Let $X\colon \Omega\to\mathcal X$ and $Y\colon \Omega\to \mathcal Y$ be two random variables. In M.S. Pinkser's *Information and information stability of random variables and processes* mutual information is defined as:

$$I(X; Y) = \sup \sum_i P_{(X,Y)}(E_i) \log \frac{ P_{(X, Y)}(E_i)}{ P_{X\times Y}(E_i),} ,$$ where the supremum is taken over all finite partitions $\{E_1, \dotsc, E_n\}$ of $\mathcal X\times\mathcal Y$ and one uses the usual conventions $0\cdot \log 0/a = a\log a/0 = 0$ for $a\ge 0$.

In particular, if $P_{(X, Y)}$ is not absolutely continuous w.r.t. $P_{X\times Y} = P_X\otimes P_Y$, then $I(X; Y) = \infty$. In the other case, mutual information becomes the well-known formula involving the Radon–Nikodym derivative:

$$I(X; Y) = \int_{\mathcal X\times \mathcal Y} \log \left( \frac{ \mathrm{d}P_{X,Y} }{ \mathrm{d}(P_X\otimes P_Y) } \right)\, \mathrm{d}P_{X,Y}$$ and is finite.

One can prove that if $f\colon \mathcal X\to \mathcal X'$ and $g\colon \mathcal Y\to \mathcal Y'$ are measurable bijective mapping with measurable inverses, then $I(X; Y) = I(f(X); g(Y))$.

I wonder if this assumption can be weakened in the smooth case:

Assume that $\mathcal X = \mathbb R^m$ and $\mathcal Y = \mathbb R^n$ and $I(X; Y)$ is finite. Let $f\colon \mathbb R^m\to \mathbb R^{m+k}$ and $g\colon \mathbb R^n\to \mathbb R^{n+l}$ be

smooth embeddings.Is this true that $I(f(X); g(Y))$ is finite and that $I(f(X); g(Y)) = I(X; Y)$?

I know that for discrete random variables the situation is simpler and I tried to ask this question at MathSE, but it has not been answered there.

Any intuitions, counterexamples, or literature suggestions would be very much welcome.