the convergence in probability of the stochastic integral In Jacod's Limit Theorems for Stochastic Processes:page 47,thm 4.31 (iii)
$X$ is a semimartingale , $H_n$ are predictable process converge pointwise to $H$ , and $|H^n|\le K$ , where $K$ is a locally bounded predictable process, then $\forall t:H^n\cdot X_t\to H\cdot X_t$ in probability.
First consider when $X\in\mathcal V$ (cadlag , adapted , $X_0=0$ , finite variation on $[0,t],\forall t$ )
I want to prove the conclusion above by markov inequality:
$$\mathbb P\left(|(H^n-H)\cdot X_t|\ge \varepsilon\right)\le\varepsilon^{-1}\mathbb E\left[|(H^n-H)\cdot X_t|\right]$$
since $H^n-H\to 0 $ pointwise and $|H^n-H|\le 2K$  , so I want to use DCT,but I can't prove that $\mathbb E\left[|K\cdot X_t|\right]<\infty$
$$\mathbb E\left[|K\cdot X_t|\right]\le\mathbb E\left[\int_0^t|K_s|d| X_s|\right]\le C\mathbb E| X_t| $$
I can't get $\mathbb E| X_t|<\infty$ from $X\in\mathcal V$.
Thanks a lot！
 A: In general, $X \in \mathcal{V}$ does not imply $\mathbb{E}(|X_t|)<\infty$. Just consider the process
$$X_t(\omega) := \begin{cases} 0, & t < 1, \\ \frac{1}{\omega}, & t \geq 1 \end{cases}, \qquad \omega \in \Omega := (0,1).$$
on the measurable space $((0,1),\mathcal{B}((0,1)))$ endowed with the Lebesgue measure. Obviously, $X \in \mathcal{V}$, but $\mathbb{E}(|X_t|)=\infty$ for all $t>1$.

This means that your argumentation doesn't work this way; however, we can modify it to make it work. Define a sequence of stopping times by
$$\tau_k := \inf\{t>0; |X_t| > k\}.$$
Then
$$\begin{align*} \mathbb{P}(|(H^n-H) \bullet X_t|>\epsilon) &= \mathbb{P}(|(H^n-H) \bullet X_t|>\epsilon, \tau_k> t) + \mathbb{P}(|(H^n-H) \bullet X_t|>\epsilon, \tau_k \leq t) \\ &\leq \mathbb{P}(|(H^n-H) \bullet X_t|>\epsilon, \tau_k > t) + \mathbb{P}(\tau_k \leq t)\\ =: I_1+I_2\end{align*}$$
By Markov's inequality
$$I_1 \leq \frac{1}{\epsilon} \mathbb{E}(1_{\{\tau_k>t\}} |(H^n-H) \bullet X_t|).$$
Proceeding as in your argumentation and using $\mathbb{E}(1_{\{\tau_k>t\}} |X_t|) \leq k$, we get $I_1 \to 0$ as $n \to \infty$. Consequently,
$$\limsup_{n \to \infty} \mathbb{P}(|(H^n-H) \bullet X_t|>\epsilon)\leq \mathbb{P}(\tau_k \leq t).$$
Letting $k \to \infty$, the claim follows.
