Bishop's theorem on analytic sets

Remarks

No condition on the dimension of $E$ is needed; it is replaced by the condition that $A$ has locally finite $2k$-dimensional volume near $E$. The idea of the proof is to use the volume bound to control generic linear projections to $\C^k$. Near each point of $E$, the closure $\overline{A}$ then projects properly, with a bounded number of sheets. The symmetric functions of the sheets are bounded holomorphic functions outside a thin set, and they extend. Example: let $E = \set{z_1 = 0}$ in $\C^2$ and $A = \set{z_2 = 0,\ z_1 \ne 0}$; the area of $A$ in each ball is finite, and indeed $\overline{A}$ is the $z_1$-axis. The volume hypothesis is needed: for $A = \set{(z_1, e^{1/z_1}) : z_1 \ne 0}$ and the same $E$, the closure $A \cup E$ is not analytic, so by the theorem $A$ has infinite area near every point of $E$. The theorem also holds with $X$ a complex manifold, since the statement is local.

Going deeper

For a student

Bishop's theorem strengthens the Remmert–Stein theorem, in which the exceptional set must have smaller dimension than $A$. Volume bounds of this kind come naturally from Wirtinger's inequality and Stokes' theorem, which makes the theorem useful for limits of analytic cycles. It is a main step in proving that limits of complex analytic cycles with bounded volume are analytic.