How to compute $\lim_{\beta\to \infty} \beta \mu(f \geq \beta)$?

Let $(X, \Omega, \mu)$ be a measure space.

Given that $f(x) = \frac{1}{x(1-\log x)}$, on $[0,1]$, how can I compute $\lim_{\beta\to \infty} \beta \mu(f \geq \beta)$?

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I assume $\mu(f\geq \beta)$ is the Lebesgue measure of the set $\{x:f(x)\geq \beta\}$? –  Alex Becker Apr 4 '12 at 21:55
It's safe to assume that:) –  Jonjo Apr 4 '12 at 22:05

We claim that $f$ is strictly decreasing on $(0,1]$. Indeed $$f'(x)=\frac{\log(x)}{x^2(\log(x)-1)^2}$$ so for all $x\in(0,1]$ we have $f'(x)<0$. Hence $f$ is strictly decreasing on $(0,1]$. Also we notice $$\lim\limits_{x\to +0}f(x)=+\infty,\qquad f(1)=1$$ From this three facts we see that for all $\beta\geq 1$ there exist unique solution $x_\beta$ of the equation $f(x)=\beta$. Note that $$x_\beta=\frac{1}{\beta(1-\log(x_\beta))}$$ Since $f$ is strictly decreasing $\{x\in(0,1]:f(x)\geq \beta\}=(0,x_\beta]$, so $$\lim\limits_{\beta\to+\infty}\beta\mu(\{x\in(0,1]:f(x)\geq \beta\})= \lim\limits_{\beta\to+\infty}\beta\mu((0,x_\beta])= \lim\limits_{\beta\to+\infty}\beta x_\beta= \lim\limits_{\beta\to+\infty}\frac{1}{1-\log(x_\beta)}$$ Since $f$ is strictly decreasing and $\lim\limits_{x\to 0+}f(x)=+\infty$ we have $\lim\limits_{\beta\to+\infty}x_\beta=+0$, so $$\lim\limits_{\beta\to+\infty}\beta\mu(\{x\in(0,1]:f(x)\geq \beta\})= \lim\limits_{x_\beta\to+0}\frac{1}{1-\log(x_\beta)}=0$$
Good work (+1). A simpler way to show that $f$ is decreasing is to consider $1/f=x(1-\log(x))$. The derivative of $1/f$ is $-\log(x)$ which is positive on $(0,1)$. –  robjohn Apr 5 '12 at 2:15