Definition of continuous semi-martingale According to William's Diffusion Markov process

X is a continuous semi-martingale if
\begin{align*}
X=X_0+M+A
\end{align*}
where $M$ is a continuous local martingale null at 0, and $A$ is a continuous Finite variation process null at zero.

However, I feel $A$ should be restricted only as a locally finite variation process.
Let $H$ be a locally bounded previsible process. Define $H\cdot X := H\cdot M + H\cdot A$. where
\begin{align*}
(H\cdot A)_t = \int_0^ t H \ dA
\end{align*}
William claims that $H\cdot M+H\cdot A$ is the semi-martingale decomposition. But this seems not true if $A$ is required to be FV process.
i.e. $H\cdot A$ is not necessarily finite variation if only assume $H$ is locally bounded process? (For example, just take $H$ as the process $t$.)
Does anyone have any comments?
Thanks!
 A: Semimartingales are usually introduced when you define the stochastic integral for càglàd integrands. Roughly spoken they are defined as the class of càdlàg adapted processes $X$ such that the stochastic integral $Y\bullet X$ is well-defined for every càglàd adapted process $Y$. It can be shown that this class of processes are exactly the processes which can be decomposed as the sum of a local martingale and a finite variation process. This is called the Bichteler-Dellacherie theorem. Thus continuous semimartingales are exactly the calss of processes as you quoted.
I think the thing with taking  processes which are locally of finite variation instead of finite variation is, that being a finite variation process is already a kind of local thing. Because an adapted càdlàg process is called to be a 'finite variation process' if it has finite variation on every compact intervall. If you now take a sequence of stopping times $T_n$ ($T_n\nearrow\infty$) so that $Y^{T_n}$ is a finite variation process, then $Y$ is already a finite variation process. So the class of finite variation processes and the one which are locally finite variation processes coincide.
Regarding the second thing you quoted of William, he is right too. An alternative proof of this can be found for example here. Your (counter)example is not ture, which can also be seen by this:
Let $X$ be a continuous finite variation process. Then the stochastic integral
$$I_t:=\int_0^t s\, \mathbb{d}X_s$$
coincides with the pathwise Stieltjes inegral. Thus we can easily calculate the variation process. As every finite variation function $f$ can be decomped into a difference of positive monotonically increasing functions $f^1,f^2$ so that $f=f^1-f^2$ (to be exactly we can choose $f^1_t:=|f|_t$ as the total variation of $f$ on $[0,t]$ and $f^2:=f^1-f$), we get
\begin{align*}
|I|_t=\sup_{\tau\text{ is partition}}\sum_{t_n,t_{n+1}\in\tau}\left|\int_{t_n}^{t_{n+1}} s\, \mathbb{d}X_s\right|&=\sup_{\tau\text{ is partition}}\sum_{t_n,t_{n+1}\in\tau}\left|\int_{t_n}^{t_{n+1}} s\, \mathbb{d}X_s^1-\int_{t_n}^{t_{n+1}} s\, \mathbb{d}X_s^2\right|\\
&\leq\sup_{\tau\text{ is partition}}\sum_{t_n,t_{n+1}\in\tau}\left|\int_{t_n}^{t_{n+1}} s\, \mathbb{d}X_s^1\right|+\left|\int_{t_n}^{t_{n+1}} s\, \mathbb{d}X_s^2\right|\\
&=\sup_{\tau\text{ is partition}}\sum_{t_n,t_{n+1}\in\tau}\int_{t_n}^{t_{n+1}} s\, \mathbb{d}(X_s^1+X_s^2)\\
&=\int_{0}^{t} s\, \mathbb{d}(X_s^1+X_s^2)\\
&=2\int_{0}^{t} s\, \mathbb{d}|X|_s-\int_{0}^{t} s\, \mathbb{d}X_s\\
&<\infty.
\end{align*}
