Let $f:\mathbb{R}\to\mathbb{C}$ be a function such that $f(x+1)=f(x)$ almost everywhere.

I want to prove that exists a function $F$ such that $F(x+1)=F(x)$ always holds and $F(x)=f(x)$ almost everywhere.

I think the intuition here is simple: let $x_0$ be a real number, if every $x\in x_0\mathbb{Z}$ satisfies $f(x+1)=f(x)$, then we define $F(x)=f(x)$. If there are a couple exceptions, we fix them and keep going for every $x_0\in \mathbb{R}$. However I am failling to formalize this.

Let $A=\{x\in\mathbb{R} : f(x+1)\neq f(x)\}$. Since $\mathbb{R}\setminus A$ is dense, for every $x\in\mathbb{R}$ there is a sequence in $\mathbb{R}\setminus A$ that converges to $x$. Then maybe we could define $F(x)$ as $$\lim_{n\to\infty}f(x_n).$$

I am lost.


Since $f(x+1) = f(x)$ for almost all $x \in \mathbb{R}$ you have that $\mu(A) = 0$ where $\mu$ is the Lebesgue measure and $A = \{x: f(x+1) \neq f(x)\}$.

Notice that for every $n \in \mathbb{Z}$, $n+A$ also has Lebesgue measure $0$ since the Lebesgue measure is translation invariant. In particular, $B = \bigcup_{n \in \mathbb{Z}} (n+A)$ has Lebesgue measure $0$ as a countable union of Lebesgue nullsets. Now define e.g. $F(x) = f(x)$ for $x \not \in B$ and $F(x) = 0$ for $x \in B$ and check that this works.


Define $F(x)=f(x\bmod 1)$ where $x\bmod 1$ is the unique element of $(x+\Bbb Z)\cap [0,1)$. Then $F(x)\ne f(x)$ only if at least one of the numbers $x+n$, $n\in \Bbb Z$ is in $A$. Hence $$\{\,x\in\Bbb R\mid F(x)\ne f(x)\,\}\subseteq \bigcup_{n\in\Bbb Z}(A+n) $$ As $A$ is a nullset, so is the countable union of translated copies of it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.