto xy+(1-x)z$ of Lebesgue measure Is there any easy way to see that the map $f(x,y,z) = xy+(1-x)z$  pushes forward the (three-dimensional) Lebesgue measure on the unit cube to the (one-dimensional) Lebesgue measure on the unit interval?
I'm pretty sure I can prove it by showing directly that $\lambda^{3}\circ f^{-1}[0,b) = b$ for every $b \in [0,1]$, but the calculuation is a bit tedious, involving cross-sectional areas, Fubini, and whatnot. I suspect there's an elegant way of doing it. 
 A: There isn't, because it doesn't. You are taking three uniformly distributed independent random variables $X,Y,Z$ and forming a weighted average of $Y$ and $Z$ using $X$ as a weight. The result of averaging will fall near $1/2$ more often than near $0$ or $1$. I took  $10^6$ samples for illustration: 

And here is a mathematical derivation of the pushforward density. By symmetry, it suffices to consider $0<x<1/2$ and double the result. For a fixed $x\in (0,1/2)$ the probability density function of $xY$ is $\frac{1}{x}\chi_{[0,x]}$, because $xY$ is uniformly distributed on $[0,x]$. Similarly, the pdf of  $(1-x)Z$ is $\frac{1}{1-x}\chi_{[0,1-x]}$. Since the two random variables are independent, the pdf of their sum  $xY+(1-x)Z$ is the convolution of pdfs:
$$p_x(t)=\frac{1}{x(1-x)}\chi_{[0,x]}*\chi_{[0,1-x]} = 
\frac{1}{x(1-x)} \times\begin{cases} t,\quad &0\le t\le x \\ x,\quad &x\le t\le 1-x  \\ 1-t,\quad &1-x\le t\le 1 \end{cases}  $$
This is a trapezoidal function which stays flat on $[x,1-x]$ and drops to $0$ at $t=0,1$.
We get  the pdf of $XY+(1-X)Z$ by integrating $p_x$ over $x$: 
$$p(t) = 2\int_0^{1/2} p_x(t)\,dx = (2t-2)\log(1-t)-2t\log t  $$
Looks pretty neat; I never saw this distribution but this does not say much. Here is the plot, for comparison with the above histogram. 

