Is limiting density of discrete points (LDDP) equivalent to negative KL-divergence? Is limiting density of discrete points (LDDP), which is a corrected version of differential entropy, equivalent to the negative KL-divergence (or relative entropy) between a density function $m(x)$ and a probability distribution $p(x)$?
What are the maths of this equivalence, and the conditions that the density and probability functions have to meet? First of all, i thought densities are probability functions.
 A: This is an old question, but I just came across it and thought I'd give it a shot for any future visitors to the post.

First, some intuition:
As you mentioned, LDDP is a "corrected" version of Shannon's proposal for a continuous version of entropy. LDDP is more appropriate than Shannon's differential entropy because LDDP is the limit of discrete entropy as we approximate a continuous distribution with increasingly dense discrete distributions.
To formalize the above, consider a set $X_N = {x_i}$, where $|X_N| = N$. If we continue increasing $N$, our points $x_i$ "fill" some space. When $N \rightarrow \infty$, $X_N$ approaches some continuous space.
LDDP is given by:  $$H(X) = -\int p(x) \log \frac{p(x)}{m(x)} ~\mathrm dx $$
where $$\lim_{N \rightarrow \infty}\{\textrm{number of } x_i \textrm{ in } (a,b)\} = \int_a^bm(x)$$
So $m$ is just the density of the space that $X_N$ approaches.
KL-Divergence is: $$D_{KL}(p(x)\Vert q(x)) = \int p(x)\log \frac{p(x)}{q(x)} ~\mathrm dx$$
So clearly, $H(X) = -D_{KL}(p(x)\Vert m(x))$.
KL-Divergence (roughly) measures how different the two distributions are. So $H(X)$ is measuring how similar $p$ is to $m$. For many familiar spaces, like $[0,1]^d$, we can sample the space continuously and have $m(x) = \textit{U}([0,1]^d)$.
Because uniform distributions maximize entropy, it is natural that the continuous version of entropy would give the KL-divergence between the probability distribution in question and a uniform distribution over the same support.
References:
See page 201-202 here
