How to compute the probability of the union of an infinite number of events? Consider the event "I am currently being seen by anyone" to be $s$. I am interested in estimating the probability $P(s|\vec{x})$ (where $\vec{x}$ is where I am at the moment), from the conditional probabilities of someone to be seeing me from the position $\vec{y}$. (Imagine that I am playing the hide and seek game, that I am optimizing my location for winning at hiding, and I have some clues of where the seekers might try to go).
My current approach is this: 


*

*define the event "I am currently being seen by someone at $\vec{y}$" as $s(\vec{y})$.

*Assume that events are independent, i.e. $P(s(\vec{y}_i),s(\vec{y}_j)|\vec{x}) = P(s(\vec{y}_i)|\vec{x})P(s(\vec{y}_j)|\vec{x})$

*conclude that the event $s$ is the union of the all events $s(\vec{y}_i)$.

*Thus, $$P(s(\vec{y})|\vec{x}) = \int P(s(\vec{y})|\vec{x})d\vec{y}$$


My issue is that this equation does not look right to me. I mean, for a numerable set of events, P(union) = sum of P(individuals), but I am not sure how this works for unions of uncountable events.
Is the formalization above correct? If not, what is the way to formalize this problem in continuous space?
 A: If an uncountable family of events are mutually independent (any finite subfamily of them is mutually independent), and each has a non-zero probability of occurring, then the probability of at least one of them occurring is $1$: flip an uncountable number of coins and you'll always get heads eventually.
Let $(A_i)$ be your family of events, indexed over some uncountable set $I$. Then for some $\epsilon>0$, there are infinitely many $A_i$ with $P(A_i)\geq\epsilon$. Now apply Borel-Cantelli or some other such argument to show that the probability of one of those events occuring is $1$. To see that my claim is true for some $\epsilon$, let $\mathcal A_n=\{A_i|P(A_i)\geq\frac1n\}$. The union of the $\mathcal A_n$ is the family of all $A_i$, thus one of the $\mathcal A_n$ must be infinite, otherwise we would have expressed an uncountable family as the union of countably many finite families.
You may also find this question and my answer to it interesting.
For a reasonable model for this situation, maybe this will give you some ideas. Assume the population is very densely packed, with a population density function $\rho(x)$. Suppose that the probability a person does not see you depends only on their position, according to the function $f(x)$. What is the probability that nobody in a given small area around a point $x$ sees you? If $f$ is continuous, it's basically $f(x)^n$, where $n$ is the number of people in that area. Taking a logarithm, we get
$$\log P(\text{nobody sees you}) = n\log f(x) = a\rho(x)\log f(x)$$
where $a$ is the area of the small region we're interested in. Taking the logarithm was important so that we could divide by $a$ and obtain a density:
$$\frac {\log P(\text{nobody sees you})} a = \rho(x) \log f(x)$$
thus, for any region $A$
$$\log P(\text{nobody in $A$ sees you})=\int_A \rho(x) \log f(x) dx$$
and in particular
$$P(\text{somebody sees you})=1-\exp\left(\int_{\mathbb R^2} \rho(x)\log f(x)dx\right)$$
