I don't understand the accepted answer to the following question:
Let $M$ be a positive, continuous martingale that converges a.s. to zero as $t$ tends to infinity. I now want to prove that for every $x>0$ $$ P\left( \sup_{t \geq 0 } M_t > x \mid \mathcal{F}_0 \right) = 1 \wedge \frac{M_0}{x}. $$
The accepted answer shows that $$\forall k:M_0 1_{\{M_0>x\}} + 1_{\{M_0 \leq x\}} x P(\tau < k \mid F_0) + E(M_k 1_{\{\tau \geq k\}} \mid F_0) = M_0$$ with $\tau := \inf\{t\geq 0 : M_t>x\}$. It then says that we obtain the desired equation by considering the limit $k\to\infty$ and this is the part I don't understand.
Edit: I think that I solved the problem thanks to a hint in the comments. It looks like the problem is much less trivial than I expected, maybe someone can confirm this:
On the one hand, we have $$\lim_{k\to\infty}E(M_k 1_{\{\tau \geq k\}} \mid F_0)=0$$ since "M converges to zero" (DCT). Thus we obtain: $$M_0 1_{\{x<M_0\}} + 1_{\{M_0 \leq x\}} x P(\tau < \infty \mid F_0)= M_0$$ Since $0<x$, this is equivalent to: $$ 1_{\{x<M_0\}}\frac{M_0}{x} + 1_{\{M_0 \leq x\}} P(\tau < \infty \mid F_0)=\frac{ M_0}{x}$$ And this is equivalent to: \begin{equation}\tag{1} 1_{\{M_0 \leq x\}} P(\tau < \infty \mid F_0)=\frac{ M_0}{x}-1_{\{M_0>x\}}\frac{M_0}{x} =1_{\{M_0\leq x\}}\frac{M_0}{x} \end{equation} On the other hand, we can use $$1\wedge\frac{M_0}{x}=1_{\{x<M_0\}}+1_{\{M_0\leq x\}}\frac{M_0}{x}$$ and $$P(x<\sup_{t}M_t\mid F_0)=P(\tau<\infty\mid F_0)$$ (see the comments) to rewrite the desired result: $$P(\tau<\infty\mid F_0)=1_{\{x<M_0\}}+1_{\{M_0\leq x\}}\frac{M_0}{x}$$ But this equation follows from $(1)$ and the fact that $1_{\{x<M_0 \}} P(\tau < \infty \mid F_0)=1_{\{x<M_0 \}}$: \begin{equation} P(\tau < \infty \mid F_0)=1_{\{M_0 \leq x\}} P(\tau < \infty \mid F_0)+1_{\{x<M_0 \}} P(\tau < \infty \mid F_0)=1_{\{M_0 \leq x\}} \frac{M_0}{x}+1_{\{x<M_0 \}} \end{equation}
