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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!

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  • $\begingroup$ $$p=q{}{}{}{}$$ $\endgroup$ – 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
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    $\begingroup$ 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
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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

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

and

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

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 $p=q$.

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  • $\begingroup$ 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
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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 $0\leq\mu\left(\Omega\right)<\infty$

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