Conditional probability is not a probability measure, but it does satisfy each of the requisite axioms with probability 1. I dont quite understand what the statement in the question means, which appears in this paragraph of a textbook I am reading. How can it not be a probability measure (not even almost surely) but satisfies each of the requisite axioms with probability 1? Why are the two things not the same?

 A: The usual confusion with conditional expectations is that they are defined in a "consistent" way for a given random variable, but not for the collection of them. Namely, for any set $A\in\mathscr F$, the conditional probability $\mathsf P(A|\mathscr G)(\omega)$ is a $\mathscr G$-measurable  random variable, which is defined up to equivalence $\mathsf P$-a.s. It means, that given each $\omega$ such that $\mathsf P(\{\omega\}) =0$, the value of $\mathsf P(A|\mathscr G)(\omega)$ is not defined precisely and in fact we can put it to be anything - even a strictly negative number, for example $\mathsf P(A|\mathscr G)(\omega) =-10$. So, you shall think of $\mathsf P(A|\mathscr G)(\cdot)$ as a collection of possible random variables that are equivalent up to $\mathsf P$-a.s.
Now, if you are given the whole $\sigma$-algebra $\mathscr F$ of such sets, for each single $A\in \mathscr F$ you have to make a choice how to pick up the value of $\mathsf P(A|\mathscr G)(\omega)$ from the collection we discussed above. Hence, in general it does not mean that for any fixed $\omega$ the function
$$
  \mathsf P(\cdot|\mathscr G)(\omega):\mathscr F\to [0,1]
$$
is a probability measure: e.g. we observed that nothing prevents it even of taking strictly negative values. The question is hence, whether there does exist a simultaneous choice of $\mathsf P(A|\mathscr G)(\omega)$ for all $A$ and $\omega$ which defines probability measures. This it alternatively called regular conditional probabilities, and under certain assumptions they do exist - see e.g. here.
A: Consider for example the summability identity $S(\mathbf a)$ for a given sequence $\mathbf a=(A_n)_n$ of disjoint events, namely the fact that
$$
\sum_nP(A_n\mid\mathcal G)=P\left(\bigcup_nA_n\mid\mathcal G\right).
$$
Then each $S(\mathbf a)$ holds except on a negligible event $N(\mathbf a)$ hence all these identities hold simultaneously on the complement of
$$
\mathcal N=\bigcup_\mathbf aN(\mathbf a).
$$
If there are uncountably many different sequences $\mathbf a$, the set $\mathcal N$ may not be negligible but, for $P(\ \mid\mathcal G)(\omega)$ to be a probability measure for some $\omega$ in $\Omega$, one needs $\omega$ to be in $\Omega\setminus N(\mathbf a)$ for every $\mathbf a$, that is, to be in $\Omega\setminus\mathcal N$.
