Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,\mathcal F,\phi)$ and $(Y,\mathcal G,\psi)$ be two probability spaces and $(Z,\mathcal H,\mu)$ be their product space: i.e. we have that $Z = X\times Y$. Given an arbitrary set $F\in\mathcal H$ and any point $x\in X$ let us define $$ \pi_x(F) = \{y\in Y:(x,y)\in F\}. $$ I wonder if the following formula is true: $$ \mu(F) = \int\limits_X \psi(\pi_x(F))\phi(dx).\tag{1} $$

I think it is true and the proof goes like this:

  1. For any $A\in \mathcal F$ and $B\in\mathcal G$ we have $$ \int\limits_X \psi(\pi_x(A\times B))\phi(dx) = \int\limits_X \psi(B)1_A(x)\phi(dx) = \phi(A)\psi(B) = \mu(A\times B) $$ so $(1)$ is true for the class $\mathcal C$ of measurable rectangles $A\times B$.

  2. If $(1)$ holds for $F\in \mathcal H$ it also holds for $F^c$.

  3. If $(1)$ holds for the disjoint sequence $(F_n)_{n\geq 0}$ in $\mathcal H$ then it holds for $F = \bigcup\limits_{n\geq 0}F_n$.

As a result, if $\mathcal H'$ is the class of sets for which $(1)$ holds then $\mathcal C\subset \mathcal H'$ and it is closed under taking complements and countable unions, so $\mathcal H'$ is a $\sigma$-algebra and hence $\mathcal H\subset \mathcal H'$ hence $(1)$ is true for any $F\in \mathcal H$.

My question are the following: first, please tell me if the proof is correct, especially I am not sure if I need to show measurability of $\psi(\pi_x(F))$ or is it implicitly proved already. Second, if the result is true I'm pretty sure it is known - so I wonder about a reference.

share|cite|improve this question
That is the Fubini Theorem, see for example Real Analysis by Stein and Sakarchi! – checkmath Apr 17 '12 at 11:01
You can also check Measures, Integrals and Martingales by R. L. Schilling page 123. – Stefan Hansen Apr 17 '12 at 11:45
@chessmath: thanks for the reference, but I was rather looking for the general result, not related to $\mathbb R^n$ and Lebesgue measure – Ilya Apr 17 '12 at 13:18
up vote 2 down vote accepted

In order for (1) to well-defined you need to show that $x\mapsto \psi(\pi_x(F))$ is $(\mathcal{F},\mathcal{B}(\mathbb{R}))$-measureable for all $F\in \mathcal{H}$. You can do this the following way:

  • Convince yourself that $\pi_x(F)\in \mathcal{G}$ for all $F\in \mathcal{H}$ and $x\in X$.
  • Define $\mathcal{D}=\{H\in \mathcal{H}\mid x\mapsto \psi(\pi_x(H)) \text{ is measureable}\}$ and show that this is a Dynkin-system, which contains $\{F\times G\mid F\in\mathcal{F},\, G\in\mathcal{G}\}$.
  • Use Dynkin's Lemma to conclude that $\mathcal{D}=\mathcal{H}$.

Besides that your proof seems correct to me.

share|cite|improve this answer
Thank you, I've done this way. It appeared that I was looking exactly for the result Lemma 1.7.4 – Ilya Apr 17 '12 at 14:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.