Lebesgue measure of the level set of sum of two nonnegative functions

Question

Let $f, g:\mathbb{R}^n\to \mathbb{R}$ be nonnegative functions such that $g$ is a strictly positive homogeneous function. As commented by Fedor Petrov below, one may not have that for any $\lambda>0$, $$\operatorname{Leb}\{x\in\mathbb{R}^n: f(x)+g(x)=\lambda\}=0. $$ If we assume that $f$ is also a strictly positive homogeneous function, possibly of different degree, does the above statement hold?

Note that the level sets of $g$ have zero Lebesgue measure as for any $\lambda>0$, $E(\lambda)=\lambda E(1)$ and $$E(\lambda)=\operatorname{Leb}\{x\in\mathbb{R}^n: g(x)\le \lambda\}.$$ Since $E(\lambda)$ is continuous in $\lambda$, we have $$\operatorname{Leb}\{x\in\mathbb R^n: g(x)=\lambda\}=0.$$

Edit: I have edited the question based on the comments below.

Answer 1

Let $\lambda\ne 0$ and both $f, g$ be non-negative measurable homogeneous functions of degrees $d_1,d_2$ respectively, and either $d_1d_2\ne 0$, or $d_1\ne 0$ and $f$ is strictly positive on the unit sphere. These restrictions may be further weaken as is seen from the below argument.

Then every ray $\{tx_0\colon t>0\}$ (for $x_0$ on the unit sphere) contains at most two solutions of the equation $f(x)+g(x)=\lambda$. Indeed, this reads as $f(x_0)t^{d_1}+g(x_0)t^{d_2}=\lambda$ which has at most two positive roots by Rolle theorem (the derivative of LHS is either identical 0,in which case there are no solutions, or has at most one positive root). Now the result follows from Fubini in spherical coordinates.