Product measure problem 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!
 A: 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 $p=q$.
A: 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 $0\leq\mu\left(\Omega\right)<\infty$
