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Fubini's Theorem on periodic function.

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

Mathematics Asked by Marios Gretsas on January 1, 2021

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