In my lecture notes, there is an interesting example for a strict local martingale: For its construction we consider Brownian motion $(W_t)_{t \geq 0}$ (w.r.t. the fitration $(\mathcal{F}_t)_{t \geq 0}$), a number $a < 0$ and the stopping time $\tau_a := \{t \geq 0 | W_t = a\}$. Then the stopped process $W^{\tau_a}$ is also a martingale. Now consider $$ M_t := \begin{cases} W^{\tau_a}_{t/(1-t)} - a,& \quad t \in [0,1) \\ 0,& \quad t \geq 1 \end{cases} $$ with the filtration $$ \mathcal{F}_t := \begin{cases} \mathcal{F}_{t/(1-t)},& \quad t \in [0,1) \\ \mathcal{F}_\infty,& \quad t \geq . \end{cases} $$ Note that $M$ is not a martingale since $\mathbb{E}[M_t] = 0$ for $t \geq 1$ and $\mathbb{E}[M_t] = -a$ for $t \in [0,1)$. It is stated that one can construct a localizing sequence in the following manner. Consider $\tau_n = \{t \geq 0|W_t = n\}$. If we remove the null set $\{\tau_a = \infty\}$ from our probability space, we can assume in the follwoing that $\tau_a < \infty$. Next, define the stopping time $$ \tau_n' := 1_{\{\tau_n < \tau_a\}}\left(\frac{\tau_n}{1+\tau_n}\right) + 1_{\{\tau_n \geq \tau_a\} }\cdot \infty $$ Now, I am trying to verify that $M^{\tau_n'} - M_0$ is a martingale (we define a local martingale by requiring that $M^{\tau_n'} - M_0$ is a martingale and not that $M^{\tau_n'}1_{\{\tau_n' > 0\}}$ is a martingale). In particular I want to verify the martinagle property. For $s < 1 \leq t$ we have (since $\left(\frac{\tau_n}{1+\tau_n}\right) < 1$ and $M_t = 0$) $$ \mathbb{E}[M_t^{\tau_n'}|\mathcal{F}_s'] - M_0 = \mathbb{E}[M_{\frac{\tau_n}{1+\tau_n}} 1_{\{\tau_n < \tau_a\}} + 1_{\{\tau_n \geq \tau_a\} }\cdot M_t|\mathcal{F}_s'] + a = \mathbb{E}[(W_{\tau_n}^{\tau_a}-a) 1_{\{\tau_n < \tau_a\}}|\mathcal{F}_s'] + a $$ Adding $(W_{\tau_n}^{\tau_a} - a) 1_{\{\tau_n \geq \tau_a\}} = 0$ in the expectation, we have $$ \mathbb{E}[M_t^{\tau_n'}|\mathcal{F}_s'] - M_0 = \mathbb{E}[(W_{\tau_n}^{\tau_a} - a) 1_{\{\tau_n < \tau_a\}} + (W_{\tau_n}^{\tau_a} - a) 1_{\{\tau_n \geq \tau_a\}}|\mathcal{F}_s'] + a = \mathbb{E}[W_{\tau_n}^{\tau_a}|\mathcal{F}_s'] $$ Using optional stopping for $W_{\tau_n}^{\tau_a}$ we have $$ \mathbb{E}[M_t^{\tau_n'}|\mathcal{F}_s'] - M_0 = W^{\tau_n \land \tau_a}_{s/(1-s)}. $$ However, if I am not mistaken, this should be $W^{\tau_n' \land \tau_a}_{s/(1-s)}$ and I am not seeing how one can get $\tau_n'$ instead of $\tau_n$ in the above calculation? Where is my mistake?
Thanks in advance!