show $E\int^\infty_0e^{-A_t}dA_t$ is bounded above

Question

show $E\int^\infty_0e^{-A_t}dA_t$ is bounded above by constant, where $A_t$ is an increasing stochastic process of locally integrable variation starting from 0. $A$ is possible to be purely discontinuous.

I feel it should be bounded like in the case when $A_t=t$, since exponent "kills" the process, but I can't show this formally.

Answer 1

Because $e^{-x}$ is convex, we know that

$$(y-x)e^{-x}\leq f(x)-f(y)$$

for $x\leq y$. So, noting that $A$ is increasing

$$\int_0^T e^{-A_t}dA_t=\lim\sum_{t_k}(A_{t_{k+1}}-A_{t_k})e^{-A_{t_k}}\leq \lim\sum_{t_k}e^{-A_{t_{k}}}-e^{-A_{t_{k+1}}}=e^{-A_0}-e^{-A_T}\leq e^{-A_0}$$

which is independent of $T$.