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Given two measures $\mu$ and $\nu$ on some measurable space $X$, is there a way to multiply them to get $\mu \cdot \nu$, another measure on $X$ (and not $X \times X$, as for the usual notion of product measure)?

Here's a case where I know how to give a definition: if both $\mu$ and $\nu$ are absolutely continuous with respect to some common measure $\lambda$, then we can take their Radon–Nikodym derivatives with respect to that measure to obtain two functions $f_\mu$ and $f_\nu$, so that $\mu = \int f_\mu d\lambda$, $\nu = \int f_\nu d\lambda$ which we can then multiply, to give us $\mu \cdot \nu = \int (f_\mu \cdot f_\nu) d\lambda$.

This came up in the context of Monte–Carlo integration, and in particular Monte–Carlo path tracing. In this case, the measure space could be, say, the set of angles at which an incoming light ray bouncing off an object could be reflected, $\mu$ would be a probability measure describing the probability of outgoing angles, and $\nu$ would be a measure describing the light sources in the scene visible from that point of reflection. The idea of the multiplication $\mu \cdot \nu$ is to produce something which describes the sampling of light sources at that point, depending on the incoming ray (and $\mu$, on top of just $\nu$).

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What plays the role of the BRDF in this formalization of path tracing? Is it not a measure on $X\times X$? – Rahul May 10 '12 at 17:45
I'm fixing the point of contact and the incident angle. What is left then is simply a measure on the set of outgoing angles, and is what I'm taking to be the (somewhat generalised notion of the) BRDF. – Will May 10 '12 at 17:50
I don't think your approach works that well. If we replace the measure $\lambda$ by $2\lambda$ we would get the derivatives $1/2 f_\mu$ and $1/2f_\nu$ instead. Now $\int (1/2 f_\mu)(1/2 f_\nu)d2\lambda$ equals $1/4\int f_\mu f_\nu d2\lambda$ equals $1/2\int f_\mu f_\nu d\lambda$, so the product depends on the underlying measure. – Michael Greinecker May 10 '12 at 17:51
I guess I don't really understand your approach. If the incident direction is fixed, there is only one light source visible in that direction, and there is no $\nu$. Perhaps you want to multiply the distribution of outgoing directions with that of the light sources, but I don't understand why you would want that. – Rahul May 10 '12 at 18:09
I think the best starting point would be to try to figure out what the product of two finite measures on a finite discrete space should be. There is a conceptual issue that has nothing to do with sophisticated machinery. – Michael Greinecker May 10 '12 at 18:32
up vote 11 down vote accepted

The case you describe is the general case since every measures $\mu$ and $\nu$ are absolutely continuous with respect to $\mu+\nu$. More precisely, there exists $h_{\mu,\nu}$ with $0\leqslant h_{\mu,\nu}\leqslant1$ everywhere such that $\mu=h_{\mu,\nu}(\mu+\nu)$ and $\nu=(1-h_{\mu,\nu})(\mu+\nu)$. Thus one can define an intrinsic product $\mu\odot\nu$ by $$ \mu\odot\nu=h_{\mu,\nu}(1-h_{\mu,\nu})(\mu+\nu). $$ When $\mu$ and $\nu$ are absolutely continuous with respect to the Lebesgue measure (or any other measure of reference) with densities $f$ and $g$ respectively, then $\mu\odot\nu$ is absolutely continuous with respect to the Lebesgue measure with density $f\odot g$ defined as follows: on $[f+g=0]$, $f\odot g=0$, and, on $[f+g\ne0]$, $$ f\odot g=\frac{fg}{f+g}. $$ This product $\odot$ on measures is commutative (good), associative (good?), the total mass of $\mu\odot\nu$ is at most $\frac14$ times the sum of the masses of $\mu$ and $\nu$, in particular the product of two probability measures is not a probability measure (not good?), $\mu\odot\mu=\frac12\mu$ for every $\mu$, and finally $\mu\odot\nu=0$ if and only $\mu$ and $\nu$ are mutually singular (good?) since $\mu\odot\nu$ is always absolutely continuous with respect to both $\mu$ and $\nu$.

Edit To normalize things, another idea is to consider $\mu\Diamond\nu=2(\mu\odot\nu)$. In terms of densities, this corresponds to a harmonic mean, since $\mu\Diamond\nu$ has density $f\Diamond g$, where $$ \frac1{f\Diamond g}=\frac1{2(f\odot g)}=\frac12\left(\frac1f+\frac1g\right). $$ In particular, this new intrinsic product $\Diamond$ is idempotent (good?), commutative (good), and not associative (not good?).

Edit A canonical product concerns probability measures and transition kernels. That is, one is given a measured space $(X,\mathcal X,\mu)$, a measurable space $(Y,\mathcal Y)$ and a function $\pi:X\times\mathcal Y\to[0,1]$ such that, for every $x$ in $X$, $\pi(x,\ )$ is a probability measure on $(Y,\mathcal Y)$. Then, under some regularity conditions, the product $\mu\times\pi$ is the unique measure on $(X\times Y,\mathcal X\otimes\mathcal Y)$ such that, for every $A$ in $\mathcal X$ and $B$ in $\mathcal Y$, $$ (\mu\times \pi)(A\times B)=\int_A\mu(\mathrm dx)\pi(x,B). $$ In particular, $B\mapsto(\mu\times\pi)(X\times B)$ is a probability measure on $(Y,\mathcal Y)$.

When $\mu$ has density $f$ with respect to a measure $\xi$ and each $\pi(x,\ )$ has density $g(x,\ )$ with respect to a measure $\eta$, $\mu\times\pi$ has density $(x,y)\mapsto f(x)g(x,y)$ with respect to the product measure $\xi\otimes\eta$.

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I need to think about my problem more to see if this completely addresses my issues, but this is really as thorough an answer as is possible. Thanks. – Will May 12 '12 at 20:49
@LVK Thanks for (the appreciation underlying the award of) this bounty. – Did Aug 25 '12 at 21:05

No, if you mean to multiply them in the freshman sense. Given two measures $u$ and $v$ on some measurable space $X$, define the "product" $w = u \cdot v$ as $w(E) = u(E) \cdot v(E)$ for all $E \in X$. We would like to see if $w$ is a measure or find a counterexample.

Consider the interval $E = (0,2)$ on $X = \mathbb R$, and let $u$ be the standard length measure and $v$ be the measure of the area under the function $f = |x|$. Suppose $w$ is the "product" of $u$ and $v$, $w = u \cdot v$, as above.

Then, $$w((0,2)) = u((0,2))\cdot v((0,2)) = 2\cdot 2 = 4$$ I can split up the interval $(0,2)$, and a measure stays the same, but: $$w((0,2)) = w((0,1]\cup (1,2)) = w((0,1]) + w((1,2)) = u((0,1])\cdot v((0,1]) +u((1,2))\cdot v((1,2)) = 1\cdot \frac 1 2 + 1 \cdot \frac 3 2 = 2 \not= 4$$

So $w$ is not a measure. Of course, you probably knew that.

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