Integration by parts for measures Let $\mu_t$ be the law of a random process $X_t$. Let $\nu$ and $\nu_t$ be arbitrary measures in $\Omega$ with the sole restriction that $\mu_t$ be absolutely continuous w.r.t. $\nu_t$, and $\nu_t$ be absolutely continuous w.r.t. $\nu$.
Then (according to a paper I'm reading)
$$\int \partial_t \log \left(\frac{\nu_t(x)}{\nu(x)} \right) d\nu_t = \int \frac{\mu_t(x)}{\nu_t(x)} \partial_t \left(\frac{\nu_t(x)}{\nu(x)} \right)d\nu = - \int \partial_t \left(\frac{\mu_t(x)}{\nu_t(x)}\right) d \nu_t.$$
The second inequality looks like an integration by parts, but I don't know what it means to take "$dv$" = $\partial_t \left(\frac{\nu_t(x)}{\nu(x)} \right)d\nu $ where the "$dv$" derivative is with respect to a different variable (t) than the integral is taken over (\nu) -- usually "dv" is something like $\frac{d}{dx}v(x) dx$. And I don't understand the first inequality at all.
I'm also not sure what the notation $\frac{\nu_t(x)}{\nu(x)}$ is when we're talking about measures - I know it's not simply a fraction but something like a Radon-Nikodym derivative.
 A: As I have mentioned, I would not trust every single part of the paper you are referring to, as there a probably some unchecked typos or even mistakes: since it is on arXiv, it may be not peer reviewed yet. But some things I can help you with clarifying. I think it is more formally correct to write $g_t(x) = \frac{\mathrm d\mu_t}{\mathrm d\nu_t}(x)$ rather that $\frac{\mu_t(x)}{\nu_t(x)}$ since here $\nu_t(x)$ is not even defined, as a measure it is a function of sets, not of points.
One thing that makes is easier to work with RN derivatives is the following: if $\mu, \nu \ll \lambda$ then
$$
\frac{\mathrm d\mu}{\mathrm d\nu} = \frac{\mathrm d\mu}{\mathrm d\lambda}/\frac{\mathrm d\nu}{\mathrm d\lambda}. \tag{1}
$$
Note that it makes total sense symbolically, but still on the left hand side of $(1)$ you have a RN derivative, not any true ratio, whereas on the right hand side you have indeed a usual ratio of (density) functions. You can always find this $\lambda$, e.g. take $\lambda = \frac12(\mu + \nu)$. For example, it means that
$$
\frac \partial {\partial t}\left(\log\frac{\mathrm d\nu_t}{\mathrm d\nu}\right) = \frac \partial {\partial t}\left(\log\left(\frac{\mathrm d\nu_t}{\mathrm d\lambda}/\frac{\mathrm d\nu}{\mathrm d\lambda}\right)\right) = \frac \partial {\partial t}\left(\log\frac{\mathrm d\nu_t}{\mathrm d\lambda} - \log\frac{\mathrm d\nu}{\mathrm d\lambda}\right) = \frac{\frac\partial{\partial t}\frac{\mathrm d\nu_t}{\mathrm d\lambda}}{\frac{\mathrm d\nu_t}{\mathrm d\lambda}} = \left(\frac\partial{\partial t}\frac{\mathrm d\nu_t}{\mathrm d\lambda}\right)\frac{\mathrm d\lambda}{\mathrm d\nu_t}.
$$
Here we used two facts. First of all, $\lambda$ should be $\lambda_t$ in general, but since all $\nu_t$ are said to be dominated by the same measure, we were able to pick a single $t$-independent dominating measure $\lambda$. Moreover,
$$
\frac1{\frac{\mathrm d\nu_t}{\mathrm d\lambda}}  = {\frac{\mathrm d\lambda}{\mathrm d\nu_t}}.
$$
Some of these equalities are guaranteed to hold only a.s. but again, everything here is dominated by $\nu$, so that's not a problem. As a result, for every function $f(x)$ it holds that
\begin{align}
\int_X f(x)\frac \partial {\partial t}\left(\log\frac{\mathrm d\nu_t}{\mathrm d\nu}(x)\right)\nu_t(\mathrm dx) &= \int_Xf(x)\left(\frac\partial{\partial t}\frac{\mathrm d\nu_t}{\mathrm d\lambda}(x)\right)\frac{\mathrm d\lambda}{\mathrm d\nu_t}(x)\nu_t(\mathrm dx)
\\
&=\int_X f(x)\left(\frac\partial{\partial t}\frac{\mathrm d\nu_t}{\mathrm d\lambda}(x)\right)\lambda(\mathrm dx)
\\
&= \frac\partial{\partial t}\left(\int_X f(x)\frac{\mathrm d\nu_t}{\mathrm d\lambda}(x)\lambda(\mathrm dx)\right)
\\
&= \frac\partial{\partial t}\left(\int_X f(x)\nu_t(\mathrm dx)\right).
\end{align}
Note also that working with this you have to be accurate in something of the kind $b_t(x)\frac\partial{\partial t}a_t(x)$ which is
$$
b_t(x)\left.\left(\frac\partial{\partial s}a_s(x)\right)\right|_{s = t}
$$
cause otherwise you may get confused. I hope I didn't.
