# Does the LHS of $E[X_n | \mathscr F_{n-1}]$ make sense even if $X_n$ is not integrable or adapted?

Let $(\Omega, \mathscr F, \{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$ be a filtered probability space. Then $X_n$ is a $(\{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)-$martingale if:

1. $X_n$'s are integrable.

2. $X_n$'s are $\mathscr F_n-$measurable.

3. $E[X_n | \mathscr F_{n-1}] = X_{n-1}$

If we are able to show (3), do we still have to show (1) and (2)? Well maybe (2), but I don't think the LHS of (3) is defined if we don't have (1) so proving (1) seems like a formality.

If we assume $X_n$'s are integrable and then evaluate $E[X_n | \mathscr F_{n-1}]$ and then get $X_{n-1}$, then what's the problem? It seems to me that if not all $X_n$'s are integrable then we will not be able to evaluate $E[X_n | \mathscr F_{n-1}]$ in the first place.

This seems like that formality in basic calculus where we had to prove 3 conditions for continuity:

Let $I \subseteq \mathbb R$. A function $f: I \to \mathbb R$ is continuous at $c \in \mathbb R$ if

1. $f(c)$ is defined.

2. $\lim_{x \to c} f(x)$ exists.

3. $\lim_{x \to c} f(x) = f(c)$

I might be remembering incorrectly, but (3) will not make sense if (1) or (2) does not hold. So if we assume (1) and (2) are true and then show (3), we can already conclude $f$ is continuous right?