there must be something im missing here, can anyone figure out what it is? Let $\mu_j,\nu_j , j=1,2$ be measures on measurable spaces $(\mathcal{X}_{j},\Sigma_{j}), j=1,2$ and suppose $\nu_j << \mu_j, j=1,2$, then without $\sigma$-finiteness we shouldn't be guaranteed $\nu_1 \otimes \nu_2 << \mu_1 \otimes \mu_2$, but i seem to get the wrong answer and find that it holds without $\sigma$-finiteness as follows:
consider $\{A \in \Sigma_1 \otimes \Sigma_2 : \mu_1 \otimes \mu_2 (A) = 0 \rightarrow \nu_1 \otimes \nu_2(A) = 0 \}$, as far as I can see, by definition of $<<$, if we can show this set is $\Sigma_1 \otimes \Sigma_2$, we get the required claim
so this set contains all product sets of the form $P_1 \times P_2$, $P_i \in \Sigma_i$, because if $\mu_1(P_1)\mu_2(P_2) = 0$, then $\nu_1(P_1) = 0$ or $\nu_2(P_2) = 0$, and so $\nu_1 \otimes \nu_2 (P_1 \times P_2) = 0$
the set of such product sets generates $\Sigma_1 \otimes \Sigma_2$, so it remains to show our set for which the desired property holds is a $\sigma$-algebra, i.e. we need to show it is closed under countable unions
oh, ignore my question, i figured out my mistake
uh wait, no i didn't, maybe there isn't a mistake: Let $A_j$ be in our set. Then if $\mu_1 \otimes \mu_2( \bigcup_{j \geq 1} A_j) > 0$, $\bigcup_{j \geq 1} A_j$ is vacuously in our set. So assume $\mu_1 \otimes \mu_2(\bigcup_{j \geq 1} A_j) = 0$, hence each $\mu_1 \otimes \mu_2(A_j) = 0$, and each $\nu_1 \otimes \nu_2(A_j) = 0$, thus $\nu_1 \otimes \nu_2 ( \bigcup_{j \geq 1} A_j) = 0$, and $\bigcup_{j \geq 1} A_j$ is in our set
but then our set contains the empty set, and is closed under countable unions, and is therefore all of $\Sigma_1 \otimes \Sigma_2$, because it contains all product sets
so we have shown that if $A \in \Sigma_1 \otimes \Sigma_2$, and $\mu_1 \otimes \mu_2(A) = 0$, then $\nu_1 \otimes \nu_2(A) = 0$, hence $\nu_1 \otimes \nu_2 << \mu_1 \otimes \mu_2$, by definition of $<<$
so where is the error here?
oh great, I forgot the definition of sigma algebra
we need $\sigma$-finiteness to deal with the closure under complementation
uhh and also for intersections actually, we'd need tonelli here
Let $R$ be a relation with column domains $\{A,B,C\}$ and $S$ be a relation with column domains $\{A,B,C\}$.
Give for the following expressions of relation algebra the equivalent expression in relational calculus.
$\sigma_{B=A}(R(A,B,C))$ $ \ \ \ \ $ [Selection from $R$ subject to conditio...
