Measure theory, measurable function, constant almost everywhere

Let f be an integrable function on the line such that for every real number x, we have $$f(x) =f(x+\sqrt{2})$$. Show that f is constant almost everywhere.
I am looking for a hit to start with, thank you all.

Here is my first try, Let $E=\{x \in R : f(x) \neq K\}$, where $K$ is a constant, and I am trying to prove measure of $E$ is zero: $m(E)=0$. To do so split $E$ to $E’=\{x \in R: f(x)>k\}$, and $E”=\{x \in R: f(x)<k\}$, and I am trying to prove $m(E’)=m(E”)=0$.

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Here’s a sketch (the details are easy): Suppose $f$ isn’t constant almost everywhere and get two disjoint sets $A, B$ of positive measure such that $f[A] \cap f[B] = \phi$. Next observe that the set $G$ of integer linear combinations of $1, \sqrt{2}$ is a dense additive subgroup of $\mathbb{R}$. It follows that $A + G$ and $B + G$ are conull subsets of $\mathbb{R}$ on which the ranges of $f$ are disjoint which is impossible.

A tentative for a proof.

1) Suppose in addition to the hypothesis that $f$ is a bounded function (a.e). Then $\displaystyle F(x)=\int_0^x f(t)dt$ is a continuous function on $\mathbb{R}$. If $a\in E=\mathbb{Z}+
\mathbb{Z}\sqrt{2}$, then we get easily as $f(t+a)=f(t)$ that for all $x$:
$$F(x+a)=F(x)-F(-a)$$
Putting $x=0$ gives $F(a)=-F(-a)$. Hence $F(x+a)=F(x)+F(a)$. As $E$ is dense in $\mathbb{R}$ and $F$ continuous, we get $F(x+y)=F(x)+F(y)$ for all $x,y$. It is well known that this functional equation with $F$ continuous imply that there exists a constant $m$ such that $F(x)=mx$ for all $x$. Hence $\displaystyle \int_0^x f(t)dt=mx$ for all $x$. Then (Lebesgue differentiation theorem ?, or the fact that $\displaystyle \int_u^v(f(t)-m)dt=0$ for all $u,v$ ?), we get $f(x)=m$ a.e.

2) Now we go to the general case. Let $n\in \mathbb{N}$, and put $\displaystyle F_n=\{x; |f(x)|< n\}$. Then $F_n\subset F_{n+1}$, and $\cup F_n=\mathbb{R}$(We have used that $f(x)\in\mathbb{R}$ for all $x$). There exists a $n_0$ such that $F_{n_0}$ has measure $>0$. Put $g(x)={\rm Min}\{|f(x)|, n_0\}$. We immediately see that $g$ verify the properties of the part 1). So $g$ is a constant $m$ a.e. Suppose that $G=\{x; |f(x)|>n_0\}$ is not of measure $0$. As $g(x)=n_0$ on $G$, we get that $m$ is $n_0$. Hence $g(x)= n_0$ a.e, and $|f(x)|\geq n_0$ a.e, in contradiction with the definition of $n_0$. We have then that $|f(x)|\leq n_0$ a.e, and we can use for $f$ the results of part 1), and we are done.

Ok, so let me introduce the idea of a period and the fundamental period. A function $f$ has period $t$ if for all $x$ in the domain it is true that $f(x+t) = f(x)$. A function is called periodic if it has (at least one) period. Take any periodic function $f$ and define its fundamental period $T=T(f)$ as $T=\inf \{t>0:$ t is a period of $f\}$.

Take the fundamental period of the function $f$. If $T>0$ then T must be a period of $f$ (see this question for the proof). Now use the fact that if $T>0$ then the set of all possible periods of $f$ is composed by $\{nT: n\in\mathbb{Z}\backslash\{0\}\}$, this comes also on the proof mentioned.

Now consider $f$. The function has as periods two numbers that are not multiples of each other, and therefore it should be the case that $T=0$. If $T=0$ then it is easy to show that the function must be constant.