Probability of picking a number from a set with a uncountably infinite number of positive numbers and a countably infinite number of negative numbers? Sorry if this is a stupid question. I couldn't see anything else quite the same as this, although I found some similar questions. Please remove this if it is a duplicate. (I hardly understand sets so please take the spirit of the question into account and give me ways to phrase it better. Thank you!)
Note: I am assuming a uniform distribution. (If you can even do that for an uncountably infinite set.)
Note 2: I have realized that this problem doesn't work, as I cannot assign a  uniform distribution to an uncountable infinite set. If anyone knows a way to somehow bypass this, please comment. Thank you!
Note 3: Someone told me the exact opposite thing that the earlier person said, so I now have no idea what is true or not.
Note 4: I have come to the conclusion that you $can$, in fact, assign a uniform distribution to an uncountably infinite set, and the method of picking the random number doesn't matter since this is a theoretical exercise. Thank you.
My question is essentially this: If you had the set $S$, $[0,1] \cup \Bbb Z$, and you randomly picked a number from the set, is it more likely for the selected number to be in between $0$ and $1$ or an integer $\Bbb Z$? I feel like it's logical that the number should be more likely to be in between $1$ and $0$, as there are infinitely many more numbers in between $1$ and $0$ than all integers $\Bbb Z$, but I also know that technically the probability of selecting any number at all is $0$. Would it just be an equal probability of $0$, or would it be a $100%$ chance of something in $[0,1]$ being selected? Or would it be something else? Thank you!
Thank you!
Edit 1: Clarified the set more, and changed it from strictly negative $\Bbb I$ into a general $\Bbb I$.
Edit 2: Added Note 2.
Edit 3: Added Note 3.
Edit 4: Question has been properly cleaned up, Note 4 has been added.
 A: If your space of probability is $\bar{\mathbb{R}}$ with Borel's $\sigma$-algebra, I'd say that you do have a grater chance of picking a positive number: $-\mathbb{Z}$ is a nullset while $[0,1]$ is not. How ever, I am no Probability Theory expert and you should probably listen to someone who knows better than I do, but I reckon that having a nullset and a set of positive measure is key to this.
A: Not exactly your problem, but it is well known and understood that one can define things such that we have an uncountably infinite probability space equipped with a uniform probability measure.  More specifically, an example of this is the real interval $[0,1]$ equipped with the Lebesgue measure.
The Lebesgue measure here is the intuitive "how long is the interval" measure you are used to but formalized in such a way as to cover all of the frustrating cases that might arise.

(Do note however that in attempting to formalize things, one does open the door for some rather bizarre outcomes such as non-measurable sets so care must be taken)

Let $A = \{\frac{1}{2^n}~:~n\in\Bbb N\}$.  We do indeed in this situation have $\Pr([0,1]\setminus A) = 1$ and $\Pr(A)=0$ as any countable set is a nullset according to the Lebesgue measure.
Indeed, in any scenario in which you attempt to define a "uniform" distribution (whatever that means in your context) using whatever specific probability measure it is over an uncountable sample space, the probability of landing in a countable subset of that will be zero.  This follows from the probability space axioms.
Specifically, the third axiom states that the probability measure must satisfy $\sum\limits_{n=1}^\infty \Pr(A_n) = \Pr(\bigcup\limits_{n=1}^\infty A_n)$ for any countable sequence of disjoint events $\{A_n\}_{n=1}^\infty$.
In particular with a "uniform" distribution you would surely be insisting that $\Pr(\{x\})=\Pr(\{y\})$ for any $x,y$ which along with the third axiom of sigma-additivity imply that probability of any individual element must be zero and further that probability of any countable subset of elements must be zero.
