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 $\lim_{\beta\to \infty} \beta \mu(f \geq \beta)$?
 A: 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
$$
\lim\limits_{\beta\to+\infty}\beta\mu(\{x\in(0,1]:f(x)\geq \beta\})=
\lim\limits_{x_\beta\to+0}\frac{1}{1-\log(x_\beta)}=0
$$
