Let $W=\{W_t\}_{t\in[0;1]}$ be a real-valued Brownian motion, $\{F_t\}_{t\in [0;1]}$ the filtration generated by $W$, augmented with the nullsets. Let $\{\sigma_t\}_{t\in[0;1]}$ be a continuous and bounded Ito process (**EDIT**: w.r.t. $W$) with bounded drift and volatility coefficients (I could also live with more restrictions). I want to prove that
\begin{align*}
A
:=
\mathbb E\bigg( \exp\bigg( \sup_{t\in[0;1]}\int^t_0 \sigma_s \mathrm dW_s\bigg) \bigg)
<
\infty.
\end{align*}

If $\sigma$ was constant, this would be obvious since the distribution of the running maximum of the Brownian motion is known and the above expectation can be computed in closed integral form.

**Question:** Any ideas how to bound $A$ in a more general case than "$\sigma$ constant"? Or can such a result be found in the literature? Thank you!

This question has been posted on math.stackexchange a few days ago, but got no answers so far. (The "partially solved" in the title refers to the fact that meanwhile I managed to bound $B$ (see below), which was not the case when I asked the question on math.stackexchange originally. Thus, I have edited the question there. The main problem to bound $A$ is still unsolved.)

**What I have tried:** As an easier exercise, first I tried to prove the following:
\begin{align*}
B
:=
\mathbb E\bigg( \exp\bigg( \int^1_0 \sigma_s \mathrm dW_s\bigg) \bigg)
<
\infty.
\end{align*}
This is also easy since the stochastic exponential is a martingale and then
\begin{align*}
1
=&
\mathbb E\bigg( \exp\bigg(
\int^1_0 \sigma_s \mathrm dW_s
- \frac 1 2 \int^1_0 (\sigma_s)^2 \mathrm ds
\bigg) \bigg)
\\\ge&
\mathbb E\bigg( \exp\bigg(
\int^1_0 \sigma_s \mathrm dW_s
\bigg) \bigg)
e^{- \frac 1 2 \Vert \sigma\Vert^2}
\\=&
B e^{- \frac 1 2 \Vert \sigma\Vert^2}.
\end{align*}

Moreover, I thought we could $L^2$-approximate the Ito integral by \begin{align*} \sum_{k=1}^n \sigma_{(k-1)/n} ( W_{k/n} - W_{(k-1)/n} ), \end{align*} but I don't know how to approximate the supremum and anyway we cannot pull $L^2$-convergence into the $\exp$ function because $\exp$ increases faster than $\operatorname{id}^2$.