I have problems trying to prove one implication of the following proposition
Let $f:[0,1] \to \mathbb R$ be a measurable function and let $h: [0,1]^2 \to \mathbb R$ defined as $h(x,y)=f(x)-f(y)$. Show that $h$ is integrable iff $f$ is integrable.
I could prove that $h$ integrable implies $f$ integrable:
Suppose $h$ is integrable, by Fubini-Tonelli we have $$\infty >\int_{[0,1]^2}|h|=\int_{[0,1]}(\int_{[0,1]}|f(x)-f(y)|dy)dx$$
This means $\int_{[0,1]}|f(x)-f(y)|dy < \infty$ for almost all $x \in [0,1]$, Pick $x_0$ for which the inequality is satisfied. Then $$\int_{[0,1]}|f(y)|dy \leq \int_{[0,1]}|f(y)-f(x_0)|dy+\int_{[0,1]}|f(x_0)|dx < \infty$$
From the chain of inequalities above it follows that $f$ is integrable.
I got stuck trying to show $f$ integrable $\implies$ $h$ is integrable
Again we have the equality $$\int_{[0,1]^2}|h|=\int_{[0,1]}(\int_{[0,1]}|f(x)-f(y)|dy)dx,$$ and for each $x$, it is clear that $g(x)=\int_{[0,1]}|f(x)-f(y)|dy< \infty$. I would like to show that $\int_{[0,1]}g(x)dx< \infty$, how could I show this? Any help would be greatly appreciated.
$|h(x,y)| \le |f(x)| + |f(y)|$.
$\int |h(x,y)| dx \le \int |f(x)| dx + |f(y)|$.
$\int \int |h(x,y)| dx dy \le \int |f(x)| dx + \int |f(y)|dy$.