Suppose that Q and R are two probability distributions on $\Omega$ equipped with $\mathcal{F}$ sigma algebra. Moreover, suppose that $Q<<R$. By Radon-Nykodym derivative we know that there exists a signed measure $f$ such that $$Q(A)=\int_{A}fdR, \forall A\in \mathcal{F}.$$
Since $Q$ is nonnegative, we can conclude that f is nonnegative $R$ a.e. My question is related to the well definedness of KL-divergence which is $$\mathbb{E}^{Q}[\log{\frac{dQ}{dR}}]$$
we know that $\frac{dQ}{dR}$ is nonnegative $R$ a.e. . But here we are taking an integral with respect to $Q$. How can we show that $\frac{dQ}{dR}$ is also nonnegative $Q$ a.e. ?
Furthermore, we know that $\frac{dQ}{dR}\in L^1({R})$. How can we show that $\log{\frac{dQ}{dR}} \in L^1(Q)$?
Let $E = \{dQ/dR < 0\}$ be the set where $dQ/dR$ is negative. By definition, $R(E) = 0$. Since $Q \ll R$, $Q(E) = 0$, by definition (or one of the equivalent characterizations) of absolute continuity of measures. Hence $dQ/dR$ is nonnegative $Q$-a.e.
It isn't always the case that $\log(dQ/dR)\in L^1(Q)$. Here is an example to illustrate. Let our space be $[1,\infty)$ with the Borel $\sigma$-algebra, and $$ R(dx) = \frac{dx}{x^2(\log x)^2},\quad Q(dx) = \frac{dx}{x(\log x)^2}. $$ It's straightforward to check that both $R$ and $Q$ are finite measures on $[1,\infty)$, and that $Q\ll R$ with $dQ/dR = x$. We compute \begin{align*} \mathbb E^Q[\log\frac{dQ}{dR}] &:= \int_{1}^\infty \log(x)\,Q(dx)\\ &= \int_1^\infty \log(x)\,\frac{dx}{x(\log x)^2}\\ &=\int_1^\infty \frac{dx}{x(\log x)} \stackrel{u=\log x}{=} \int_0^\infty \frac{du}{u} = \infty, \end{align*} so $\log(dQ/dR)\notin L^1(Q)$.