Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$

Question

I need help understanding the proof of this question. Basically it involves Fubini's theorem and I can't seem to get a grasp on the proof (and I am pretty sure this question uses techniques from the proof). The question is as follows:

Assume $f:[0,1] \rightarrow \mathbb{R} $ is measurable and define $g: [0, 1]^2 \rightarrow \mathbb{R}$ by $g(x, y) := |f(x) - f(y)|$. Prove that $f$ is integrable if and only if $g$ is integrable.

Proof. The slice of $g$, $g_x(y) = |f(x) - f(y)|$ is integrable iff $h(y) = f(x) - f(y)$ is integrable and $h(y)$ is integrable iff $f$ is integrable. $(*)$

Now, assume $g$ is integrable. If $g$ is integrable, then by Fubini's theorem, the slice $g_x$ is integrable for almost all $x$. If the slice is integrable then $(*)$ shows that $f$ is integrable.

Now assume $f$ is integrable. Then again by $(*)$ $g_x$ is integrable $\forall x$. Also, $$\int f - |f(x) \leq \int |f(x) - f(y)| dy \leq |f(x)| + \int f \qquad (**)$$

Then the function $\alpha(x) = \int |f(x) - f(y)| dy$ is bounded and measurable and therefore integrable on $[0, 1]$. By Fubini we have $\int g = \int \int |f(x) - f(y)| dy dx = \int \alpha(x) dx$ hence $g$ is integrable.

Now my question(s) are:

• How did $(*)$ come about? It seems intuitive to me to claim that but can someone explain to me why that is the case?
• In the forward direction (assuming f is true), how can the claim that $g_x$ be integrable hold for ALL x?
• And where is the $(**)$ coming from?

Answer 1

For your first point, suppose first that $f(y)$ is integrable, then $h(y)$ is integrable because it is the difference of two integrable functions (note that $f(x)$ is a constant w.r.t. $y$ and constants are integrable since $[0,1]$ has finite measure). Conversely, if $h(y)$ is integrable, then $f(y)=h(y)+f(x)$ is integrable, because it is the sum of two integrable functions.

For the second point, if $f$ is integrable then $h(y)$ is integrable for every $x$ (for what we have showed before) so that $g_x(y)$ is integrable for every $x$.

The third point follows from mookid's answer, but you need to substitute $\int f$ with $\int |f|$.

Answer 2

• Both first points come from the characterization: $$ f\text{ is measurable} \iff \forall c: \{ x:f(x)<c \}\text{ is measurable} $$
• the last one comes from the inequality $$ |f(y)| - |f(x)| \le |f(y) - f(x)| \le |f(y)| + |f(x)| $$