Conditional expectation of a finite variation process

A simple question:

Let $H$ be a cadlag, adapted process and $A$ a process of finite variation. Then also

$\int_t^T HdA_t$ is a finite variation process (see "Limit Theorems... "Jacod&Shiryaev Prop. 3.5 p. 28). Is $\mathbb{E}\left[\int_t^T HdA_t\right]$ also a finite variation process?

I think one can refute that:

Suppose $H\equiv1$. Then $\mathbb{E}\left[\int_t^T HdA_t\right]=\mathbb{E}\left[A_T-A_t\right]=\mathbb{E}\left[A_T\right]-A_t$

$\mathbb{E}[A_T|\mathcal{F}_t]$ is a true martingale, if $A_T$ is integrable. Now if $\mathbb{E}[A_T|\mathcal{F}_t]$ is predictable (esp. if it is continuous) then it can be of finite variation only, if it is constant. In other words, $A$ must be deterministic, if $\mathbb{E}[A_T|\mathcal{F}_t]$ is of finite variation.

Do you agree?

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Your example works fine for me. You can specify it a little further if you like by taking any non integrable random variable $X$ (e.g. a Cauchy r.v.), and make it a process $X_t$ defined by $X_t=0$ for all $t\in [0,1[$ and $X_1=X$ . Your $H_t=1$ $\forall t$ process is still fine for your purpose and as the integral $Z_t=\int_t^1H_s dX_s=X$ is well defined, has finite variation almost surely but as it has no finite expectation for all t, it is not a finite variation process.
You may want to check Cor. 3.16 in the same book as above. Note that the process $M_t:=\mathbb{E}[A_T|\mathcal{F}_t]$ is a martingale and can thus either be constant or it is not of finite variation. Sorry for creating confusion. –  user13655 Sep 28 '12 at 6:52