It is shown when constructing the Itô-integral that if:
$E[\int_0^T X_t^2dt]< \infty$. Then we have that Itô-isomtry:
$E[\int_0^T X_t^2dt]=E[(\int_o^TX_tdB_t)^2]$.
In the extended Itô integral, we relax the assumtion that $E[\int_0^T X_t^2dt]< \infty$, and instead have that: $P(\int_0^T X_t^2dt< \infty)=1$. If we are in this case, and not in the first case we have that: $E[\int_0^T X_t^2dt]= \infty$.
But I am wondering if we have the Itô-isometry in the extended case as well?(Where both sides must now be infinity?) I have only seen the Itô-isometry proved in the elementary $L^2$-setting.
A way to frame my question is this:, we know that we have the implication $E[\int_0^T X_t^2dt]< \infty \rightarrow E[(\int_o^TX_tdB_t)^2] <\infty$. But do we have the other implication?
Here is what I mean:
Part 1(this one is clear): $E[\int_0^T X_t^2dt]< \infty$ Now we know that:
$E[(\int_o^TX_tdB_t)^2=E[\int_0^T X_t^2dt]< \infty$
Part 2: $E[(\int_o^TX_tdB_t)^2]<\infty$
If we know that $E[(\int_o^TX_tdB_t)^2]<\infty$, as an extended Itô-integral, do we know that we are in the elementary setting and that also $E[\int_0^T X_t^2dt]< \infty$? Or can we have the case that $P(\int_0^T X_t^2dt< \infty)=1 , E[\int_0^T X_t^2dt]= \infty$, and $E[(\int_o^TX_tdB_t)^2]<\infty$, where the last is an Itô-integral in the extended case?
My attempt of proof: When I tried proving this I use that in the extended case instead of $L^2$-convergence of simple functions, we have convergence in probability. By taking subsequences we are working with convergence almost surely. So I have a sequence of simple functions such that $\int_o^T|\phi_n-X_t|^2dt\rightarrow 0$ a.s. And $\int_o^T\phi_ndB\rightarrow \int_0^TX_tdB$ a.s. We know that we can use the Itô-isometry on the simple functions, so we have that $E[(\int_o^T\phi_ndB)^2]=E[\int_0^T\phi_n^2dt]$. But in order to finish I need that $E[\int_0^T\phi_n^2dt]\rightarrow E[\int_0^T X_t^2dt]$, and that $E(\int_o^T\phi_ndB)^2 \rightarrow E(\int_o^TX_tdB)^2$. But since we started with a.s.-convergence and not $L^2$-convergence, I can't see that we have this.
Any ideas? Do you know if the statement holds?
As suggested by you, I put my comment as an answer.
I suggest you have a look at lemma 3 (the iff part of the assertion) here .
Best regards.