Suppose that there is a finite $c$ such that $\int_0^1 | f(a+t) - f(b+t) | dt \le c$ for all $a$ and $b$. Show that $f \in L(0, 1)$.

Question

Please help me understand the following proof.

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Q) Let $f$ be measurable and periodic with period $1$, that is, $f(t+1)=f(t)$. Suppose that there is a finite $c$ such that $$\int_0^1 | f(a+t) - f(b+t) | dt \le c$$ for all $a$ and $b$. Show that $f \in L(0, 1)$.

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A)

1. \begin{align} c &\ge \int_0^1 \int_0^1 |f(x+t) - f(-x+t)| ~dt~dx \\ &= \frac12 \int_0^1 \int_{-\xi}^\xi |f(\xi) - f(\eta)| ~d\eta~d\xi + \frac12 \int_0^1 \int_{\xi-1}^{1-\xi} |f(\xi)-f(\eta)| ~d\eta~d\xi \end{align}
2. \begin{align} \int_0^2 \int_{-1}^1 |f(\xi)-f(\eta)|~d\eta~d\xi &\le 2c \\ \int_0^1 \int_0^1 |f(\xi)-f(\eta)|~d\eta~d\xi &\le\frac c2 \end{align}
3. Thus, $|f(\xi)-f(\eta)|$ is integrable over the square $[0, 1]\times[0, 1]$.
4. There is a theorem: for measurable $f$ on $(0, 1)$, if $f(x)-f(y)$ is integrable over $[0, 1]\times[0, 1]$, then $f\in L(0, 1)$.
5. By this theorem, we can have $f$ is integrable over $(0, 1)$.

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Can someone explane about the process $1$ and $2$? I do not understand them.