> To be more precise: I have a function yielding
> only 0 or 1 depending on the membership to a given set.

I think the implicit assumption of the iso_surface is that
the set f(x) == threshold actually represents a 2d surface
and the numerical solver can creep up to that solution from
both sides using the 3d function gradient.

What you do seems more like a description of an "inside"
volume and you expect to render the surface of the volume.
I would not be surprised at unintuitive behavior here, I'm
more surprised you got a reasonable result at all :)

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