Identity theorem for $\mathbb{R}^n$

Question

This question is follow up to an interesting question I found here.

The results of this question states the following:

If $U$ is a domain, and $f,g$ are two real-analytic functions defined on $U$, and if $V\subset U$ is a nonempty open set with $f\lvert_V \equiv g\lvert_V$, then $f \equiv g$. If the domain is one-dimensional (an interval in $\mathbb{R}$), then it suffices that $f\lvert_M \equiv g\lvert_M$ for some $M\subset U$ that has an accumulation point in $U$.

I have a few question about this theorem:

1. I was looking for a reference for this. However, in the previously pointed out reference A Primer of Real Analytic Functions by Krantz and Parks, I was not able to locate this theorem. I would appriciate if someone could point me to proper reference of this theorem.
2. My main question, is about the second part of the theorem. Specifically, I am interested in why there is such a difference going from $\mathbb{R}$ to $\mathbb{R}^n$. That is in one-dimension we can assume that $M$ is just a set with an accumulation point, but in $\mathbb{R}^n$ we have to assume that $M$ is an open set. I would really like see a counter example that demonstraes that assuming that $M$ is a set with accumulation points is not sufficient in $\mathbb{R}^n$.
3. I would really like for you to speculate or suggest an extra assumptions on functions $f$ and $g$ such that it suffices to consider $M$ to be only a set with an accumulation point.

Answer 1

For 2 and 3: the functions $0$ and $x$ are a counterexample if all you want is an accumulation, and since a general analytic function will have an $n-1$ dimensional $0$ set, the best you can do is what Mark van Leeuwen suggests here.

Answer 2

$M$ being open is not a necessary condition for $M$ to be a uniqueness set in higher dimensions. For example, if $f,g$ are real analytic on $\mathbb R^2$ and $f = g$ on $\cup_{k=1}^\infty \{(x,y):x=1/k\},$ then $f\equiv g.$ Also any set of positive measure, in any dimension, will give the result.