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?

Mathematics Asked by Marios Gretsas on January 1, 2021

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