I wonder if you can help me out with this problem that I'm trying to understand (from an old exam, not homework):

Let $(E,\mathcal{P}(E))\,$ (where $E= \lbrace 0,1 \rbrace )$ be a measurable space, $\mu_1$ and $\mu_2\,$ two measures on $(E,\mathcal{P}(E))$ (i.e., $\mu_{i}(E) = 1$ for $i = 1,2$) such that $\mu_1(\lbrace0\rbrace)=p \,, \mu_2(\lbrace0\rbrace)=q \,$ and $ p,q \in [0,1]$. Give a necessary and suffiecient condition for $\mu_1 \otimes \mu_2 = \mu_2 \otimes \mu_1$

I'm thinking along the lines: "$\mu_1 \otimes \mu_2$ has to be a product measure on the product space $(E^2,\mathcal{P}(E)^2)$ then $\mu_1 \otimes \mu_2=\mu_1\mu_2=\mu_2\mu_1=\mu_2 \otimes \mu_1$. For this to be true $\forall A \in\mathcal{P}(E), \mu_1(A)$ has to be a measurable function on $(E,\mathcal{P},\mu_2)$ and vice versa. Is this true? Can someone help me sort this out? Thank you in advance!

  • $\begingroup$ $$p=q{}{}{}{}$$ $\endgroup$ – Did Apr 24 '16 at 13:46
  • $\begingroup$ "...then $\mu_1 \otimes \mu_2=\mu_1\mu_2$..." This is the point where you jump into the void. What is $\mu_1\mu_2$ already? $\endgroup$ – Did Apr 24 '16 at 13:54
  • $\begingroup$ There is no reason to assume the OP meant $\mu_1$ and $\mu_2$ to be probability measures. The original statement of question did NOT say $\mu_1$ and $\mu_2$ were probability measures and the question makes perfect sense for measures in general. $\endgroup$ – Ramiro Apr 24 '16 at 14:31
  • $\begingroup$ @drhab I agree. I removed the tag. $\endgroup$ – Ramiro Apr 24 '16 at 14:32
  • 1
    $\begingroup$ I re-read the question and it acctually says that $\mu_1$ and $\mu_2$ are probability measures. What difference does it make in this case? $\endgroup$ – user202542 Apr 24 '16 at 14:57

Note that $E\times E$ has only 4 elements. For any $x\in E$ we have

$$\mu_1\otimes\mu_2(\{(x,x)\}) =\mu_1(\{x\})\mu_2(\{x\})= \mu_2\otimes\mu_1(\{(x,x)\})$$

So we need to inspect only the two other points: $(0,1)$ and $(1,0)$.

Let $p=\mu_1(\lbrace0\rbrace)$ and $q=\mu_2(\lbrace0\rbrace)$. Then we have

$$p(\mu_2(E)-q) = \mu_1\otimes\mu_2(\{(0,1)\}) = \mu_2\otimes\mu_1(\{(0,1)\}) =q(\mu_1(E)-p) \tag{1}$$


$$(\mu_1(E)-p)q = \mu_1\otimes\mu_2(\{(1,0)\}) = \mu_2\otimes\mu_1(\{(1,0)\}) =(\mu_2(E)-q)p \tag{2}$$

From $(1)$ and $(2)$, we have that the necessary and suficient condition for $\mu_1\otimes\mu_2= \mu_1\otimes\mu_2$ is $$p\mu_2(E)= q\mu_1(E)$$

Remark: If $\mu_1$ and $\mu_2$ are probability measures, then the condition reduces to $p=q$.

  • $\begingroup$ Thank you! The fact that one can 'split' $\mu_1 \otimes \mu_2(\lbrace (x,x)\rbrace) $ into $\mu_1(x)\mu_2(x)$ is because we're dealing with a product measure? Can I always say $\mu \otimes \nu(A \times B)=\mu(A)\nu(B)$? $\endgroup$ – user202542 Apr 24 '16 at 15:40
  • $\begingroup$ @user202542 , Yes. For retangles $A \times B$, we always have $\mu\otimes\nu (A\times B)= \mu(A)\nu(B)$. $\endgroup$ – Ramiro Apr 24 '16 at 15:44

In general let $\mu,\lambda$ be measures $\langle\Omega,\mathcal{A}\rangle$ with $0<\lambda\left(\Omega\right)<\infty$ and $0<\mu\left(\Omega\right)<\infty$.

For $\nu\in\left\{ \mu,\lambda\right\} $ prescribe probability measure $P_{\nu}$ by $A\mapsto\frac{\nu\left(A\right)}{\nu\left(\Omega\right)}$ for $A\in\mathcal{A}$.

Then $\mu\otimes\lambda=cP_{\mu}\otimes P_{\lambda}$ and $\lambda\otimes\mu=cP_{\lambda}\otimes P_{\mu}$ for $c=\left[\mu\left(\Omega\right)\lambda\left(\Omega\right)\right]^{-1}$ so that $$\mu\otimes\lambda=\lambda\otimes\mu\iff P_{\mu}\otimes P_{\lambda}=P_{\lambda}\otimes P_{\mu}$$ Also we have $$P_{\mu}\otimes P_{\lambda}=P_{\lambda}\otimes P_{\mu}\iff P_{\mu}=P_{\lambda}$$

For this observe the necessity of $$P_{\mu}\left(A\right)=P_{\mu}\left(A\right) P_{\lambda}\left(\Omega\right)=P_{\mu}\otimes P_{\lambda}\left(A\times\Omega\right)=P_{\lambda}\otimes P_{\mu}\left(A\times\Omega\right)=P_{\lambda}\left(A\right)P_{\mu}\left(\Omega\right)=P_{\lambda}\left(A\right)$$ for each $A\in\mathcal{A}$.

Our final conclusion is that: $$\mu\otimes\lambda=\lambda\otimes\mu\iff\lambda\left(\Omega\right)\mu=\mu\left(\Omega\right)\lambda$$

If $\lambda(\Omega)=0$ or $\mu(\Omega)=0$ then this also appears to work, so the original conditions can be weakened to become: $0\leq\lambda\left(\Omega\right)<\infty$ and $0\leq\mu\left(\Omega\right)<\infty$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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