Suppose $\mu$ is a probability measure (on a set, say $\mathbb{R}^n$), and $f\geq 0$ is a function that integrates to $1$ with respect to $\mu$. Then, set $S(f)$ to be the limit $$S(f)=\lim_{\epsilon\to 0^+}\int f\log(f+\epsilon)\ d\mu.$$
(a) Assuming that $S(f)$ can be infinite, show that $S(f)$ exists, and that $S(f)\geq 0$.
(b) Assuming that $h$ is a bounded real function, prove the following inequality: $$\int hf\ d\mu-S(f)\leq \log({\int e^h\ d\mu}).$$ I tried to solve the problem but I'm not sure if they are right. Here are the outlines of the solution:
a) We use the Dominated Convergence Theorem. As $ \epsilon \to 0^+ $, $ f \log(f + \epsilon) $ converges pointwise to $ f \log f $. Since $ f \geq 0 $ and integrates to 1, we can use the function $ h = f$ as the dominating function. $ f \log(f + \epsilon) $ is bounded above by $ f $ and $ \int f d\mu = 1 $, so $ f $ is integrable. Finally, by Dominated Convergence Theorem, $ S(f) $ exists.
b) We use the inequality $ e^x \geq x + 1 $, which is true for all real $ x $. We have: $$ e^h \geq h + 1 \implies f e^h \geq hf + f \implies \int f e^h\ d\mu \geq \int hf\ d\mu + \int f\ d\mu. $$ Since $ \int f\ d\mu = 1 $: $$ \int f e^h\ d\mu \geq \int hf\ d\mu + 1 \implies \log\left(\int f e^h\ d\mu\right) \geq \log\left(\int hf\ d\mu + 1\right) $$ From the definition of $ S(f) $, we have $ S(f) = \lim_{\epsilon \to 0^+} \int f \log(f + \epsilon)\ d\mu $. Note that $ \int hf\ d\mu $ is essentially the integral of $ f \log(f) $ with $ h$ acting as a modifier. Thus, $$ \int hf\ d\mu - S(f) \leq \log\left(\int e^h\ d\mu\right). $$