Integrability of a periodic function based on $\int_0^1 |f(a+t)-f(b+t)| dt$ 
Let $f$ be a measurable function with period $1$ on the real line such $\int_0^1 |f(a+t)-f(b+t)| dt$ is bounded uniformly for all $a, b \in \mathbb{R}$. Show that $f$ is integrable on $[0, 1]$. [Hint: Use $a = x$, $b = −x$, integrate with respect to $x$, and change variables to $ξ=x+t$, $η=−x+t$.]

First of all what does it is bounded uniformly for all $a, b \in \mathbb{R}$? Does it mean that for all $a, b \in \mathbb{R}$, $\int_0^1 |f(a+t)-f(b+t)| dt \le M$ for a single $M$? And how does it help to solve the exercise?
How the hint is useful when nothing cancels out to reach $\int_0^1 |f(t)| dt$ with the use of the hint?
 A: Suppose $M:=\sup_{a, b}\int^1_0|f(t+a)=f(t+b)|\,dt<\infty$.
If you follow the hint, and for instance, integrate over $-1\leq x\leq 0$ and $0\leq t\leq 1$ one gets
$$\begin{align}\int^0_{-1}\int^1_0|f(t+x)-f(t-x)|\,dr\,dt&=\frac{1}{2}\left(\int^0_{-1}\Big(\int^{\xi+2}_{-\xi}|f(\xi)-f(\eta)|\,d\eta\Big)\,d\xi\right.\\&\qquad+\left.\int^1_0\Big(\int^{-\xi+2}_\xi|f(\xi)-f(\eta)|\,d\eta\Big)\,d\xi\right)<M\end{align}$$
for some $M>0$. Indeed, the linear transformation $T(x,t)=(x+t,t-x)$ transforms the cell $[-1,0]\times[0,1]$ into to a rhombus whose vertices are $P_1(0,0),\,P_2(1,1),\,P_3(0,2),P_4(-1,1)$, and the Jacobian determinant $J_T(x,y)=2$.
A consequence of Fubini-Tonelli's theorem is that for almost all $0\leq\xi\leq 1$, for example,  the integral $\int^{-\xi+2}_{\xi}|f(\xi)-f(\eta)|\,d\eta <\infty$. For such fixed $\xi$, the integrand $\eta\mapsto|f(\xi)-f(\eta)|$ is $1$-periodic. Take $0\leq \xi<\frac12$ so that $2(1-\xi)\geq1$ (the length of $[\xi,2-\xi]$). Then, the periodicity of $f$ implies that
$$\begin{align}
\int^1_0|f(\xi)-f(\eta)|\,d\eta=\int^{1+\xi}_\xi|f(\xi)-f(\eta)|\,d\eta\leq \int^{-\xi+2}_{\xi}|f(\xi)-f(\eta)|<\infty\end{align}$$
whence one concludes that
$$\int^1_0|f(\eta)|\,d\eta\leq |f(\xi)|+\int^1_0|f(\xi)-f(\eta)|\,d\eta<\infty$$

The last two bits follow from the fact that if $f$ is $T$ periodic and integrable, then $\int^T_0f = \int^{T+a}_af$ for all $a$.
Whether one integrates $x$ over $[-1,0]$, or $[\alpha,\alpha+1]$ (notice that the length of interval of integration is conveniently chosen to be the same as the period) is not that important. The change of variables transforms a cell parallel to the axis into some rhombus. The main idea is then to exploit Fubini's theorem. For example, had we  integrated $x$ over $[0,1]$ we would have obtained
$$\begin{align}
\int^1_0\int^1_0|f(x+t)-f(t-x)|\,dt\,dx&=\frac{1}{2}\left(\int^1_0\Big(\int^\xi_{-\xi}|f(\xi)-f(\eta)|\,d\eta\Big)\,d\xi\right.+\\
&\qquad\qquad\left.\int^2_1\Big(\int^{-\xi+2}_{\xi-2}|f(\xi)-f(\eta)|\,d\eta\Big)\,d\xi\right)<M
\end{align}
$$
Fubini's theorem implies for instance that for almost all $\xi\in[0,1]$,
$\int^\xi_{-\xi}|f(\xi)-f(\eta)|\,d\eta<\infty$. Taking a typical $\tfrac12\leq \xi\leq 1$ and using the periodicity iff the integrand gives
$$
\int^1_0|f(\xi)-f(\eta)|\,d\eta=\int^{1/2}_{-1/2}|f(\xi)-f(\eta)|\,d\eta\leq \int^\xi_{-\xi}|f(\xi)-f(\eta)|\,d\eta<\infty
$$
Similar argument if one chooses the other piece. So no matter how once dices it and slice it, the idea remains the same: exploit Fubini's theorem.
