TITLE: Weak continuity of K-L divergence function
QUESTION [10 upvotes]: If $P_n$ and $Q_n$ are two pmf's of a discrete set (say $A$) with common support and $P_n \to P$ and $Q_n \to Q$ where the convergence is pointwise here (even weak would be fine here I guess), then
$$ D(P_n\|Q_n) \to D(P\|Q)$$ 
follows from continuity and the fact that it is a finite sum. To recall, for finite/countable sets, KL Divergence is given by
$$D(P\|Q) = \sum_{k=1}^{|A|}P(k)\log_2\left(\frac{P(k)}{Q(k)}\right)$$
Now for uncountable sets, one would use a slightly different definition of divergence (if $A$ here were uncountable but a measurable space with a Borel Sigma algebra). For any two measures $P$ and $Q$ on $A$ where $P \ll Q$ ($P$ is absolutely continuous wrt $Q$), we have
$$D(P\|Q) = \int_A \frac{dP}{dQ}\log\left(\frac{dP}{dQ}\right)dQ$$
where $\frac{dP}{dQ}$ is the Radon Nikodym derivative of $P$ wrt $Q$.
My question is, Suppose $P_n \to^w P$ and $Q_n \to^w Q$, and $P_n \ll Q_n ~ \forall n$, does it follow that 
$$ D(P_n\|Q_n) \to D(P\|Q)$$
? (Here "$\to^w$" denotes weak convergence). 
My attempt: My guess was that it would not hold every time but in certain cases. To characterize them, I started with understanding $D(P_n\|Q)$. This is simply 
$$D(P_n\|Q) = \int_A \frac{dP_n}{dQ}\log\left(\frac{dP_n}{dQ}\right)dQ$$.
Only now I am not sure how to show $\frac{dP_n}{dQ} \to \frac{dP}{dQ} ~ a.e$
Source and References: While one would encounter these terms in any good information theory book, I would recommend Csiszar and Korner "Coding theorems for discrete memoryless systems". There are also quite a few papers that deal with this kind of divergence. However the source of the problem is not from these books but was on a research problem I was tackling years back and am now revisiting.
Hence I appreciate ideas and tips on proceeding rather than outright answers. Feel free to request further clarification.
Update1: I had mentioned above that $P \ll Q$. If it helps we may also let $Q \ll P$.

REPLY [4 votes]: @Ashok comment resolved this completely, so I am rewriting it as an answer. Posner, "Random Coding Strategies for Minimum Entropy", IEEE Trans Info Theory, 1975 showed that KL divergence is lower-semicontinuous in the topology of weak convergence. In particular, if $P_n \to P$ and $Q_n \to Q$ weakly, then
$\lim_n D(P_n\Vert Q_n) \ge D(P\Vert Q)$. A very nice summary is provided in the excellent Y. Polyanskiy, Y. Wu, "Lecture notes on Information Theory" (see Theorem 3.6)
Note that the first claim in your question is incorrect: it is not the case that $\lim_n D(P_n \Vert Q_n) = D(P\Vert Q)$ for all countable/discrete alphabets, even if all the distributions have the same support, because the continuity property you state only holds for finite alphabets. Here is a counterexample.
Let $P(i)=\frac{1}{Z}e^{-i^2}$, where $Z=\sum_{i=1}^\infty e^{-i^2}$ is finite normalization constant, and let $P=Q=Q_n$, so that $D(P\Vert Q)=0$. Let $\alpha_n=\sum_{i=1}^n P(i)$ indicate partial sums, and define $P_n$ in the following manner:
\begin{align}
P_n(i) = \begin{cases}
(1-n^{-2}) P(i) & \text{for $i\in\{1,\dots,n-1\}$}\\
1-(1-n^{-2})\alpha_{n-1} & \text{for $i=n$}\\
0 & \text{otherwise}
\end{cases}
\end{align}
It is easy to see that $P_n$ is a valid probability distribution and that $P_n \to P$ in total variation, hence also in the topology of weak convergence.
Now write
$$D(P_n\Vert Q_n)=D(P_n\Vert P)=\alpha_{n-1}(1-n^{-2})\ln(1-n^{-2})+P_n(n)\ln\frac{P_n(n)}{P(n)}.$$
Since $\lim_{n} \alpha_{n-1}=1$ and $\lim_n (1-n^{-2})\ln(1-n^{-2})=0$,
\begin{align}
\lim_n D(P_n\Vert Q_n) &= \lim_n P_n(n)\ln\frac{P_n(n)}{P(n)}\\
&= \lim_n -P_n(n)\ln{P(n)}\\
&=\lim_n P_n(n)(n^2 + \ln Z)= \lim_n \frac{P_n(n)}{n^{-2}}
\end{align}
where we used that $\lim_n P_n(n)=0$ and the definition of $P(n)$. Finally, a simple application of L'Hospital's rule shows that  $\lim_n P_n(n)/n^{-2}=1$.
Combining gives
$$\lim_n D(P_n\Vert Q_n) =1 > 0 = D(P\Vert Q).$$