Girsanov density as a functional on $C[0,1]$

I'll formulate the question via an example.

On $$( C[0,1], \mathcal{C} )$$, where $$C[0,1]$$ is the set of continuous functions on $$[0,1]$$ and $$\mathcal{C}$$ the Borel $$\sigma$$-algebra given by uniform topology, consider Wiener measure $$\mathbb{P}$$. Denote by $$t \mapsto W_t$$ the Brownian paths given by $$\mathbb{P}$$.

Let $$X$$ be specified via $$dX = \mu(X) dt + dW,$$ where $$\mu: \mathbb{R} \rightarrow \mathbb{R}$$ is fixed.

Girsanov's theorem says, under the measure $$\mathbb{Q}$$ given by (assume, e.g., Novikov's condition holds) $$\frac{d \mathbb{Q} }{ d\mathbb{P} } = e^{ \int_0^1 \mu(X) dW - \frac{1}{2} \int_0^1 \mu(X)^2 dt },$$ the $$\mathbb{Q}$$-law of $$W$$ is the $$\mathbb{P}$$-law of $$X$$.

Question

The Radon-Nikodym derivative $$\frac{d \mathbb{Q} }{ d\mathbb{P} }$$ is specified via stochastic integral. But in principle, a Radon-Nikodym derivative is an object one must be able to define $$\omega$$-by-$$\omega$$, $$\mathbb{P}$$-almost surely in this case. So what is $$\frac{d \mathbb{Q} }{ d\mathbb{P} }$$ as a functional on $$C[0,1]$$ (strictly speaking on the support of $$\mathbb{P}$$)?

Conjecture

Ignore that Brownian paths do not have finite variation, etc. Formally, it is the functional $$\phi : C[0,1] \rightarrow \mathbb{R}$$ given by $$f(\cdot) \stackrel{\phi}{\mapsto} e^{\int_0^1 \mu(x(t)) df(t) - \frac{1}{2} \int_0^1 \mu^2(x(t)) dt }$$ where $$dx = \mu(x) dt + df$$, and $$\int_0^1 \mu(x(t)) df(t)$$ is a Riemann-Stieltjes integral with respect to $$df$$. In other words, given $$f \in C[0,1]$$, one acts as if $$f$$ is a realization of Brownian path and substitute formally into the expression for $$\frac{d \mathbb{Q} }{ d\mathbb{P} }$$. Is this correct in some sense---e.g. discretize into step functions and taking weak limit in the Skorohod space $$D[0,1]$$...?

Suggestive Example (or not)

Suppose $$dX = a \, dt + dW$$ where $$a$$ is a real number. Then, $$\omega$$-by-$$\omega$$, the Radon-Nikodym derivative $$\frac{d \mathbb{Q} }{ d\mathbb{P} } = e^{ \int_0^1 a dW - \frac{1}{2} \int_0^1 a^2 dt },$$ is given by the functional $$f(\cdot) \stackrel{\phi}{\mapsto} e^{ a \int_0^1 df(t) - \frac{1}{2} \int_0^1 a^2 dt }$$ where $$\int_0^1 df(t)$$ is interpreted as $$f(1) - f(0)$$.

• How can you "ignore that Brownian paths do not have finite variation"? If, say, $\mu(x) = x$, then how is $\phi$ defined? – Mateusz Kwaśnicki Aug 9 at 19:43