If $\Omega\subset\mathbb C^n$ is an open subset of $\mathbb C^n$ and $\varphi$ is a plurisubharmonic function in $\Omega$, for a point $x\in \Omega$, we define the complex singularity exponent of $\varphi$ at $x$ as following: $$c_x(\varphi):=\sup\{c\geq 0\: e^{-2c\varphi} \text{ is } L^1 \text{ on a neighborhood of }x\}.$$
Demailly said in his paper, semi-continuity of complex singularity exponents and Kahler-Einstein metrics on Fano orbifolds, that we have another definition for this: $$c_x(\varphi)=\sup\{c\geq 0: r^{-2c}\mu_U(\{\varphi<\log r\}) \text{ is bounded as } r\to 0, \text{ for some } U\ni x\},$$ where $\mu_U$ is the measure de Lebesgue on $U$.
To see the equivalence of these two definitions, we have the following inequality: $$r^{-2c}\mu_U(\{\varphi<\log r\})\leq \int_U e^{-2c\varphi}\leq \mu_U(U)+\int_0^1 2c r^{-2c}\mu_U(\{\varphi<\log r\})\frac{dr}{r}.$$
My question: How to show the equivalence of the definitions par this inequality? I thought that $1/r$ is not integrable on $[0,1]$. Therefore, even $r^{-2c}\mu_U(\{\varphi<\log r\})$ is bounded as $r\to 0$ for some $U\ni x$, we can not obtain that the right hand of the inequality is finite.
Maybe I made some stupid mistakes here. Any comments are welcome. Thanks.