Riemann extension theorem for dimension greater than 1 case This is a theorem in P.Griffiths Principles of Algebraic Geometry. I do not get a condition and the relation between two functions. $C^n$ below is the complex n-dimensional space.
Theorem: Suppose $f(z,w)$ is holomorphic in a disc $\Delta\subset C^n$ and $g(z,w)$ is holomorphic in $\bar{\Delta}-\{f=0\}$ and bounded. Then g extends to a holomorphic function on $\Delta$. 
Here are my questions.


*

*What is the exact relationship between f and g? It seems that they are only related by the isolated singularities of f. 

*Why g needs to be defined on the $\bar{\Delta}-\{f=0\}$ instead of $\Delta-\{f=0\}$?

*Can one give a more intuitive explanation of this theorem as this seems more relaxed version of one-variable riemann extension theorem which simply replace g with f?
 A: *

*There is no relation between $f$ and $g$. The theorem just says that if $g$ is holomorphic and bounded on the complement of a zero set for holomorphic functions, then $g$ extends to a holomorphic function. 
The corresponding thing in one variable would be to say that if $g$ is holomorphic and bounded on the complement of a discrete set, then it extends (so just a variation of the usual formulation of Riemann's theorem on removable singularities).

*This isn't really necessary. It's enough to assume that $g$ is holomorphic on $\Delta^2 \setminus \{ f=0 \}$ and locally bounded near every point in $\{ f=0 \}$. See for example Theorem 4.7.2 in Wiegerinck's notes on several complex variables

*This is hopefully explained by 1. The theorem looks a little stronger like this: zero sets of holomorphic functions are removable for (locally) bounded holomorphic functions. Of course there are other very strong theorems about removable singularities in several complex variables. For example we have Hartogs' theorem: If $K$ is compact in $\Omega$ and $\Omega \setminus K$ is connected, then every function holomorphic on $\Omega \setminus K$ (bounded or not) extends to $\Omega$. Or another result (also due to Hartogs I think) in the same spirit as your question: If $V$ is a subvariety of $\Omega$ of codimension at least $2$, again every function holomorphic on $\Omega \setminus V$, bounded or not, extends across $V$.
