Is Boltzmann entropy well-defined for arbitrary probability density function?

Question

$\newcommand{\bR}{\mathbb{R}}\newcommand{\diff}{\mathop{}\!\mathrm{d}}$ We define a continuous function $\varphi : \bR_+ \to \bR$ by $$ \varphi (s) := \begin{cases} 0 &\text{if} \quad s =0 , \\ s \ln s &\text{if} \quad s > 0. \end{cases} $$

Let $D$ be the set of probability density functions on $\bR^d$, i.e., $f \in D$ if and only if $f : \bR^d \to \bR_+$ is measurable with $\int f (x) \diff x =1$. Let $D_2$ be the subset of $D$ that contains those $f$ such that $\int |x|^2 f (x) \diff x$ is finite. For $s \in \bR$, we define $s^+ := \max \{s, 0 \}$ and $s^- := \max \{-s, 0 \}$.

By Proposition 15.6 in the book Lectures on Optimal Transport by Ambrosio/Brué/Semola, $(\varphi \circ f)^-$ is integrable (w.r.t. Lebesgue measure) for every $f \in D_2$. As such, Boltzmann entropy is well-defined (in extended real number line) for every $f \in D_2$.

I would like to ask if any of the below statements is true:

1. $(\varphi \circ f)^+$ is integrable for every $f \in D$.
2. $(\varphi \circ f)^-$ is integrable for every $f \in D$.

Thank you for your elaboration.

Answer 1

Counterexamples, with $d=1$:

$$f(x)=\frac{\ln2}{x\ln^2 x}\,1(0<x<1/2)$$ for all real $x$ -- for your statement 1;

$$f(x)=\frac{\ln2}{x\ln^2 x}\,1(x>2)$$ for all real $x$ -- for your statement 2.