joint differential entropy $h(X,Y)$, when $Y=g(X)$ It's well known that if X & Y are discrete random variables X & Y (r.v.s), and $Y=g(X)$, then
$$H(X,Y)=H(X)+H(Y\mid X)=H(X),$$
where the last equality is due to $H(Y\mid X)=0.$  It also has a very intuitive interpretation - since Y is a deterministic function of X, the entropy/uncertainty of (X,Y) is the same as that of X, i.e. having Y on top of X doesn't increases uncertainty.
However, it's not very clear to me if above equalities & interpretation still hold or how to modify them, when X & Y are continuous r.v.'s?  In particular, how should $h(X,Y)$ and $h(Y|X)$ be defined in this case, $-\infty$?  
Thanks a lot!
 A: It is important to notice here that differential entropy has a wildly different interpretation from the "standard" entropy of discrete variables. In particular, I would avoid putting any uncertainty-related interpretations on it.
Now, on to your questions:

Do the  above equalities & interpretation still hold when X & Y are continuous r.v.'s?

As @syeh_106 points out in the comments section, the equality $h(X,Y) = h(X) + h(Y|X)$ holds perfectly fine. The equality $h(X) + h(Y|X) = h(X)$, however, doesn't. Here's a quick proof by reductio ad absurdum that $h(Y|X) \neq 0$:
We know that $I(X, Y) \geq 0$. At the same time, $I(X, Y) = h(X) - h(X|Y)$. So, if $h(Y|X)$ were $0$, we would have $h(X) \geq 0$. We know this is not the case (e.g. if X is uniformly distributed in an interval of length $L$, then $h(X) < 0$ if $L < 1$), therefore in general $h(Y|X) \neq 0$.

How should $h(X,Y)$ and $h(Y|X)$ be defined in this case, $-\infty$?

Indeed $h(X,Y) = -\infty$, although to be honest I'm not entirely sure about $h(Y|X)$ -- I think it's also $-\infty$, but it could well be the case that it's just not defined.
In any case, the fact remains that your intuition that "since $Y$ is a deterministic function of $X$, the entropy/uncertainty of $(X,Y)$ is the same as that of $X$," while correct for discrete variables, just doesn't apply to continuous variables.
