I am studying a proof of Doob-Meyer decomposition in Kallenberg's "Foundations of Modern Probability", which goes by a discrete approximation based on Dunford's characterisation of weak compactness, combined with Doob's approximation of inaccessible times. The setting is that of a right-continuous and complete filtration on $R_+$.
I have a problem with a gap in the proof of the latter, which is (in 1997 ed.; 10.8 in 2021 ed.):
Lemma 22.8 (approximation of inaccessible times, Doob) $\qquad$For a totally inaccessible time $\tau$, put $\tau_n=2^{-n}[2^n\tau]$, and let $X^n$ be a right-continuous version of the process $P\{\tau_n\le t|\mathcal{F}_t\}$. Then $$ \lim_{n\to\infty} \sup_{t\ge 0} \left| X^n_t - 1\{\tau\le t\} \right| = 0 \;\; a.s. $$
Now, as the Author's hint is to use optional sampling to fill a gap in the proof, I wonder if one can show that $X^n$ is a martingale; also the proof of the next lemma (on accessibility of jumps) suggests that this is the case.
Here is my line of thought. Take $n=0$, so that $\tau_0=[\tau]$ (for positive $n$ we scale by $2^n$). Note that $X^0_t$ is a martingale iff so is $1-X^0_t=Y_t=P\{ \tau_0 > t | \mathcal{F}_t \}$. Now, $\tau_0$ takes only integer values. If we take $t\in[0,1)$, we have $$ \{\tau_0 > t\} = \{ \tau_0 \ge 1 \} = \{ \tau \ge 1 \}, \qquad\text{so}\quad Y_t = P\{ \tau \ge 1 | \mathcal{F}_t \}. $$ So far, so good: on $[0,1)$, $Y_t$ is a martingale. What about $t$ crossing $1$? Is $$ Y_t = E\,[Y_1 |\mathcal{F}_t] \; a.\!s. \quad \text{for}\; t\in[0,1)? $$ By definition, $Y_1=P\{ \tau_0 > 1 | \mathcal{F}_1 \}=P\{ \tau\ge2 | \mathcal{F}_1 \}$, so $E\,[Y_1 |\mathcal{F}_t] = P\{ \tau\ge2 | \mathcal{F}_t \}$.
So, the question is: If $\tau$ is totally inaccessible, do we have $$ P\{\tau\ge 1 | \mathcal{F}_t\} = P\{\tau\ge 2 | \mathcal{F}_t\} \; a.\!s. \quad\text{for}\;\; t\in[0,1)? $$