5

Let $X,Y$ be some nice measureable spaces (i'm interested in $[0,1]$ so we can assume compact, etc.). let $\mu$ be a measure on $X\times Y$.(again, assume it's nice, i.e. probability measure. anything else needed?

Are there necessarily measures $\mu_1,\mu_2$ on $X,Y$ resp. such that $\mu=\mu_1\times\mu_2$?

I thought it is a reasonable conjecture, but i couldn't prove it. My intuition is the lebesgue [/haar] measure, for which the answer is obviously true. I thought one could define $\mu_1(A)=\mu(A\times Y)$ and reps. $\mu_2$, but I couldn't show it would be a good construction (it is good for lebesgue, though).

EDIT - only this is now relevant:

thanks guys. But now i'm slightly confused, as I saw in a paper the terminology "projection of a measure", which is supposed to be some measure on XX derived from the measure on $X\times Y$. this must be something like the pushforward measure by projection, which is μ1μ1 that I defined here if i'm not wrong. I think that the paper does not really require the measure to be the product of those (which I now understand is not true), but perhaps satisfy some kind of a fubini identity $\mu(A)=\int\mu_1(A^y)d\mu_2(y)$? is that necessarily true?, (under nice enough conditions)

The goal is to deduce properties of μ from the projections, so what could be true?

J. Doe
  • 71
  • 10
    Counterexample: Let $X$ and $Y$ both be the 2-element set ${0,1}$, and give $X\times Y$ a probability measure in which 3 of the 4 singleton subsets have measure $1/3$ while the remaining one has measure 0. – Andreas Blass Jul 23 '16 at 15:50
  • I suppose the difficulty lies in, once having identified a measure on the product space, verifying that there are or are not measures on the individual component spaces such that that is the product. My intuition says no, since the product of measures satisfies a universal property: it in some sense the "minimal" measure on the product space which has well defined projections onto the measures on the individual component spaces, see for example here: http://math.stackexchange.com/questions/276632/is-my-understanding-of-product-sigma-algebra-or-topology-correct – Chill2Macht Jul 23 '16 at 15:50
  • 2
    Lebesgue measure on $[0,1]\times[0,1]$ is the completion of the product of Lebesgue measure on $[0,1]$ with itself, not the product itself. – Aweygan Jul 23 '16 at 15:50
  • @Aweygan True, but the OP's misunderstanding is much more basic. – Did Jul 23 '16 at 15:59
  • thanks guys. But now i'm slightly confused, as I saw in a paper the terminology "projection of a measure", which is supposed to be some measure on $X$ derived from the measure on $X\times Y$. this must be something like the pushforward measure by projection, which is $\mu_1$ that I defined here if i'm not wrong.

    I think that the paper does not really require the measure to be the product of those (which I now understand is not true), but perhaps satisfy some kind of a fubini identity $\mu(A)=\int \mu_1 (A^y) d\mu_2 (y)$. Does this sound more likely to you?

     – J. Doe Jul 23 '16 at 16:06
  • The goal is to deduce properties of $\mu$ from the projections, so what could be true? – J. Doe Jul 23 '16 at 16:06
  • In order to answer this question, we need a precise definition of "nice." – Math1000 Jul 23 '16 at 17:41
  • I wouldn't know. Im interested in understanding when it's true. some suffice conditions (not neccessarily the strongest possible) should be fine i guess. @Math1000 – J. Doe Jul 23 '16 at 18:03

0 Answers0