If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.
Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Since Thus $f$ is analytic and the has zero Taylor expansion at some pointvanishes, we have hence $f\equiv 0$.
If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.
Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes on $Z$ too. And so on. Since $f$ is analytic and the Taylor expansion at some point vanishes, we have $f\equiv 0$.