Dose "optional stopping theorem" imply "optional sampling theorem"?

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Suppose $X$ is a martingale,$\tau$ and $\sigma$ are two stopping times which satisfy (a)$\sigma\le\tau$ and (b)the "optional stopping theorem" holds,that is to say:

$$\mathbb E[X_\sigma]=\mathbb E[X_0],\mathbb E[X_\tau]=\mathbb E[X_0] ---(1)$$

Could we get the "optional sampling theorem":

$$\mathbb E[X_\tau|\mathcal F_\sigma]=X_\sigma ---(2)$$ by(1)?


What I really want to say is the following :

There are many conditions leads to (1) such as :

$\tau$ is bounded

or $X$ is UI

or $\mathbb E[\tau]<\infty $ and bounded increments

or $|X_{\tau\wedge n} | $ is bounded

or $|X|$ is bounded.

etc.

(apply bounded convergence theorem /dominate convergence theorem to interchange the limit and expectation in $ \lim\mathbb E[X_{\tau\wedge n}]=\mathbb E[X_0]$ )

But I have only seen two conditions leads to (2):

$\tau$ is bounded or$X$ is UI

I wonder the other conditions can leads to (2),moreover whether any conditions lead to (1) can lead to (2) and whether (1) itself can lead to (2).