# A reduction problem from $\mathbb{R}^2$ to $\mathbb{R}$

Let $$f,g \in L^1_\text{loc}(\mathbb{R})$$, with $$g \geq 0$$, and such that for almost every $$(x,y) \in \mathbb{R}^2$$, at least one of the following equations is true : \begin{align*} f(x) + f(y) + g(x) + g(y) & = 0 \tag{E_1} \\ f(x) + f(y) + g(x) - g(y) & = 0 \tag{E_1} \\ f(x) + f(y) - g(x) + g(y) & = 0 \tag{E_1} \\ f(x) + f(y) - g(x) - g(y) & = 0, \tag{E_1} \end{align*} we call $$S_i \subset \mathbb{R}^2$$ the subset on which equation $$E_i$$ holds, so $$S_1 \cup S_2 \cup S_3 \cup S_4 = \mathbb{R}^2$$. How to prove that almost everywhere in $$\mathbb{R}$$, one of the following equations is true : \begin{align*} f- g = 0, \textit{ or } f = 0, \textit{ or } f+g = 0 \textit{ ?} \end{align*}

It is true when $$f$$ and $$g$$ are continuous : from the four equations we can take the product \begin{align*} (f(x) + f(y) + g(x) + g(y))(f(x) + f(y) + g(x) - g(y))\\ \times (f(x) + f(y) - g(x) + g(y))(f(x) + f(y) - g(x) - g(y)) = 0 \end{align*} a.e. in $$\mathbb{R}^2$$, and we let $$x$$ go to $$y$$ to get \begin{align*} 8f(x)^2(f(x) + g(x))(f(x) - g(x)) = 0, \end{align*} therefore we can conclude. But when $$f$$ and $$g$$ are just $$L^1$$ I don't manage to finish. I tried to do it by contradiction. Let \begin{align*} X := \{f-g \neq 0, f \neq 0, f+g \neq 0 \} \subset \mathbb{R}, \end{align*} and assume that it has positive measure. If $$X_+ := \{f > g \} \cap X$$ has positive measure, then there is some $$i$$ such that $$(X_+)^2 \cap S_i$$ has positive measure, but it's not possible since \begin{align*} f(x) + f(y) \pm g(x) \pm g(y) > 0 \end{align*} on this subset. So we can conclude that on $$X$$, we have essentially $$\{-g and without loss of generality, we can assume that $$\{0 has positive measure. Now from $$E_2$$ we can easily then show that $$g(x) < g(y)$$ a.e. on $$Y^2 \cap S_2$$. I'm here and I don't know how to continue...

• You may take sequences of continuous functions $(f_k)_k$ and $(g_k)_k$ converging a.e. to $f$ resp. to $g$, so taking the limit you also get $(f-g)(f+g)f=0$ a.e. – Pietro Majer May 22 at 18:26
• @PietroMajer: But why would $f_k$ and $g_k$ satisfy one of the four equations? – Mateusz Kwaśnicki May 22 at 20:54
• Don't you mean $\bigcup S_i$ is the complement of a measure-$0$ set (rather than necessarily all of $\mathbb R^2$)? – LSpice May 22 at 21:13
• @Mateusz Kwaśnicki, right, this is going to become too technical. Then say by Lusin theorem there is a sequence of compact sets $K_j$ whose union is $\mathbb{R}^2$ up to a null set, and both f and g restricted to $K_j$ are continuous, so one concludes again. – Pietro Majer May 22 at 21:15
• @Johanna: I just posted an answer, but I find this question so much weird! May I ask how did this problem arise? – Mateusz Kwaśnicki May 22 at 22:43

There must be a smarter way, but here is the best I could find.

Part 1.

Let $$A$$ be the set of those pairs $$(x, y)$$ for which (at least) one of the four equations $$f(x) + f(y) \pm g(x) \pm g(y) = 0$$ is satisfied (whenever we write $$\pm$$, we allow an arbitrary sign).

By Fubini's theorem, there is a set $$B$$ of full Lebesgue measure such that for every $$x \in B$$, we have $$(x, y) \in A$$ for almost all $$y$$. For $$x \in B$$, let $$B_x$$ be the set those $$y$$ such that $$(x, y) \in A$$. Thus, $$B_x$$ has full Lebesgue measure.

Fix any $$x \in B$$ and $$y \in B \cap B_x$$, and suppose that $$z \in B_x \cap B_y$$, so that $$(x, y)$$, $$(x, z)$$ and $$(y, z)$$ all lie in $$A$$. Since $$\begin{gathered} f(x) + f(y) \pm g(x) \pm g(y) = 0 \\ f(x) + f(z) \pm g(x) \pm g(z) = 0 \\ f(y) + f(z) \pm g(y) \pm g(z) = 0 , \end{gathered}$$ we find that $$2 f(z) = \pm g(x) \pm g(x) \pm g(y) \pm g(y) \pm g(z) \pm g(z) .$$ Note that $$\pm a \pm a$$ is equal to $$-2 a$$, $$0$$ or $$2 a$$. It follows that there is a finite set (with at most 9 elements) $$F = \{\pm g(x) \pm g(x) \pm g(y) \pm g(y)\}$$ such that $$f(z) \in \{-g(z), 0, g(z)\} + F$$ for almost all $$z$$.

