[proofplan]
The proof is an application of the Skoda Integrability Theorem for plurisubharmonic weights, which gives the sharp relationship between the [Lelong number](/page/Lelong%20Number) of a plurisubharmonic function and local integrability of its exponential. The first assertion follows directly from the low-Lelong-number half of Skoda's theorem. For the second assertion, we argue by contradiction: a holomorphic germ not vanishing at $p$ is bounded below near $p$, so its membership in the [multiplier ideal](/page/Multiplier%20Ideal) would force $e^{-\phi}$ itself to be locally integrable, contradicting the high-Lelong-number half of Skoda's theorem.
[/proofplan]
[step:Apply Skoda's theorem to obtain local integrability when the Lelong number is below $2$]
Assume $\nu(\phi,p)<2$. Since $\phi: U \to [-\infty,\infty)$ is plurisubharmonic on the open neighbourhood $U$ of $p$, the Skoda Integrability Theorem applies: if a plurisubharmonic function has Lelong number strictly less than $2$ at a point, then its exponential weight is locally integrable near that point. Thus there exists an open neighbourhood $V \subset U$ of $p$ such that
\begin{align*}
\int_V e^{-\phi(z)} \, d\mathcal L^{2n}(z) < \infty.
\end{align*}
Here $\mathcal L^{2n}$ denotes Lebesgue measure on $\mathbb C^n \cong \mathbb R^{2n}$.
The constant germ $1 \in \mathcal O_{\mathbb C^n,p}$ satisfies
\begin{align*}
|1|^2 e^{-\phi} = e^{-\phi}.
\end{align*}
Hence $|1|^2e^{-\phi}$ is locally integrable near $p$, and therefore $1 \in \mathcal I(\phi)_p$.
[guided]
We use the following form of the Skoda Integrability Theorem: for a plurisubharmonic function $\psi$ near a point $q \in \mathbb C^n$, if $\nu(\psi,q)<2$, then $e^{-\psi}$ is locally integrable near $q$; if $\nu(\psi,q)\ge 2n$, then $e^{-\psi}$ is not locally integrable near $q$. This theorem is the external input of the proof.
Apply this theorem with $\psi=\phi$ and $q=p$. The hypotheses are exactly satisfied: $\phi: U \to [-\infty,\infty)$ is plurisubharmonic on the open neighbourhood $U$ of $p$, and the present assumption gives $\nu(\phi,p)<2$. Therefore there exists an open neighbourhood $V \subset U$ of $p$ such that
\begin{align*}
\int_V e^{-\phi(z)} \, d\mathcal L^{2n}(z) < \infty.
\end{align*}
By definition, the multiplier ideal stalk $\mathcal I(\phi)_p$ consists of holomorphic germs $f \in \mathcal O_{\mathbb C^n,p}$ for which $|f|^2e^{-\phi}$ is locally integrable near $p$. For the constant germ $1$, the corresponding density is
\begin{align*}
|1|^2 e^{-\phi} = e^{-\phi}.
\end{align*}
The local integrability just obtained is precisely the membership condition for $1$. Hence $1 \in \mathcal I(\phi)_p$.
[/guided]
[/step]
[step:Show that a nonvanishing holomorphic germ is bounded below near $p$]
Let $f \in \mathcal O_{\mathbb C^n,p}$ be a holomorphic germ with $f(p)\ne 0$. Choose a representative, still denoted
\begin{align*}
f: W \to \mathbb C,
\end{align*}
where $W \subset U$ is an open neighbourhood of $p$ and $f$ is holomorphic on $W$.
Since $f$ is continuous at $p$ and $|f(p)|>0$, there exists an open neighbourhood $W_0 \subset W$ of $p$ such that
\begin{align*}
|f(z)| \ge \frac{|f(p)|}{2}
\end{align*}
for every $z \in W_0$.
[guided]
Let $f \in \mathcal O_{\mathbb C^n,p}$ be a germ that does not vanish at $p$. To speak about pointwise bounds, choose a representative
\begin{align*}
f: W \to \mathbb C
\end{align*}
on an open neighbourhood $W \subset U$ of $p$. The representative is holomorphic, hence continuous.
Because $f(p)\ne 0$, the number $|f(p)|$ is strictly positive. Continuity of the map $z \mapsto |f(z)|$ at $p$ implies that there exists an open neighbourhood $W_0 \subset W$ of $p$ such that
\begin{align*}
\bigl||f(z)|-|f(p)|\bigr| < \frac{|f(p)|}{2}
\end{align*}
for every $z \in W_0$. Rearranging gives
\begin{align*}
|f(z)| > |f(p)|-\frac{|f(p)|}{2}
= \frac{|f(p)|}{2}
\end{align*}
for every $z \in W_0$. This lower bound is the bridge between integrability of $|f|^2e^{-\phi}$ and integrability of $e^{-\phi}$ itself.
