What is the difference between $\mathbb E[Z|\mathcal G]=Y$ and $\mathbb E[Z|\mathcal G]\stackrel{\text{a.s.}}{=}Y$? I'm somewhat confused by the definition of martingale:

Let $(\Omega, \mathcal F, \mathcal F_n, \mathbb P)$ be a filtered probability space. We call $(X_n)_{n\in\mathbb N}$ martingale if for all $n\in\mathbb N$ holds:

$X_n \in \mathcal L^1(\mathbb P)$

$X_n$ is $\mathcal F_n$-measurable

$\mathbb E[X_{n+1}|\mathcal F_n]\stackrel{\text{a.s.}}{=}X_n$



Why do we need &quot;a.s.&quot; in the 3rd part? Isn't $\mathbb E[Z|\mathcal G]$ always unique only up to &quot;a.s.&quot; anyway? How would we check the 3rd part without knowing that the 2nd part holds? (I.e. doesn't the 3rd imply the 2nd?) Why do we separate the points 2 and 3? Wouldn't it be enough to just ask for $\mathbb E[X_{n+1}|\mathcal F_n]=X_n$ instead?
In general, 3 does not imply 2

Suppose $g$ is $\mathcal{F}$-measruable and $h=g$ almost surely, it does not imply $h$ is also  $\mathcal{F}$-measurable.

Take a set $A$ which is contained in $B \in \mathcal{F}$ with $P(B) = 1$, but $A$ is not itself in $\mathcal{F}$. By definition, $h = 1_{A}$ is alomst surely equal to the constant variable $g \equiv 0$. But $h$ is not $\mathcal{F}$-measurable, since $h^{-1}(\{1\}) = A \not\in \mathcal{F}$.

If $\mathcal{F}$ has been completed and the filtration is augmented by $\mathbb{P}$, then (2) is indeed unnecessary. That's one of the reason of completion and augmentation: to make sure modification on a null set doesn't cause measurability problem.
Let's generalise how 3 doesn't imply 2:


  All versions of a conditional expectation are almost surely equal to each other, but any version is free to be equal to another random variable that is not a version of said conditional expectation.


Precisely:


  Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space with random variable $X \in \mathscr L^1$ and a sub-$\sigma$-algebra $\mathcal G$. Versions of $\mathbb E[X|\mathcal G]$ are almost surely equal to each other, but any version is free to be equal to another random variable, $Y$, that is not $\mathcal G$-measurable.


Consider $X=c$, $Y=c1_A$, $P(A)=1$ and $A \notin \mathcal G$. Then

$$\mathbb E[X|\mathcal G] := \mathbb E[c|\mathcal G] = c \stackrel{\text{a.s.}}{=}c1_A=:Y,$$ but $Y$ is not $\mathcal G$-measurable. You could say that $Y$ is a non-$\mathcal G$-measurable version of $\mathbb E[X|\mathcal G]$, i.e. it satisfies $E[Y1_G] = E[X1_G] \ \forall G \ \in \ \mathcal G$ but not $\sigma(Y) \subseteq \mathcal G$, if that's even a thing in probability.