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 $\{-g<f<g \}$ and without loss of generality, we can assume that $\{0<f<g \} =: Y$ 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...