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I'm tring to prove the following statement:

Suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is continuously differentiable, and for any $(x_0,y_0)\in \mathbb{R}^2$, we have $$ \frac{\partial f}{\partial x}(x_0,y_0)+\frac{\partial f}{\partial y}(x_0,y_0)\neq 0 $$ then show that: $$ E=\{(x,y):f(x,y)=0\} $$ is a zero-measured set in $\mathbb{R}^2$.

It's easy to see that $0$ in the definition of $E$ can be replaced by any real number $c$. I have tried to integrate the two variable function $\frac{\partial f}{\partial x}(x_0,y_0)+\frac{\partial f}{\partial y}(x_0,y_0)$ and then applied the Fubini theorem but with no valuable findings. Can anybody give me some hints on proving this statement?

Robin
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1 Answers1

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The desired conclusion can be obtained under the weaker assumption that $$ f(x_0,y_0)=0 \; \Longrightarrow \; \frac{\partial f}{\partial x}(x_0,y_0)+\frac{\partial f}{\partial y}(x_0,y_0)\neq 0 \,. $$

For every real $b$, consider the function $g_b:{\mathbb R} \to {\mathbb R}$ defined by $g_b(t)=f(t,t+b)$. For every $t$, the chain rule and the hypothesis give $$g_b(t)=0 \; \Longrightarrow \; g_b'(t)=\frac{\partial f}{\partial x}(t,t+b)+\frac{\partial f}{\partial y}(t,t+b)\ne 0 \,,$$ so the zeros of $g_b$ are isolated, whence they form a countable set (at most). Thus the set $E$ of zeros of $f$ intersects every line of the form $\{y=x+b\}$ in a set of zero length, so Fubini's theorem implies that $E$ has zero area.

To formalize the last step, you may want to consider the (area preserving) rotation $R$ by $45$ degrees, observe that $R(E)$ intersects every vertical line in a set of zero length, and deduce that $R(E)$ (and hence also $E$) has zero area.

Yuval Peres
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