Stochastic integral under stopping

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Let $\nu^+$ (resp. $\nu$) be the set of all real valued processes $A$ that are càdlàg, adapted with $A_0=0$ and whose each path is non-decreasing (resp. has finite variation over each finite interval $[0,t]$).

Let $\mathcal{A}$ denote the set of all stochastic processes $A\in\nu$ which satisfy $$E[\text{Var}(A)_\infty]<\infty$$ and $\mathcal{A}^+$ the set of all processes $A\in\nu^+$ which satisfy $$E[A_\infty]<\infty$$ where $\text{Var}(.)$ is the variation of a stochastic process. Furthermore we know that $$\mathcal{A}_{loc}=\mathcal{A}^+_{loc}\ominus\mathcal{A}^+_{loc}$$.

Let $A\in\mathcal{A}_{\text{ loc}}$. Then there is a sequence of stopping times $(T_n)_{n\in\mathbb{N}}$ s.t. $A^{T_n}\in\mathcal{A}$ and processes $B,C\in\mathcal{A}^+_{loc}$, uniquely determined up to evanescent, s.t. there are localizing sequences of stopping times $(T_n')_{n\in\mathbb{N}}$ and $(T_n'')_{n\in\mathbb{N}}$ s.t. $$A^{T_n}=B^{T_n'}-C^{T_n''}$$ for $n\in\mathbb{N}$.

Now let $H$ be a predictable process and $H\cdot A\in\mathcal{A}_{\text{ loc}}$. Then $H^+\cdot (B^{T_n'})$ is clearly increasing and since $(B^{T_n'})\in\nu^+$, $H^+\cdot (B^{T_n'})\in\nu^+$.

But my aim is to show $H^+\cdot B\in\mathcal{A}_{loc}^+$. So the question is if $H^+\cdot (B^{T_n'})=(H^+\cdot B)^{T_n'}$ holds and how to proof that $E[(H^+\cdot B)^{T_n'''}_\infty]=E[(H^+\cdot B)_{T_n'''}]<\infty$ for an appropriate stopping time $T_n'''$?

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Please use the following easy to verify equalities \begin{gather} (B^T)_t=B_{T\wedge t}=(1_{ \,{[\!\![}\, 0,T\; ]\!\!]}\cdot B)_t.\\ (H\cdot B)^T=(H 1_{\,{[\!\![}\, 0,T\;]\!\!]})\cdot B= (H 1_{\,{[\!\![}\, 0,T\;]\!\!]})\cdot B^T=H\cdot (B^T).\\ \text{Var}(H\cdot A)= |H|\cdot \text{Var}(A). \tag{1} \end{gather} From (1), if $(H\cdot A)\in\mathcal{A}_{\text{loc}}$, then there exists a sequence $\{T_n, n\ge 1\}$ of stopping times such that, $T_n\uparrow +\infty$ and $(|H|\cdot \text{Var}(A))^{T_n}\in \mathcal{A}^+$. Since $H^+\cdot B\le |H|\cdot \text{Var}(A)$, then $(H^+\cdot B)^{T_n}\in \mathcal{A}^+$ too.