Let $f(t)$ be a progressively measurable process wrt Brownian motion $B(t)$ so that $$P\left(\int_0^Tf^2(s)ds<\infty\right)=1$$ Is it true then that the exponential
defines a density on Wiener space? I know the Novikov condition
implies that the exponential defines a density. But what is you just have a.s. $L^2$?