real analysis - Show that if $f(x+y)=g(x)+g(y)$ a.e. then functions are linear

Let $f(x), g(x), h(x)$ be measureable functions such that

$$\begin{align} f(x+y)=g(x)+h(y), \end{align}$$
almost everywhere $(x,y) \in \mathbb{R}^2$.

Can we show that

$$\begin{align} g(x)&=ax+b_1,\\ h(x)&=ax+b_2, \end{align}$$
almost everywhere $(x,y) \in \mathbb{R}^2$ for some $a,b_1,b_2$.

My attempt:
Which I think is problematic.

Since, $f(x+y)=g(x)+h(y)$, we have that

$$\begin{align} f(x+0)=g(x)+h(0)\\ f(0+y)=g(0)+h(y) \end{align}$$

Now adding the two equations and use our identity we get

$$\begin{align} f(x+0)+f(y+0)=g(x)+h(y)+h(0)+g(0)= f(x+y)+f(0). \end{align}$$

So, we have that

$$\begin{align} f(x)+f(y)=f(x+y)+f(0) \end{align}$$
Next, difine a function $\phi(x)=f(x)-f(0)$ and we get

$$\begin{align} \phi(x)+\phi(y)=\phi(x+y). \end{align}$$

The above is know as Cauchy's functional equation and if $\phi(x)$ is measureable (which it is since $f$ is measurable) then it must be linear (i.e., $\phi(x)=ax$).

Going backward this shows that $g(x)$ and $h(x)$ are given by

$$\begin{align} g(x)&=ax+b_1,\\ h(x)&=ax+b_2, \end{align}$$

for some $a,b_1,b_2$.

Questions:

1) Is this proof correct?

2) Since, we are assuming that equation $f(x+y)=g(x)+h(y)$ holds almost everywhere and not everywhere. Is this problem? For example, in the very first step of the proof

$$\begin{align} f(x+0)=g(x)+h(0)\\ f(0+y)=g(0)+h(y) \end{align}$$

can we do this? Is there problem of choosing $(x,0)$ and $(0,y)$? I think there is a possibility that $(x,0)$ and $(0,y)$ might belong to a set of measure zero on which the equations don't hold.

3) Another, question very similar to 2). Since, $\phi(x)+\phi(y)=\phi(x+y)$ holds a.e. can we apply Cauchy equation?

Answer

For 2), you're right, this could be a problem. For example, if $f(x) = x$ for $x \neq 0$ and $f(0) = 1$, and $g = h = f$, then the functional equation holds whenever $x, y, x+y$ are all nonzero (so a.e.), but the two equations you get from setting $x = 0$ and $y = 0$ are never true. Since the set of $(x, y)$ such that the equation fails has measure zero, there are some fixed $a, b \in \mathbb{R}$ such that

$$f(x+a) = g(x) + h(a)$$
for a.e. $x$ and
$$f(b+y) = g(b) + h(y)$$

for a.e. $y$.

This allows you to write $f(x + a) + f(y + b) = f(x+y) + h(a) + g(b)$ for a.e. (x, y) (this only fails when either of the two previous equations fail or where the original functional equation fails). From here we can replace $x$ with $x+b$ and $y$ with $y+a$ to get

$$f(x+a+b) + f(y+a+b) = f(x+y+a+b) + h(a) + g(b)$$

for a.e. (x, y), so if we define $\phi(x) = f(x + a + b) - h(a) - g(b)$, then

$$\phi(x) + \phi(y) = \phi(x+y)$$

for a.e. (x, y).

Now for 3), De Bruijn has shown that we must have $\phi = \phi_0$ a.e., where $\phi_0$ is a solution to Cauchy's functional equation (for all $x, y$). But since $f$ is measurable, $\phi$ and $\phi_0$ must be measurable, and thus $\phi_0$ must be linear. From here it follows that $f, g, h$ are linear (or rather affine) almost everywhere.