Can we infer a set has positive measure from the conditional probability formula? First question: Suppose that I know $P(x\in A|\;y)>0$ where $y$ is a realization of some random variable. Then, in general can I infer using the conditional probability formula $P(x\in A|\; y)=\frac{P(x\in A,y)}{P(y)}$ that $P(y)>0$? This seems obvious but I am still note sure.
Second question: If I know $P(x\in A,y\in B)>0$ is it true that there exists a measureable set $B_{1}\subset B$ such that $y\in B_{1}$ implies that $P(x\in A|y)>0$? My proof follows as:
Let $B_{0}=\{y\in B:P(x\in A|\;y)=0\}$ and $B_{1}=\{y\in B:P(x\in A|\; y)>0\}$. Then $B=B_{0}\sqcup B_{1}$ and therefore, 
$$
P(x\in A, y\in B)=\int_{y\in B} P(x\in A|y)dP=\int_{y\in B_{0}} 0\cdot dP+\int_{y\in B_{1}} P(x\in A |y)\cdot dP=\int_{y\in B_{1}} P(x\in A| y)dP>0
$$ 
and the last inequality implies that $B_{1}$ is measurable.
I am dealing with an uncountable probability space, say $\Omega=[0,1)$, with the Borel $\sigma-$algebra, and the Lebesgue measure. If you don't know the answer but some references it would be much appreciated.
 A: First question
No. If $y\in A$ then clearly $P(X\in A|X=y)=1$, also if $P(X=y)=0$. Here $y$ is a realization of r.v $X$ and in that sense you are dealing with a special case, however it is not hard to find other counterexamples. You always have $P(X\in \mathbb R|Y=y)=1$ but not necessarily $P(Y=y)>0$. So the answer on your edit is again: no.
Second question
Yes. In general if $f$ is nonnegative then $\int fd\lambda>0$ can only be true if $f>0$ on some measurable set $C$ that satisfies $\lambda(C)>0$. 
So $0>P\left\{ X\in A\wedge Y\in B\right\} =\int_{B}P\left\{ X\in A|Y=y\right\} P\left(dy\right)$
implies the existence of a measurable set $B_{1}\subset B$
with $P\left(B_{1}\right)>0$ and $P\left\{ X\in A|Y=y\right\} >0$
for $y\in B_{1}$. Note here that $P\left(B_{1}\right)>0$ implies
$B_{1}\neq\emptyset$. The empty set satisfies your condition vacuously,
but is not interesting.

Further explanation:
$\int fd\lambda=\sup\left\{ \int td\lambda\mid0\leq t\leq f\wedge t\text{ has a finite image and is measurable}\right\} $.
Then $\int fd\lambda>0$ implies $\int td\lambda>0$ for some $t$
that has finite image and is measurable. Here $t$ can be written as $t=\sum_{i\in I}r_{i}\chi_{A_{i}}$
where $I$ is finite and $r_{i}>0$ and the $\chi_{A_{i}}$ are characteristic
functions of measurable sets $A_i$. 
So $0>\int td\lambda=\sum_{i\in I}r_{i}\lambda\left(A_{i}\right)$
leading to $\lambda\left(A_{i_{0}}\right)>0$ for some index $i_{0}$.
