TITLE: A necessary condition for differential entropy to be finite
QUESTION [7 upvotes]: An ensemble corresponding to a probability distribution usually has finite free energy so it is not a great loss of generality to assume that the ensemble also has finite energy in following discussions.
It is known that when the probability distribution $\mu$(assuming it is also dominated by Lebesgue measure and its density $f_{\mu}(x)$ for simplicity.) has a finite (compact) support then the classical differential entropy $$En(\mu) = \int_{\mathbb{R}} -\log[f_{\mu}(x)]f_{\mu}dx$$
can be bounded by Bekenstein bound (Wiki) and hence finite.
However, when the support of $\mu$ is not finite (compact), there exists counter example that the entropy could be infinite (Math.SE).
Question:
(1)Is finite support of $\mu$ also a necessary condition to make sure the $\mu$ has a finite entropy?
Update: It is clear that a $\mu$ with infinite support can have finite entropy. Thanks Anthony Quas for pointing out.
(2)Is there a characterization(sufficient and necessary condition) of probability distributions/statistical ensembles with finite entropy?
(3)Also, it is known that Bekenstein bound may also be applied(with volumes defined only for atoms) to entropy defined for $\sigma$-algebras(MO.post), so can we translate the characterization in (2) onto $\sigma$-algebras?
(4)From physicists' viewpoint, what will a finite entropy system looks like?

REPLY [5 votes]: A sufficient condition would be that the tails of $f$ decay as $O(x^{-1-\epsilon})$, which is most distributions you might encounter over $\mathbb{R}$.
That being said, the differential entropy of a continuous pdf isn't really a meaningful physical, or information theoretical, quantity. The equation isn't homogeneous: changing the units in which you measure $x$ changes the differential entropy (dimensional analysis can lead to a surprising amount of mathematical insights, even outside of physics).
What does make sense is the KL-divergence of one distribution with respect to another. For a pdf with a compact support, you're implicitly looking at the divergence with respect to the uniform distribution. However, there is no "uniform" distribution over the real line and thus the concept isn't meaningful.