Main theorem (Section 2)

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Source. Section 2, printed p. 60 (PDF p. 2).

Statement. Let $\lambda$ be Lebesgue measure on $\mathbb R$, and let $\lambda^2$ denote two-dimensional Lebesgue measure on $\mathbb R^2$, equivalently the completion of the product measure $\lambda\times\lambda$. Suppose $f:\mathbb R\to\mathbb R$ and there is a Lebesgue-measurable set $N\subseteq\mathbb R^2$ with $\lambda^2(N)=0$ such that

$$f(x+y)=f(x)+f(y)$$

whenever $(x,y)\notin N$. Then there is a function $h:\mathbb R\to\mathbb R$ such that

$$h(x+y)=h(x)+h(y)$$

for every $x,y\in\mathbb R$, and $f(x)=h(x)$ for $\lambda$-almost every $x$.

Proof. For $x\in\mathbb R$, write

$$N_x=\{y\in\mathbb R:(x,y)\in N\}.$$

Fubini's theorem gives a null set $M\subseteq\mathbb R$ such that $N_x$ is null for every $x\notin M$.

Fix $x\in\mathbb R$. Both $M$ and $x-M$ are null, so their union cannot be all of $\mathbb R$. Choose $x_1$ outside that union. Thus $x_1\notin M$ and $x-x_1\notin M$. The corresponding vertical sections are null, so

$$f(x_1+y)-f(y)=f(x_1)$$

for almost every $y$, and

$$f(x-x_1+z)-f(z)=f(x-x_1)$$

for almost every $z$. Translation invariance of Lebesgue measure permits the substitution $z=x_1+y$ in the second almost-everywhere identity. Adding the resulting two identities gives

$$f(x+y)-f(y)=f(x_1)+f(x-x_1)$$

for almost every $y$. Hence, for each $x$, the function $y\mapsto f(x+y)-f(y)$ is almost everywhere equal to a constant. That constant is unique, since two conull subsets of $\mathbb R$ have nonempty intersection. Define it to be $h(x)$. We have therefore proved that, for every $x$,

$$f(x+y)-f(y)=h(x) \tag{2}$$

for almost every $y$. Notice that the auxiliary $x_1$ above was chosen after $x$ was fixed and may depend on $x$.

If $x\notin M$, the original equation also gives $f(x+y)-f(y)=f(x)$ for almost every $y$. Uniqueness of the almost-everywhere constant in (2) yields $h(x)=f(x)$. Thus $h=f$ almost everywhere.

It remains to prove that $h$ is additive. For each $t\in\mathbb R$, choose a null set $K_t$ outside which (2) holds with $x=t$. Fix $a,b\in\mathbb R$. Consider the following five exceptional subsets of the $(w,z)$-plane:

$$\begin{aligned} E_1&=K_a\times\mathbb R,\\ E_2&=\mathbb R\times K_b,\\ E_3&=\{(w,z):w+z\in K_{a+b}\},\\ E_4&=N,\\ E_5&=\{(w,z):(a+w,b+z)\in N\}. \end{aligned}$$

Each is a two-dimensional null set. For the coordinate cylinders this follows by first intersecting the unrestricted coordinate with $[-n,n]$ and then taking a countable union. The set $E_3$ is null because the shear $(w,z)\mapsto(w,w+z)$ preserves Lebesgue measure and sends it to $\mathbb R\times K_{a+b}$. Finally, $E_5=(-a,-b)+N$, so it is null by translation invariance.

A finite union of null sets cannot cover $\mathbb R^2$. Choose $(w,z)$ outside $E_1\cup\cdots\cup E_5$. The five corresponding identities are

$$\begin{aligned} f(a+w)-f(w)&=h(a),\\ f(b+z)-f(z)&=h(b),\\ f(a+b+w+z)-f(w+z)&=h(a+b),\\ f(w+z)&=f(w)+f(z),\\ f(a+b+w+z)&=f(a+w)+f(b+z). \end{aligned}$$

Substituting the last two identities into the third and regrouping with the first two gives

$$h(a+b)=h(a)+h(b).$$

Since $a$ and $b$ were arbitrary, $h$ is additive. $\square$

Dependencies. Fubini's theorem for Lebesgue measure, translation invariance of null sets, and invariance of two-dimensional Lebesgue measure under determinant-one linear transformations.

Formalization. A Lean 4 proof formalizes this real-valued theorem in mathlib v4.29.1. The proof was located during compilation but was not built here.

Bears on. #1126