Part 2.

We fix a number $$c \ne 0$$ (we think that $$c$$ belongs to the set $$F$$ constructed above, but we will not need that).

(a) Suppose that there is a set $$Z$$ of positive Lebesgue measure such that $$f(z) = g(z) + c , \text{ but } f(z) \notin \{-g(z), 0, g(z)\}$$ for all $$z \in Z$$. That is, $$f(z) = g(z) + c , \text{ and } g(z) \notin \{0, -c/2, -c\}$$ for all $$z \in Z$$. For almost all pairs $$(p, q)$$ such that $$p, q \in Z$$ we then have $$g(p) + g(q) + 2 c = f(p) + f(q) = \pm g(p) \pm g(q) ,$$ that is, $$g(p) + g(q) \pm g(p) \pm g(q) = -2 c .$$ Since $$g(p), g(q) \ne -c$$, we necessarily have $$g(p) + g(q) = -c$$ for almost all pairs $$(p, q)$$ such that $$p, q \in Z$$. This is clearly not possible: as in part 1, we find $$p, q, r \in Z$$ such that the above equality holds for $$(p, q)$$, $$(p, r)$$ and $$(q, r)$$, and solving these equalities we get $$g(p) = g(q) = g(r) = -c/2$$, a contradiction.

(b) In a similar way we prove that the set $$Z$$ of those $$z$$ for which $$f(z) = -g(z) + c , \text{ but } f(z) \notin \{-g(z), 0, g(z)\}$$ has Lebesgue measure zero.

(c) Finally, the set $$Z$$ containing all $$z$$ such that $$f(z) = c , \text{ but } f(z) \notin \{-g(z), 0, g(z)\}$$ also has Lebesgue measure zero, but here the argument is slightly different. Indeed: $$Z$$ contains all $$z$$ such that $$f(z) = c , \text{ and } g(z) \notin \{c, -c\} .$$ For almost all pairs $$(p, q)$$ such that $$p, q, \in Z$$ we have $$2c = f(p) + f(q) = \pm g(p) \pm g(q) ,$$ or, equivalently, $$g(p) = \pm 2 c \pm g(q) .$$ Now we find $$q \in Z$$ such that the above identity holds for almost all $$p \in Z$$. It follows that almost everywhere on $$Z$$, $$g$$ only takes one of the four values $$\pm c \pm g(q)$$. In particular, there is a subset $$Y$$ of $$Z$$ with positive Lebesgue measure on which $$g$$ is constant, and since $$2 c = \pm g(p) \pm g(q)$$ for almost all pairs $$(p, q)$$ such that $$p, q \in Y$$, we conclude that $$g(p) = \pm c$$ almost everywhere on $$Y$$, a contradiction.

Part 3.

By the first part, we have $$f(z) \in \{-g(z), 0, g(z)\} + F$$ for almost all $$z$$, where $$F$$ is some finite set. On the other hand, by the second part, for every $$c \in F$$, $$c \ne 0$$, we have $$f(z) \notin \{-g(z) + c, c, g(z) + c\} \setminus \{-g(z), 0, g(z)\}$$ for almost all $$z$$. It follows that $$f(z) \notin (\{-g(z), 0, g(z)\} + F) \setminus \{-g(z), 0, g(z)\}$$ for almost all $$z$$, and consequently $$f(z) \in \{-g(z), 0, g(z)\}$$ for almost every $$z$$, as desired.

• @Johanna: You're welcome! Fedor Petrov's solution seems to by much smarter, though. – Mateusz Kwaśnicki May 23 at 10:03

Choose large $$N$$. By Luzin theorem choose a subset $$A\subset [-N, N]$$ of measure $$\mu(A)\geqslant 2N-1/N$$ such that $$f, g$$ are continuous on $$A$$. Remove from $$A$$ the points which belong to rational intervals $$\Delta$$ such that $$\mu(\Delta\cap A)=0$$. The measure of $$A$$ does not change after this step, and now any neighborhood $$\delta\ni a$$ of any point $$a\in A$$ satisfies $$\mu(\delta\cap A)>0$$. Now you may substitute $$x=y$$ on $$A$$: if $$H(x, x) \ne 0$$ for $$x\in A$$, where $$H$$ is the product of the four functions as in OP, then by continuity the inequality $$H(z, y) \ne 0$$ holds for $$z, y$$ close enough to $$x$$, but the set of such pairs has positive measure. Therefore the equality $$f(x)(f(x)+g(x))(f(x)-g(x))=0$$ holds for all points in $$[-N,N]$$ but a set of measure at most $$1/N$$. Tending $$N$$ to infinity we conclude that it holds a.e.