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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.

Any hint, please? Thanks in advance.

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Write the inequality as

$$\frac{1}{\mu(E)} \int_E \log f\,d\mu \leqslant \log \frac{1}{\mu(E)}.$$

Jensen's inequality gives you

$$\exp \left(\frac{1}{\mu(E)}\int_E \log f\,d\mu\right) \leqslant \frac{1}{\mu(E)}\int_E f\,d\mu.$$

The remaining part should be clear.

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  • $\begingroup$ 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. $\endgroup$ – Romeo Jul 23 '14 at 15:27
  • $\begingroup$ Yes, exactly so. $\endgroup$ – Daniel Fischer Jul 23 '14 at 15:28

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