Example 6.9.4.2AA6A0B94DE7 On $\mathbb C^2$ consider the plurisubharmonic functions (proposition 5.10.3)
They have the same Lelong numbers at every point. Off the origin both functions are finite, so their Lelong numbers vanish there by proposition 6.7.1. At the origin, $\nu (\psi ,0)=1$ by proposition 6.7.9. For $|z_2|\leq 1$ we have $\varphi \leq \psi $, so $\nu (\varphi ,0)\geq 1$; conversely, the supremum of $\varphi $ over the ball $|z|<\mathrm e^{t/2}$ is at least $\sup _{|z_1|<\mathrm e^{t/2}}\log |z_1|^2=t$, so $\nu (\varphi ,0)\leq 1$ by (6.7.2). Nevertheless $\mathcal I(\frac32\psi )_0\neq \mathcal I(\frac32\varphi )_0$. Indeed, $\mathrm e^{-3\psi /2}=|z|^{-3}$ is integrable near $0\in \mathbb R^4$ by polar coordinates, since $\int _0^ rt^{-3}t^3\, \mathrm dt<\infty $; so $\mathcal I(\frac32\psi )_0=\mathcal O_{\mathbb C^2,0}$. On the other hand, on the set $A_\delta =\{ 0<|z_2|<\delta ,\ |z_1|<|z_2|^2\} $ we have $|z_1|^2+|z_2|^4<2|z_2|^4$, hence $\mathrm e^{-3\varphi /2}>2^{-3/2}|z_2|^{-6}$, and by Fubini's theorem
Since every neighborhood of $0$ contains $A_\delta $ for small $\delta >0$, the function $\mathrm e^{-3\varphi /2}$ is not integrable near $0$, and $1\notin \mathcal I(\frac32\varphi )_0$.
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