Intuitions for probability on countably infinite sample spaces I've learnt over the years that measure theory is subtle and counterintuitive, and many things that seem obvious actually have counterexamples. My intuition serves me well in the world of finite sample spaces, but if I step outside that I can't trust it.
However, I'm wondering if I can safely step into the world of infinite but countable sample spaces, such as the natural numbers.
I know that in this case there is no uniform distribution, which is a significant difference. Other than that, does anything substantial change when passing from probability distributions on finite sets to probability measures on countable sample spaces? I'm wondering in particular about basic intuitions like the following, which of course are wrong for general sample spaces but true in the finite case.

*

*Unless you're doing something very weird the appropriate $\sigma$-algebra is always the power set.


*Defining a probability measure is equivalent to assigning a probability to each outcome, such that they sum to 1.


*If an outcome has probability 0, the appropriate intuitive interpretation is that it can't happen.
I ask partly because I know that the following definitely is wrong:

*

*The expectation of a function $f\colon \mathbb{N}\to\mathbb{R}$ is always given by $\sum_{k\in\mathbb{N}}p(k)f(k)$.

In fact this is only true if the series is absolutely convergent, and the expectation is undefined otherwise. This makes me suspect that there might be important subtleties that I'm missing in my other bullet points as well, but it's surprisingly hard to find anything concise on this topic that doesn't gloss over this kind of point.
(Note: there are several related questions that are about finitely additive measures on countable sets. I'm not asking about that - I'm interested in the usual countably additive kind of measure here.)
 A: Yes, if $\Omega$ is a finite or countably infinite sample space, we might as well use the sigma algebra of all subsets of $\Omega$. Then, specfiying a probability measure is equivalent to specifying probability masses for each outcome.

Probability masses:
Formally, suppose $\Omega$ is a nonempty set that is finite or countably infinite.  Suppose we have probability masses $p(\omega)$ for all $\omega \in \Omega$ such that
\begin{align}
&p(\omega) \geq 0  \quad \mbox{for all $\omega \in \Omega$} \\
&\sum_{\omega \in \Omega} p(\omega) = 1
\end{align}
Define $\mathcal{F}$ as the set of all subsets of $\Omega$. Define $P:\mathcal{F} \rightarrow\mathbb{R}$ by
$$ P[A] = \sum_{\omega \in A} p(\omega) \quad \forall A \subseteq \Omega$$
Then $(\Omega, \mathcal{F}, P)$ is a valid probability measure (it can be shown to satisfy the three axioms of probability).

Throwing away outcomes:
If $(\Omega, \mathcal{F}, P)$ is a triplet for a probability space (with $\Omega$ being either finite, countably infinite, or uncountably infinite) and if $A \in \mathcal{F}$ is such that $P[A]=0$, then we can throw away all outcomes in $A$ to produce a new probability space $(\tilde{\Omega}, \mathcal{\tilde{F}}, \tilde{P})$ where
\begin{align}
\tilde{\Omega} &= \{\omega \in \Omega : \omega \notin A\}\\
\tilde{\mathcal{F}} &= \{B \in \mathcal{F} : B\subseteq \tilde{\Omega}\} \\
\tilde{P}(B) &= P(B) \quad \forall B \in \tilde{\mathcal{F}}
\end{align}
The resulting space $(\tilde{\Omega}, \mathcal{\tilde{F}}, \tilde{P})$ is a valid probability space (it can be shown to satisfy the axioms of probability).  In particular, $\tilde{P}[\tilde{\Omega}]=1$ and none of the old probabilities change.
For example suppose $\Omega = \{blue, red, green\}$, $\mathcal{F}$ is the set of all subsets of $\Omega$,  and $P[\{blue\}] = 1/2, P[\{red\}] =1/2, P[\{green\}]=0$. We say that the outcome is surely in the set $\{blue, red, green\}$.  We also say that the outcome is
almost surely in the set $\{blue, red\}$.
We could throw away the green outcome since it has probability zero, without changing the other probability masses, to create a new probability space:
$\tilde{\Omega} = \{blue, red\}$, $\tilde{\mathcal{F}}$ is the set of all subsets of $\tilde{\Omega}$, $\tilde{P}[\{blue\}]=1/2$, $\tilde{P}[\{red\}]=1/2$. Here we can say the outcome is surely in the set $\{blue, red\}$ (so, unlike the previous case, green is not an outcome).
If the sample space is finite or countably infinite, you can always throw away all outcomes $\omega$ such that $P[\{\omega\}]=0$, without changing the probabilities of any of the remaining events.  So by reducing the sample space, you can assume all probability masses are positive.

