For $S$ and stopping time and $M_n$ a martingale on $\mathbb{N}$ with respect to the same filtration, does it always hold that $$\lim_{n \to \infty} M_{S \wedge n} = M_S,$$ or do we need some extra conditions like convergence of $M_n$ or boundedness of $S$?
I thought that if $M_n$ has an a.s. limit $M_\infty$, we can argue that $$M_{S \wedge n} = \sum_{i=1}^{n-1} \mathbf{1}_{\{S=i\}} M_i + \mathbf{1}_{\{ S \geq n \}} M_n$$ and that this converges a.s. to $$ \sum_{i=1}^{\infty} \mathbf{1}_{\{S=i\}} M_i + \mathbf{1}_{\{ S = \infty \}} M_\infty = M_S.$$ Is this correct?
I am not sure how we would even define $M_S$ when $M_n$ does not have a limit and $S$ might be infinite.
Thank you!
If $\mathbb{P}(S<\infty)=1$ or if $M_\infty$ is defined almost everywhere then obviously $M_{S\wedge n}\to M_S$ almost surely. But if at some point $M_n$ does not converge and $S=\infty$ then at this point $M_{S\wedge n}=M_n$ for all $n$ and this sequence does not converge. So at such points $M_S$ is not really defined. So the problem is when the set of such points has positive probability. Then there is no almost sure convergence of $M_{S\wedge n}$.