I'm preparing for my qualifying exam and i want some imput in my solution of this problem.
Let $f:\Bbb{R} \to \Bbb{R}$ measurable and $1-$periodic(i.e $f(x+1)=f(x),\forall x \in \Bbb{R}$)
If $\int_0^1|f(x+t)-f(x)|dx \leq 1,\forall t \in \Bbb{R}$,prove that $f \in L^1[0,1]$
My proof goes like this:
we have that $$\int_0^1 \int_0^1 |f(x+t)-f(x)|dxdt \leq 1 \Longrightarrow \int_0^1 \int_0^1 |f(x+t)-f(x)|dtdx \leq 1$$
By Fubini: we have that the function $g_x(t)=|f(x+t)-f(x)|$ is integrable on $ [0,1]$ for almost every $x \in [0,1]$ Let $A$ be the set of such $x$. By periodicity we have that $$\int_I g_x(t)dx=\int_J g_x(t)dx$$ for every interval $J,I$ with length $1$ for all $x \in A$
Now $$1 \geq \int_A \int_0^1 |f(x+t)-f(x)|dtdx=$$ $$\int_A \int_{-x}^{-x+1} |f(x+t)-f(x)|dtdx=\int_A \int_0^1 |f(z)-f(x)|dxdz$$
Since $A^c$ has measure zero we conclude the $G(x,z)=f(x)-f(z) \in L^1([0,1]^2)$ so again by Fubini it is easy to conclude that $f \in L^1[0,1]$
Is there a flaw in my argument?
Thnak you in advance