[/guided]
[/step]
[step:Use Skoda's nonintegrability threshold to force vanishing at $p$]
Assume now that $\nu(\phi,p)\ge 2n$. Let $f \in \mathcal I(\phi)_p$. Suppose, for contradiction, that $f(p)\ne0$. By the previous step, after choosing a representative
\begin{align*}
f: W \to \mathbb C
\end{align*}
on a neighbourhood $W$ of $p$, there is an open neighbourhood $W_0 \subset W$ of $p$ such that
\begin{align*}
|f(z)|^2 \ge \frac{|f(p)|^2}{4}
\end{align*}
for every $z \in W_0$.
Since $f \in \mathcal I(\phi)_p$, there exists an open neighbourhood $V \subset W_0$ of $p$ such that
\begin{align*}
\int_V |f(z)|^2 e^{-\phi(z)} \, d\mathcal L^{2n}(z) < \infty.
\end{align*}
The lower bound for $|f|$ gives, pointwise on $V$,
\begin{align*}
e^{-\phi(z)}
\le
\frac{4}{|f(p)|^2}|f(z)|^2 e^{-\phi(z)}.
\end{align*}
Therefore
\begin{align*}
\int_V e^{-\phi(z)} \, d\mathcal L^{2n}(z)
\le
\frac{4}{|f(p)|^2}
\int_V |f(z)|^2 e^{-\phi(z)} \, d\mathcal L^{2n}(z)
<\infty.
\end{align*}
Thus $e^{-\phi}$ is locally integrable near $p$.
This contradicts the nonintegrability half of the Skoda Integrability Theorem, which says that if $\nu(\phi,p)\ge 2n$, then $e^{-\phi}$ is not locally integrable near $p$. Hence $f(p)=0$. Since every $f \in \mathcal I(\phi)_p$ vanishes at $p$, we have
\begin{align*}
\mathcal I(\phi)_p \subset \mathfrak m_p.
\end{align*}
[guided]
Now assume $\nu(\phi,p)\ge 2n$, and let $f \in \mathcal I(\phi)_p$. We must prove that $f$ lies in the maximal ideal $\mathfrak m_p$, which means exactly that $f(p)=0$.
Argue by contradiction. Suppose $f(p)\ne0$. Choose a holomorphic representative
\begin{align*}
f: W \to \mathbb C
\end{align*}
on an open neighbourhood $W \subset U$ of $p$. From the preceding step, after shrinking to an open neighbourhood $W_0 \subset W$ of $p$, we have the pointwise lower bound
\begin{align*}
|f(z)| \ge \frac{|f(p)|}{2}
\end{align*}
for every $z \in W_0$. Squaring this inequality gives
\begin{align*}
|f(z)|^2 \ge \frac{|f(p)|^2}{4}.
\end{align*}
Because $f \in \mathcal I(\phi)_p$, the defining condition for the multiplier ideal gives an open neighbourhood $V \subset W_0$ of $p$ such that
\begin{align*}
\int_V |f(z)|^2 e^{-\phi(z)} \, d\mathcal L^{2n}(z) < \infty.
\end{align*}
The lower bound on $|f|^2$ lets us compare $e^{-\phi}$ with the integrable density $|f|^2e^{-\phi}$. For every $z \in V$,
\begin{align*}
\frac{|f(p)|^2}{4} e^{-\phi(z)}
\le
|f(z)|^2 e^{-\phi(z)}.
\end{align*}
Since $|f(p)|^2/4>0$, divide by this constant to obtain
\begin{align*}
e^{-\phi(z)}
\le
\frac{4}{|f(p)|^2}|f(z)|^2 e^{-\phi(z)}.
\end{align*}
Integrating this pointwise inequality over $V$ with respect to $\mathcal L^{2n}$ gives
\begin{align*}
\int_V e^{-\phi(z)} \, d\mathcal L^{2n}(z)
\le
\frac{4}{|f(p)|^2}
\int_V |f(z)|^2 e^{-\phi(z)} \, d\mathcal L^{2n}(z)
<\infty.
\end{align*}
Thus $e^{-\phi}$ is locally integrable near $p$.
This conclusion is incompatible with the nonintegrability half of the Skoda Integrability Theorem: if a plurisubharmonic function has Lelong number at least $2n$ at $p$, then its exponential weight is not locally integrable near $p$. The contradiction came from assuming $f(p)\ne0$, so $f(p)=0$.
Therefore every germ $f \in \mathcal I(\phi)_p$ vanishes at $p$. By definition of the maximal ideal,
\begin{align*}
\mathfrak m_p
=
\{g \in \mathcal O_{\mathbb C^n,p}: g(p)=0\},
\end{align*}
so $\mathcal I(\phi)_p \subset \mathfrak m_p$.
[/guided]
[/step]
