I have a probability space $\mathcal{X} \times \mathcal{Y}$. $\mathcal{X}$ and $\mathcal{Y}$ could be infinite, but they are at worst case countable.
There is a matrix $A_{xy}$ ranging over $\mathcal{X}$ for rows $\mathcal{Y}$ for columns, which is either 0 or 1 for each $x,y$.
We have the information that
$P(${$ (x,y) \mid A_{xy} = 1$}$) \le \epsilon$.
Is it possible to upper bound the probability $P(${$x \mid \exists y A_{xy} = 1$}$)$, where the bound depends on $\epsilon$ and other parameters given in the question? (say, structure of $A$ perhaps... but I'd rather it be $A$-independent.)
The problem is that we can always create an $A$ such that it has $1$ in row (for each $x$ somewhere, therefore the probability that I want to bound would be 1), but still the probability of total 1s would be $\epsilon$. So I am pretty sure I am not approaching the whole problem formulation the right way.
So maybe you have an idea what assumptions I can make to get a better bound.
One idea that I had is doing the following: assume that we have $\alpha = \sup_{x,y} \frac{P(x)}{P(x,y)}$. Then, we know that:
$P(${$x \mid \exists y A_{xy} = 1$}$) = \sum_x P(x) I(\exists y A_{xy} = 1)\le \sum_{x,y \mathrm{s.t.}\, A_{xy}=1} \alpha P(x,y) \le \alpha \epsilon $ (where $I$ is a 0 1 indicator.)
The thing is that in my case, $\alpha$ could get pretty big if I really define it that way.