By definition, a strategy $\xi:=(\xi_{i})_{i=1,...T}$ is called self-financing for a price process $(X_{i})_{i=1,...,T}$ if $\xi_{i}X_{i}=\xi_{i+1}X_{i}$ for all $i = 1,...,T$.
Background to my question:
I want to prove that:
Statement a): a bounded, self-financing strategy $\xi:=(\xi_{i})_{i=1,...T}$ for a price process $(X_{i})_{i=1,...,T} $ implies a $V^{\xi}$ is a $Q-$Martingale, where $V^{\xi}_{i}=\xi_{i} \times X_{i}$ for all ${i=1,...T}$ where $\times$ is a scalar product in $\mathbb R^{d}$
Statement b): any self-financing strategy $\xi:=(\xi_{i})_{i=1,...T}$ for a price process $(X_{i})_{i=1,...,T} $ where $E_{Q}[(V_{T}^{\xi})^{-}]< \infty$ implies $V^{\xi}$ is a $Q-$Martingale, where $V^{\xi}_{i}=\xi_{i} \times X_{i}$ for all ${i=1,...T}$
I am required to prove a)$\implies$ b).
I want to truncate $\xi$ by introducing $\xi^{n}:=(\xi_{i} 1_{ \{\vert \xi_{i}\vert \leq n\}})_{i=1,...,T}$ for $n \in \mathbb N$. Of course this sequence is bounded, but I am not sure whether any $\xi^{n}$ is indeed self-financing, considering:
$\xi^{n}_{t+1}X_{t}=\xi_{t+1}X_{t}1_{ \{\vert \xi_{t+1}\vert \leq n\}}$. How can I be sure that this equals $\xi_{t}X_{t}1_{ \{\vert \xi_{t}\vert \leq n\}}$?
For $X_t = 0$ it obviously holds $$\xi^n_{t}X_t = \xi^n_{t+1}X_t$$ so assume $X_t \not= 0$:$$\begin{align*}&&|\xi_{t+1}| &\le n \\ &\iff& |\xi_{t+1}X_t| &\le n|X_t| \\ &\iff& |\xi_{t}X_t| &\le n|X_t| \\ &\iff& |\xi_{t}| &\le n \end{align*}$$
Hence $$1_{ \{\vert \xi_{t}\vert \leq n\}} = 1_{ \{\vert \xi_{t+1}\vert \leq n\}}$$
And in total we get: $$\xi^{n}_{t+1}X_{t}=\xi_{t+1}X_{t}1_{ \{\vert \xi_{t+1}\vert \leq n\}} = \xi_{t}X_{t}1_{ \{\vert \xi_{t}\vert \leq n\}} = \xi^{n}_{t}X_{t}$$