Showing Ito integrals equal implies stochastic processes equal

Question

Suppose for $S<T$ being two real numbers and $(X_t), (Y_t)$ being two progressively measurable processes we have $$C + \int_S^T X_tdB_t \overset{a.s.}{=} D + \int_S^TY_tdB_t$$ for some $C, D \in \mathbb{R}$. We seek to show that $C=D$ and that $(X_t)_{t \in [S, T]} \overset{a.s.}{=}(Y_t)_{t \in [S, T]}$. I'm not sure how to begin so I'm considering the special case $C = D = 0$ to start with. I can only think of using either the definition (construction) of the Ito integral, or one of Ito's Lemmas here to work with. So assuming $$\int_S^T X_tdB_t \overset{a.s.}{=} \int_S^TY_tdB_t$$ I seek to show that $(X_t)_{t \in [S, T]} \overset{a.s.}{=} (Y_t)_{t \in [S, T]}$. I first tried to take expectations of $\int_S^T(X_t-Y_t)dB_t$ and square both sides to apply the Ito Isometry, but I'm not sure that helps, so I'm not sure how to proceed. I also tried a contrapositive argument assuming $(X_t) \overset{a.s.}{\neq} (Y_t)$ to show their Ito integrals cannot be equal, but wasn't sure how to make that work either. Some hints are much appreciated!

Answer 1

The martingale property of the Ito integral says that the finite integral $\int_0^K X_tdB_t$ is a martingale in $K \in \mathbb{R}$. Thus $\int_0^SX_tdB_t, \int_0^TX_tdB_t$ have the same expectation, so that $E\left(\int_S^T X_TdB_t\right) = 0$, and likewise $E\left(\int_S^T X_TdB_t\right) = 0$. Thus taking expectations makes both integrals vanish and we get $C = D$ immediately.

Then, we are given $$\int_S^T (X_t - Y_t)dB_t = 0 \quad\text{a.s.}$$ and from Ito isometry, this means $$0=\mathbb E\left[\int_S^TX_t-Y_tdBt\right]^2=\mathbb E\left[\int_S^T\left(X_t-Y_t\right)^2dt\right]$$ which implies the integrand is null almost surely, because if $E(A) = 0$ for a non-negative definite random variable $A$, then $A = 0$ almost surely. In our case, $A = \int_S^T(X_t-Y_t)^2dt$ is clearly a non-negative rv since it's the integral of a square. Thus $(X_t-Y_t)^2 = 0$ almost surely, so that we get $(X_t)_{t \in [S, T]} \overset{a.s.}{=} (Y_t)_{t \in [S, T]}$, as required to conclude.