# Inequality involving Jensen (Rudin's exercise)

Exercise (Rudin, R&CA, no. 3.25). Suppose $\mu$ is a positive measure on the space $X$ and let $f \colon X \to (0,+\infty)$ be such that $\int_X f \, d\mu=1$.

Then for every $E \subset X$ with $0<\mu(E)<\infty$ we have $$\int_E \log f \, d\mu \le \mu(E) \log \frac{1}{\mu(E)}.$$

This is not homework, it is self-studying. I think I should use Jensen's inequality, but I cannot get it.

I thought considering the positive probability measure $\nu$ given by $\nu:=f\mu$ but I do not see which are the right functions to play Jensen's inequality with. Since $\log$ is concave, I would have $$\int_E f \log f \, d\mu \le \log \int_E f^2 \, d\mu$$ but this is not helpful. Than I can try $\log \frac{1}{x}$ which is convex and I would have $$\int_E f \log\left( \frac{1}{f}\right) \, d\mu \ge \log \int_E 1\, d\mu = \log \mu(E)$$ which looks nicer since it gives $$\int_E f \log f \, d\mu \le \log\left(\frac{1}{\mu(E)}\right)$$ but now I do not know how to handle the LHS.

$$\frac{1}{\mu(E)} \int_E \log f\,d\mu \leqslant \log \frac{1}{\mu(E)}.$$
$$\exp \left(\frac{1}{\mu(E)}\int_E \log f\,d\mu\right) \leqslant \frac{1}{\mu(E)}\int_E f\,d\mu.$$
• Now we have to take $\log$ of both sides and from $\int_E f d\mu \le 1$ we conclude, right? Thank you very much for your help. – Romeo Jul 23 '14 at 15:27