If $\alpha$ is some irrational such that $f(x)=f(x+_{\mathbb{Z}}\alpha)$ for a.e. $x\in[0,1]$, then $f$ is constant for a.e. $x\in[0,1]$

Question

Let $f\in L^{1}[0,1]$ with respect to the Lebesgue measure. Suppose there exists some irrational $\alpha$ such that $f(x)=f(x+\alpha)$ for almost all $x\in[0,1]$, where $+$ is modulo $\mathbb{Z}$. Then there exists $c\in\mathbb{R}$ such that $f(x)=c$ for almost all $x\in[0,1]$.

Let $\mu$ be the Lebesgue measure and $X=[0,1]$. Suppose the assumption is true everywhere on some $Y$ with measure $1$. On $Y$, we can define an equivalence relation where $[x]=[y]$ iff $x-y\in \alpha\mathbb{Z}$. On this $Y/\alpha\mathbb{Z}$, each $[x]$ is countable (thus measurable) as $[x]=(x+\alpha\mathbb{Z}+\mathbb{Z})\cap Y$.

Then every $[x]$ is null. If $Y/\alpha\mathbb{Z}$ is countable, then this is impossible, for $Y=\bigcup_{x} [x]$ has positive measure. It follows that uncountably many $[x]$ contain no rational number. Let $Z$ be the union of all such $[x]$. Then $\mu(Z)=\mu(Y)$, by removing a countable union of null sets from $Y$. It seems that at this point I have to prove $f$ is constant on $Z$, which I have no idea how to proceed.

On the other hand, the statement to show is equivalent to exactly one of $f^{-1}(r)$ being not null for $r\in\mathbb{R}$. Also, every $f^{-1}(r)$ with positive measure is a disjoint union of uncountably many elements in $Y/\alpha\mathbb{Z}$. It's probably possible to prove any traversal of $Y/\alpha\mathbb{Z}$ is non-measurable, but I'm not sure if there's any merit to that. It doesn't seem that more than one $f^{-1}(r)$ not being null contradicts $f$ being $L^1$ either.

Answer 1

This can be proven using the generalized Lebesgue differentiation theorem stated at https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem . Namely, the generalized version says (paraphrasing compared to the original source): let $\mathcal{V}$ be a family of subsets of $\mathbb{R}^n$ of bounded eccentricity, which means there exists $c > 0$ such that each $U \in \mathcal{V}$ is contained in a ball $B$ with $\lambda(U) \ge c \lambda(B)$. Also, suppose each $x \in \mathbb{R}^n$ is contained in sets of arbitrary small measure from $\mathcal{V}$. Then for almost every $x \in \mathbb{R}^n$, $$f(x) = \lim_{\lambda(U)\to 0, x\in U\in \mathcal{V}} \frac{1}{m(U)} \int_U f\,d\lambda.$$

Now in your situation, extend $f |_{[0, 1)}$ to a function on $\mathbb{R}^1$ with period 1, and let $\mathcal{V}$ be the family of balls with center of the form $n + m \alpha$ where $m, n \in \mathbb{Z}$. It is then easy to check that this family satisfies the conditions of the generalized Lebesgue differentiation theorem (with $c = 1$) and that the right hand side does not depend on $x$, at least among the $x$ for which the limit exists. For the latter part, use the fact that every $U$ containing $x_2$ can be translated to a $U$ containing $x_1$ by some $n + m\alpha$ and the integrals of $f$ over $U$ and the translate of $U$ are equal. From there, we see that $f$ is a.e. equal to this constant.

Answer 2

Without ergodic theory: Now this is a bit like cheating, as I will just use the proof of ergodicity of the irrational shift (see here Ergodicity of irrational rotation). I will assume that we have already modified $f$ on a null set such that it is now invariant under the irrational shift $$S:\mathbb{T}\rightarrow \mathbb{T}, x\mapsto x+\alpha$$ where $\mathbb{T}$ is the torus. In the following I'll use freely that $S$ is measure-preserving (as it is just a shift).

What follows now is just a minor modification of the argument in Ergodicity of irrational rotation.

Basic idea: First, what is the idea. If we'd already know that $f$ is constant, then we could compute said constant. Namely, $$ \int_0^1 f(x)dx = \int_0^1 c\, dx =c.$$ So we need to show that $f$ is equal to its average almost everywhere. As $f$ is invariant under the shift, we know that for all $x$ and all integers $m$ we have $$ f(x)=f(x+m\alpha).$$ If $f$ was continuous, we would be done as $\{ x+m \alpha \ : \ m\in \mathbb{Z}\}$ is dense in $\mathbb{T}$. As $f$ is only in $L^1$ we will have to pass through an approximation argument by continuous functions. As the approximation is only close in $L^1$ norm we won't be able to directly show that we are pointwise close to a constant function. Hence, our objective will be to show $\Vert f-\int f \Vert_1=0$ and conclude from there.

Approximation argument: Pick a sequence $(f_n)_n\subseteq C(\mathbb{T})$ such that $\Vert f-f_n\Vert_1\leq n^{-1}$. As $f$ is invariant under $S$ we get for all integers $m$ $$ \Vert f_n \circ S^m -f_n\Vert_1 \leq \Vert f_n \circ S^m - f\circ S^m\Vert_1 +\Vert f-f_n \Vert_1=2\Vert f-f_n\Vert_1\leq 2n^{-1}.$$ Using that $f_n$ is continuous and the orbits of $S$ are dense, we get for all $s\in \mathbb{T}$ $$\Vert f_n(\cdot+s)-f_n\Vert_1\leq 2n^{-1}.$$ Thus, using Tonelli, we get \begin{align*} \Vert f_n - \int_\mathbb{T} f_n\Vert_1 &= \int_\mathbb{T} \left\vert f_n(x) -\int_\mathbb{T} f_n(y)dy\right\vert dx \\ &=\int_\mathbb{T} \left\vert \int_\mathbb{T} (f_n(x)- f_n(x+s))ds\right\vert dx \\ &\leq \int_\mathbb{T} \int_\mathbb{T} \vert f_n(x)-f_n(x+s)\vert dsdx \\ &=\int_\mathbb{T} \int_\mathbb{T} \vert f_n(x)-f_n(x+s)\vert dx ds \\ &= \int_\mathbb{T} \Vert f_n(\cdot+s)-f_n\Vert_1 ds\\ &\leq 2n^{-1}. \end{align*} $$ $$ Taking the limit $n\rightarrow \infty$ yields $$ \Vert f - \int_\mathbb{T} f \Vert_1 =0.$$ Thus, $f$ is a.e. equal to its average.

With ergodic theory: If one happens to know ergodic theory, then the argument would be:

$1.)$ the irrational shift is ergodic (i.e. measurable sets which are invariant under the shift must have either measure $0$ or $1$)

$2.)$ our function can be modified to be invariant under the irrational shift and hence all sets of the form $f^{-1}([a,b])$ have either measure $0$ or $1$

$3.)$ The total measure is one, so a bisection argument yields the claim.

People liking to nuke flies can also combine $1.)$ with the strong ergodic theorem to reach the desired conclusion.