On intuitive definitions of "possible":
Consider a probabilty space $\Omega = [0,1]$ with the standard Borel measure.  Define $X(\omega) = \omega$ so that
$$X \sim Uniform([0,1])$$
Define
\begin{align}
Y &= \left\{\begin{array}{cc}
0 & \mbox{ if $X \neq 0.2$} \\
8 & \mbox{ if $X = 0.2$} 
\end{array}\right.\\
Z &= \left\{\begin{array}{cc}
X/2 & \mbox{ if $X \neq 0.2$} \\
8 & \mbox{ if $X = 0.2$} 
\end{array}\right.
\end{align}
Then $Z \sim Uniform([0, 1/2])$, where $[0,1/2]$ is the support of the PDF of $Z$.

*

*Is $X=0.2$ possible?

*Is $Y=8$ possible?

*Is $Z=8$ possible?

These three events are literally the same:
$$ \{X=0.2\} = \{Y=8\} = \{Z=8\} $$
So any intuitive definition of "possible" or "impossible" should be consistent for all three of these.
Note that $X(0.2) = 0.2$ and so there is an outcome in the sample space that leads to $X=0.2$. Of course, $P[X=0.2]=0$.
If you want, you can throw the outcome $\omega = 0.2$ away from the sample space to define a new sample space $\tilde{\Omega} = [0, 0.2) \cup (0.2, 1]$, without affecting the probabilities for the remaining events.  In that new space $(\tilde{\Omega}, \tilde{F}, \tilde{P})$ we can say that $X$ is surely not equal to $0.2$.
A: To address your edit about expected values, I would say that is probably the only major difference between a finite sample space and a countable one: A finite random variable on a finite sample space always have a well-defined and finite expected value, while a finite random variable  on a countably infinite sample space may not have a well-defined expected value.
If we had $\Omega = \mathbb{N}$ for our sample space with probability measure $p$, the formula $\mathbb{E}[X] = \sum_{\omega \in \Omega} p(\omega)X(\omega)$ might still make intuitive sense for the expected value since there is a natural ordering for $\mathbb{N}$ that we could use to interpret the sum if it is not absolutely convergent, i.e. $$\sum_{\omega \in \Omega} p(\omega)X(\omega) = \sum_{\omega=1}^\infty p(\omega) X(\omega) = \lim_{N \rightarrow \infty} \sum_{\omega=1}^N p(\omega) X(\omega)$$ if that limit exists.  The problem is that, if the series is convergent but not absolutely convergent, then this definition is not stable under permutations: if we add the terms in a different order, we can get a different result.
This is more clearly a problem if the sample space is something a little more complex, like $\Omega = \mathbb{Q}$.  Let $p$ still be a probability measure on $\Omega$.  We define a random variable $X$ on $\Omega$ and want to compute $\mathbb{E}[X]$.  We still want $\mathbb{E}[X] = \sum_{\omega \in \Omega} p(\omega)X(\omega) = \sum_{\omega \in \mathbb{Q}} p(\omega)X(\omega),$ but now that sum is more difficult to interpret.  We can let $(q_n)_{n \in \mathbb{N}}$ be an enumeration of the rationals, so $$\mathbb{E}[X] = \sum_{\omega \in \mathbb{Q}} p(\omega)X(\omega) = \sum_{n=1}^\infty p(q_n)X(q_n) = \lim_{N \rightarrow \infty} \sum_{n=1}^N p(q_n)X(q_n)$$ if that limit exists.  The key thing is that we don't want this to depend on the enumeration of the rationals we chose.  This is why we require $\mathbb{E}[|X|] = \sum_{\omega \in \mathbb{Q}} p(\omega)|X(\omega)| < \infty$ in order to say $\mathbb{E}[X]$ is well-defined: absolute integrability implies that the expected value won't depend on our choice of enumeration $(q_n)$.  Note that if we know $X$ is non-negative (i.e. $X(\omega) \ge 0$ for all $\omega \in \Omega$) we can still define $\mathbb{E}[X]$; it just might be infinite.
