Big Rudin Exercise 3.26 - Which integral is larger This is exercise 3.26 in Rudin's Real & Complex Analysis:

If $f$ is a positive measurable function on $[0,1]$, which is larger,
  $$\int_0^1 f(x) \log f(x) \, dx$$
  or
  $$\int_0^1 f(s) \, ds \int_0^1 \log f(t) \, dt$$

I tried a bunch of functions and always got the first to be larger, which suggests that Hölder's inequality won't help here (at least not a direct application). I couldn't find an example that made the second larger. I'm stuck otherwise.
(This is self-study, not homework)
Clarification: The integral here is the Lebesgue integral. The only answer so far is only applicable to Riemann integrable functions.
 A: $$\int_0^1f(t)d(t)=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)$$ So in order to prove the inequality $$
\int_0^1 f(x) \log f(x) dx \geq \int_0^1f(s)ds \int_0^1\log f(t)dt $$
it is adequate to show  $$
\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right) \log f\left(\frac{i}{n}\right) \geq 
\frac{1}{n}\sum_{j=1}^n f\left(\frac{j}{n}\right) \cdot \frac{1}{n}\sum_{k=1}^n\log f\left(\frac{k}{n}\right) $$
Since $\log f(x)$ increases as $f(x)$ increases, we can apply Chebychev's inequality to give  $$
\sum_{j=1}^n f\left(\frac{j}{n}\right) \cdot \sum_{k=1}^n\log f\left(\frac{k}{n}\right) \leq n \sum_{i=1}^n f\left(\frac{i}{n}\right) \log f\left(\frac{i}{n}\right) $$
from which the required result follows immediately.
A: The function $x\mapsto x\log x$ is convex on $(0,\infty)$, as its second derivative is positive. Thus by Jensen's inequality,
$$\int_0^1 f(t)\log f(t) dt \geq \int_0^1 f(t) dt \log\left( \int_0^1 f(t) dt \right) .$$
The function $x\mapsto \log x$ is concave, so another application of Jensen's inequality yields
$$\log\left( \int_0^1 f(t) dt \right) \geq \int_0^1\log f(t) dt .$$
Combining these two inequalities proves the result. 
