Additivity + Measurability $\implies$ Continuity

A function $f:\Bbb R \to \Bbb R$ is additive and Lebesgue measurable. Prove that $f$ is continuous.

I know that on $\Bbb Q$, $f$ comes out to be linear. So, if $f$ is to be continuous then $f$ must be linear in $\Bbb R$. But, I'm stuck here.

• I like the proof given in Theorem 5.5 in the book Herrlich: Axiom of Choice. He shows there that every discontinuous solution of $f(x+y)=f(x)+f(y)$ in non-measurable. – Martin Sleziak Mar 9 '13 at 12:52
• I like this result because when you combine it with the Solovay Model, you get a statement like "There is a model of ZF where every additive function is linear (and of course with choice that's not the case)." – Mark S. Dec 28 '13 at 22:46
• – Martin Sleziak Jan 30 '18 at 3:20

Using Lusin's theorem, there exists a compact $K \subset [0,1]$ such that $\mu(K)>2/3$ and $f$ continuous on $K$. Let $\epsilon>0$. In fact, $f$ is uniformly continuous on $K$, so there exists $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$; without loss of generality, suppose $\delta<1/3$.

Let $h<\delta$. Notice that the intersection between $K$ and $K-h$ is nonempty; otherwise, $1+h= \mu([-h,1]) \geq \mu(K \cup K-h)= \mu(K)+\mu(K-h)=2\mu(K)>4/3$, so $h>1/3$ whereas $h<\delta<1/3$ by assumption.

Let $x_0 \in K \cap (K-h)$. We have $|f(x_0+h)-f(x_0)|<\epsilon$, hence $|f(h)|<\epsilon$ because $f$ is additive. You deduce that $f$ is continuous at $0$.

Finally, it is straightforward to conclude that $f$ is continuous on $\mathbb{R}$.

Edit: For a simple proof of Lusin's theorem, see for example: Marcus B. Feldman, A proof of Lusin's theorem, The American Mathematical Monthly Vol. 88, No. 3 (Mar., 1981) (pp. 191-192).

• You used the fact that $f$ is measurable and i understand that in any measure space , additivity and measurability would imply continuity but isn't there any simple proof for lebesgue measure space( if there wasn't , why would they especially ask for lebesgue measure) $?$ – Aang Mar 8 '13 at 7:52
• @Avatar: The complexity of the proof will depend on the proof of Lusin's theorem; but there is a simple proof using the density of continuous functions in $L^1$, so considering Legesgue measure space seems to be simpler here. – Seirios Mar 8 '13 at 9:54
• In "any measure space" I don't know what "additive" means. – GEdgar Oct 28 '13 at 12:48
• I couldn't understand the last statement that it is straightforward to prove that f is continuous on $\mathbb R$. Can you provide a hint? – bellcircle Jan 23 '17 at 19:26
• This is just because $f$ is additive and continuous at $0$. (Translations are of course continuous.) – Seirios Jan 25 '17 at 10:29

Here's how I first did it. First we make a few notes. We have $\displaystyle f(0) = f(0 +0) = f(0) +f(0)$, so $f(0)=0$. More interestingly, we have $f(2x) = f(x + x) = f(x) + f(x) = 2f(x)$, upon iterating, $f(nx) = nf(x)$. Looking at $-x$, we have $f(0) = f(x +(-x)) = f(x) + f(-x)$ so $f(-x) = - f(x)$. Finally, we have $f(x) = \displaystyle f\left(\frac x2 + \frac x2\right) = 2 f\left( \frac x2\right)$ and upon iterating, $\displaystyle f\left(\frac{x}{2^n}\right) = \frac{1}{2^n} f(x)$. Putting it all together, we have that $\displaystyle f\left(\frac{k}{2^n}x \right) = \frac{k}{2^n}f(x)$ for any $k \in \mathbb{Z}$, $n \in \mathbb{N}$. \\par Suppose $f$ is not identically $0$, for otherwise our function would be trivially continuous. Then there exists some $p$ such that $f(p) = a \neq 0$. Then by our observation above, every number of the form $\displaystyle \frac{k}{2^m} p$ is such that $\displaystyle f\left(\frac{k}{2^m} p\right) = \frac{k}{2^m}f(p) = \frac{k}{2^m} a \neq0$. Since there are countably such numbers of the form $\displaystyle\frac{k}{2^m} p$, enumerate them by $\{p_n\}_{n \in \mathbb{N}}$. Call $f(p_n) = a_n$. Since every $a_n$ is of the form $\displaystyle \frac{k}{2^m} a$ for some $k, m$, and $a\neq 0$, we have that $\{a_n\}_{n \in \mathbb{N}}$ is dense in $\mathbb{R}$, since dyadic rationals are dense in $\mathbb{R}$, and scaling a dense, infinite set in $\mathbb{R}$ will do nothing to change its density.

Now, fix $\epsilon >0$. Let $V= \displaystyle B\left(0, \frac\epsilon2\right)$. Since $\{a_n\}_{n \in \mathbb{N}}$ are dense, $\mathbb{R} = \displaystyle \bigcup_{n \in \mathbb{N}} V + a_n$. Now we have the following claim to show: $f^{-1}(V+a_n) = f^{-1}(V) + p_n$. First, suppose $x \in f^{-1}(V) +p_n$. Then $x = y+ p_n$ for some $y \in f^{-1}(V)$, so $f(x) = f(y) + f(p_n) = v + a_n$ for some $v \in V$, and thus $f^{-1}(V+a_n) \supset f^{-1}(V) + p_n$. Now we want to show $f^{-1}(V+a_n) \subset f^{-1}(V) + p_n$ where $f(p_n) =a_n$. So, suppose $f(x) = v+a_n$ for some $v \in V$. Then $f(x-p_n) = f(x) - f(p_n) = v+a_n - a_n = v$, so $x-p_n \in f^{-1}(V)$. Thus $x = (x-p_n) +p_n \in f^{-1}(V) + p_n$. Thus we have shown the equality we wanted. \\par From here, we see $$\mathbb{R} = f^{-1}(\mathbb{R}) = f^{-1}\left( \displaystyle \bigcup_{n \in \mathbb{N}} V+a_n\right) = \bigcup_{n \in \mathbb{N}} f^{-1}(V+a_n) = \bigcup_{n \in \mathbb{N}} f^{-1}(V) + p_n.$$

Now, since $V$ is open and $f$ measurable, $f^{-1}(V)$ is measurable, and by translation invariance of the Lebesgue measure (denoted by $m$), we have $m(f^{-1}(V) +p_n) = m(f^{-1}(V))$. Set $U = f^{-1}(V)$. The equality above implies $m(U) >0$, otherwise we'd have $\mathbb{R}$ as the countable union of null sets. Thus we can apply Steinhaus's theorem to $U-U$ to see that $U-U \supset B(0, \delta)$, that is, a ball of radius $\delta>0$ around the origin. Then for every $|x|< \delta$, we have $x= y-w$ for $y, w \in U$. But then $$|f(x)| = |f(y-w)| = |f(y) - f(w)| \le |f(y)| + |f(w)| < \frac\epsilon2 + \frac\epsilon2 = \epsilon$$ since $f(y), f(w) \in V$.

This shows continuity of $f$ at zero, and by the pseudo-linearity property, we can get linearity at any point $z$. For take any sequence $z_n \to z$, then $z- z_n \to 0$, so we have $f(z) - f(z_n) = f(z-z_n) \to 0$ by continuity at zero. This concludes the proof.