Let $X_{n}^{1}$ and $X_{n}^{2}$ be supermartingales w.r.t $\mathbb F$ and $\tau$ a stopping time so that $X_{\tau}^{1}\geq X_{\tau}^{2}$
Show $Y_{n}:=X_{n}^{1}1_{ \{\tau > n\}}+X_{n}^{2}1_{ \{\tau \leq n\}}$ is a supermartingale
I have already shown that this is the case for $Z_{n}:=X_{n}^{1}1_{ \{\tau \geq n\}}+X_{n}^{2}1_{ \{\tau < n\}}$
My ideas for $Y$
$E[Y_{n+1}\vert \mathcal{F}_{n}]=E[X_{n+1}^{1}1_{ \{\tau > n+1\}}\vert \mathcal{F}_{n}]+E[X_{n+1}^{2}1_{ \{\tau \leq n+1\}}\vert \mathcal{F}_{n}]$
I am stuck on what to do for the first term. My ideas for the second term are as follows: $E[X_{n+1}^{2}1_{ \{\tau \leq n+1\}}\vert \mathcal{F}_{n}]=E[X_{n+1}^{2}1_{ \{\tau < n+1\}}\vert \mathcal{F}_{n}]+E[X_{n+1}^{2}1_{ \{\tau = n+1\}}\vert \mathcal{F}_{n}]=E[X_{n+1}^{2}1_{ \{\tau \leq n\}}\vert \mathcal{F}_{n}]+E[X_{n+1}^{2}1_{ \{\tau -1 < n+1 \leq \tau\}}\vert \mathcal{F}_{n}](*)$
And I know that $\{n+1 = \tau\}=\{\tau -1 < n+1 \leq \tau\}$ is previsible since $\tau-1$ and $\tau$ are stopping times. Hence:
$(*)\leq 1_{\{\tau < n+1\}}X_{n}^{2}+1_{\{\tau=n+1\}}X_{n}^{2}=X_{n}^{2}1_{\{\tau\leq n+1\}}$
How can I manipulate the first term using the fact that $X_{\tau}^{1}\geq X_{\tau}^{2}$?
and is my assumption that $\{n+1 = \tau\}=\{\tau -1 < n+1 \leq \tau\}$ is previsible, correct?
Here:
\begin{alignat*}{2} E[Y_{n+1}\vert \mathcal{F}_{n}] & = E[X_{n+1}^{2}1_{ \{\tau \le n\}}\vert \mathcal{F}_{n}] + E[X_{n+1}^{2}1_{ \{\tau = n+1\}}\vert \mathcal{F}_{n}] + E[X_{n+1}^{1}1_{ \{\tau > n+1\}}\vert \mathcal{F}_{n}]\\ & \le E[X_{n+1}^{2}1_{ \{\tau \le n\}}\vert \mathcal{F}_{n}] + E[X_{n+1}^{\color{red}{1}}1_{ \{\tau = n+1\}}\vert \mathcal{F}_{n}] + E[X_{n+1}^{1}1_{ \{\tau > n+1\}}\vert \mathcal{F}_{n}]\\ & \le E[X_{n+1}^{2}\vert \mathcal{F}_{n}]1_{ \{\tau \le n\}} + E[X_{n+1}^{1}\vert \mathcal{F}_{n}]1_{ \{\tau > n\}}\\ & \le X_{n}^{2}1_{ \{\tau \le n\}} + X_{n}^{1}1_{ \{\tau > n\}}\\ & \le Y_n \end{alignat*}
But $\{\tau = n+1\}$ isn't predictable.