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Let $D$ be an open connected subset in $\mathbb{R}^n.$ Assume that the non-trivial/non-zero function $f \equiv f(x,y): D \times (0, \infty) \to \mathbb{R} $ is real analytic in both $x$ and $y.$ Denote by $f_y(x):=f(x,y), f_x(y)=f(x,y).$ Assume that $f_x(.), f_y(.)$ are not identically zero. I'm trying to check, if possible that the set $S:= \{y \in (0, \infty): f(x,y) = 0 \text{ for at least one $x \in D$ } \}$ has measure zero or not.

To solve the problem, I've written:

$$S= \bigcup_{x \in D}\{y \in (0, \infty): f(x,y) = 0 \}$$

Next we take the complement of $S,$ denoted by $S':$

$$S'= \bigcap_{x \in D}\{y \in (0, \infty): f(x,y)= f_y(x) \ne 0 \}$$

Next, we note that since for each fixed $y, f_y(x)$ is a continuous function of $x,$ therefore if $f_y(x) \ne 0,$ then $f_y(z) \ne 0 \forall z$ sufficiently near $x.$ Therefore, the above intersection $\bigcap_{x \in D}$ is essentially a countable intersection where $x$ have rational coordinates (this is where I'm hesitating a bit, and please check if this rationale is correct!). Therefore we have:

$$S'= \bigcap_{x \in D \cap \mathbb{Q}^n}\{y \in (0, \infty): f(x,y)=f_y(x) \ne 0 \}$$.

So taking complement again:

$$S= \bigcup_{x \in D \cap \mathbb{Q}^n}\{y \in (0, \infty): f(x,y)=f_x(y) = 0 \}$$.

We already know that the zero set of a real analytic function (of one or several variables) is of zero Lebesgue measure in the corresponding dimension, c.f. Zeros of analytic function of several real variables, https://arxiv.org/abs/1512.07276. So each $\{y \in (0, \infty): f(x,y)=f_x(y) = 0 \}$ above has measure zero in $\mathbb{R}$, since $f_x(y)$ is real analytic in $y\in \mathbb{R}.$

But this implies that $S$ is a countable union $x \in D \cap \mathbb{Q}^n$ of measure zero subsets $\{y \in (0, \infty): f_x(y) = 0 \}$ of $\mathbb{R}.$ So we conclude that $S$ has measure zero in $\mathbb{R}.$

Is my argument correct?

Mathguest
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    Are you claiming that $S$ always has measure zero? What about $f: \mathbb{R}\times(0, \infty)\rightarrow \mathbb{R}$, $(x,y) \mapsto x-y$? Then given $y$, letting $x=y$ shows $y \in S$, so $S = (0, \infty)$. Or am I missing something? – Jair Taylor Jan 30 '23 at 23:27
  • @JairTaylor Thanks - I guess I must have made a mistake somewhere in my argument! I don't think you're missing something... – Mathguest Jan 30 '23 at 23:46

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It is not true. Let $f: \mathbb{R} \times (0, \infty) \rightarrow \mathbb{R}$, $(x,y) \mapsto x-y$. Then given $y$, letting $x=y$ shows $y \in S$, so $S = (0, \infty)$ does not have measure zero. This example also shows that

$$\bigcap_{x \in D}\{y \in (0, \infty): f(x,y)= f_y(x) \ne 0 \} \neq \bigcap_{x \in D \cap \mathbb{Q}^n}\{y \in (0, \infty): f(x,y)=f_y(x) \ne 0 \}.$$

For $x = \sqrt{2}$, we have $x-y \neq 0$ for all $x \in \mathbb{Q}$ but we cannot conclude $x-y \neq 0$ for all $x \in \mathbb{R}$. (Note though that from $f(x,y) = 0$ for all $x \in D \cap \mathbb{Q}^n$ we could conclude that $f(x,y) = 0$ for all $x \in D$, by continuity; perhaps that is the confusion.)

Jair Taylor
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