Let $P$ be the Wiener measure on $\Omega=C[0,\infty)^d$ and $W=(W^{(1)},\ldots,W^{(d)})$ be the standard Brownian motion. We may then define the Brownian filtration $\{\mathscr F_t^W,t\ge0\}$. Let $X$ be a $d$-dimensional progressively measurable process with respect to $\{\mathscr F_t^W\}$. The following is an exercise in Karatzas and Shreve($\|\cdot\|$ is the $2$-norm on $\mathbb R^d$):
Suppose that $\int_0^\infty\|X_t(\omega)\|^2dt<\infty$ for every $\omega\in\Omega$. Show that $Z_t(X)$, defined as following, is a $\{\mathscr F_t^W\}$-martingale(under $P$). $$Z_t(X):=\exp\left(\sum_{i=1}^d\int_0^t X_s^{(i)}dW_s^{(i)}-\frac12\int_0^t\|X_s\|^2ds\right)$$
The result seems much sharper than Novikov and Kazamaki. Beneš condition, which specializes in dealing with Brownian functionals, seems not to work well here, either. I have really no idea where to begin with. The book gives a reference to H.J. Engelbert and R. Hohnle but I can't find relevant results of either author.
The word "every" in Italics mystifies me a lot. There are some elementary examples showing that "every" can not be replaced by "almost every", so this must be the key to a proof. But for what reason is "every" stronger than "almost every"? The only difference I can think of is that "every" is not dependent on the particular measure while "almost every" is. But how can I turn this into a proof?