Zeros of analytic function of several real variables Suppose $f$ is a non-constant analytic function on $R^n$, what can we say about its zeros set $Z(f)=\{x\in R^n|f(x)=0\}$? Does $Z(f)$ have measure 0? Is it true that $Z(f)$ is union of manifolds of dimension less than $n$?
 A: Let $m_1,m_2$ be Lebesgue measure on $\mathbb {R},\mathbb {R}^2$ respectively. If $f$ is real-analytic on $\mathbb {R}$ and $m_1(Z(f))>0$ then $f\equiv 0.$ This follows from the identity principle: If $m_1(Z(f)) > 0,$ then $Z(f)$ is uncountable, hence has an accumulation point, hence $f\equiv 0.$
Claim: If $f$ is real-analytic on $\mathbb {R}^2$ and $m_2(Z(f)) > 0,$ then $f\equiv 0.$
Proof: If $m_2(Z(f)) > 0,$ then by Fubini there is a set $E\subset \mathbb {R}, m_1(E)>0,$ such $m_1(Z(f)\cap(\{x\}\times \mathbb {R}))> 0$ for all $x\in E.$ From the above we conclude that $f=0$ on each of those vertical lines (note that for each fixed $x,$ $y\to f(x,y)$ is real analytic on $\mathbb {R}).$ Since $m_1(E)>0,$ we have $E$ uncountable, hence $E$ has an accumulation point $a\in \mathbb {R}.$ Thus there is a sequence of distinct points $a_k\to a$ such that $f =0$ on the vertical lines based in $\{a_1,a_2,\dots \}\cup\{a\}.$
Now think about any horizontal line. Because $f$ is real analytic on this line, and $f = 0$ on a set with an accumulation point on that line, we have $f=0$ on that line. This was any horizontal line, hence $f\equiv 0$ on $\mathbb {R}^2.$
Assuming the result for $\mathbb {R}^{n-1},$ the above proof is easily adapted to give the result for $\mathbb {R}^n.$
