Stopping times and the left limit So assuming $X$ is càdlàg martingale and $\tau\leq T$ is a stopping time. The stopping theorem gives us something like 
$$E[X_T| \mathcal{F}_\tau]= X_\tau.$$
But if the usual conditions hold, shouldn't there be something like 
$$E[X_T| \mathcal{F}_{\tau-} ]= X_{\tau-}$$
for $X_{t-}:= \lim\limits_{ t_n\uparrow \tau, t_n\not=t} X_{t_n}$? 
I think this is correct:
$$\mathcal{F}_{\tau-}= \bigcup_{ n\in \mathbb{N}} \left\{ A\in \mathcal{F}: A \cap\{ \tau\leq t-1/n\} \right\} $$
And because the filtrations is right-continious and must not be left-continious that could differ from $\mathcal{F}_\tau$. 
So is there something like an extended stopping theorem? 
 A: I am aware of the following result, for a slightly different definition of $\mathcal F_{\tau-}$. I'll start with a few definitions. For two $\sigma$-algebras $\mathcal F$ and $\mathcal G$, we denote the smallest $\sigma$-algebra containing both $\mathcal F$ and $\mathcal G$ by $\mathcal F \vee \mathcal G$.

Definition. The $\sigma$-algebra $\mathcal F_{\tau-}$ of events strictly prior to a stopping time $\tau$ is given by $$ \mathcal F_{\tau-} = \mathcal F_0 \vee \sigma\left(\left\{A \cap \{ t< \tau\} : A\in\mathcal F_t , t\in [0,\infty)\right\}\right). $$

In other words, $\mathcal F_{\tau-}$ is the $\sigma$-algebra generated by $\mathcal F_0$ and all the sets of the form $A \cap \{ t <\tau \}$, where $t \in [0,\infty)$ and $A \in \mathcal F_t$. I believe the spirit of this definition should be the same as the one you state. However, I think your definition of $\mathcal F_{\tau-}$ is, in general, not a $\sigma$-algebra.
I will need one more definition before I can state the result.

Definition. A stopping time $\tau$ is said to be predictable if there is a sequence $\{ \tau_n \}_{n\in\mathbb N}$ of stopping times such that:
  
  
*
  
*$\tau_n \uparrow \tau$ almost surely as $n \to \infty$; and,
  
*on the set $\{ \tau >0 \}$, $\tau_n < \tau$ a.s. for all $n$.
  

The first point above implicitly requires that the sequence $\{ \tau_n \}_{n\in\mathbb N}$ be almost-surely non-decreasing. When $\tau$ is a predictable stopping time, we call any sequence satisfying the requirements in the above definition an announcing sequence for $\tau$.
I can now state the result you allude to.

Theorem. If $\tau$ is a predictable stopping time with announcing sequence $\{ \tau_n \}_{n\in\mathbb N}$ and $M$ is a càdlàg uniformly integrable martingale, then $$ M_{\tau-} = \lim_n M_{\tau_n} = E[M_{\tau} \mid \mathcal F_{\tau-}]. $$

To obtain the equality you state, we can use the optional stopping theorem to conclude that $M_{\tau} = E[M_T \mid \mathcal F_{\tau}]$. Using the tower property of conditional expectations, we have, under the conditions of the stated theorem, that $$ M_{\tau-} = E[M_T \mid \mathcal F_{\tau-}]. $$
As with the optional stopping theorem, I believe that the requirement of uniform integrability can be relaxed when we assume that our stopping time is bounded, but I'll need to have a careful look at the proof of this result before I can say that we can do the same here.

Why do we need predictability of the stopping time?
In general, if $X$ is a random variable, and we have that $M_t = E[X \mid \mathcal F_t]$ up to indistinguishability, it does not follow that $M_{\tau-} = E[X \mid \mathcal F_{\tau-}]$. To see this, suppose $X$ is a Poisson process with parameter $\lambda$, and $M_t = E[X_1 \mid \mathcal F_t] = X_{t \wedge 1} + \lambda (1-t)^+$. If $\tau$ is the first jump time of $X$, then we have that $M_{\tau-} \neq E[X_1 \mid \mathcal F_{\tau -}]$.
Note that the jump times of a Poisson process are not predictable.

Being able to justify all this is going to make this post quite lengthy, so I simply refer you to Chapter 6 of the reference given below, which contains everything I've written here.
Reference: Cohen, S. N., & Elliott, R. J. (2015). Stochastic calculus and applications (Vol. 2). New York: Birkhäuser.
