Removable singularities for $L^2$ bounded holomorphic functions Suppose $V$ is a  analytic variety of an  open subset $U\subset ℂ^n$. Suppose that  $f:U\setminus V\rightarrow C$ is holomorphic and  that $f$ is  $L^2$-bounded in $U$. Question: Is it true that there exists a unique holomorphic extension of $f$ to all of $U$?
 A: Every analytic variety is pluripolar, according to page 2 here. Pluripolar sets are removable for $L^2$ holomorphic functions, according to page 20 here. (These are not the best references, but they are what I could find without going through paywalls.)
A: See Ohsawa's book Analysis of Several Complex Variables, Theorem 1.13 and Proposition 1.14.
The proof relies on the fact that a $L^2$ function is holomorphic if and only if $\bar{\partial} f$ is $0$ in the sense of distribution. So we need to verify $\int f\bar{\partial}_i u=0$ for any smooth function $u$ with compact support. We can use a cut-off function supported in a $\varepsilon$-neighborhood of $V$ to separate the integration into two parts. Taking $\varepsilon\to 0$ will give us the desired identity. One property used when taking limits is that the volume growth of $\varepsilon$-neighborhood is of order $\varepsilon^2$. The conclusion can be applied to any subset with this property. 
Also there is a proof following the discussion for $n=1$. We only need to deal with the case $V=\{h=0\}$ with $h$ holomorphic. Near the nonsingular points, a holomorphic transformation with $h=0$ can reduce the problem to one-dimensional case. So we have extended $f$ to the smooth part of $V$. Notice that the singular part has codimension at least $2$. A final use of Hartog-type extension theorem would give us the extension of $f$ to $U$.
