Let $g$ be a Lebesgue measurable function on $[0, 1]$ such that the function $f(x, y) := g(x) - g(y)$ is Lebesgue integrable over the square $[0, 1] \times [0, 1]$. Show that $g$ is integrable over $[0, 1]$.
Attempt: I want to apply Fubini's Theorem and say $\int_{[0, 1] \times [0, 1]} f(x, y) d(m \times m) = \int_{0}^{1} \int_{0}^{1} [g(x) - g(y)] dy dx$. For $x$ fixed, we have $g(x) - g(y) = c - g(y)$ is integrable on $[0, 1]$, $c$ constant. So are we done, or am I leaving something out?
Your idea is correct the proof is not precise. By Fubini's Theorem integrability of $f(x,y)$ implies integrability of $g(x)-g(y)$ w.r.t. $y$ for almost all $x$. In particular there is at least one $x$ for which this holds and this gives integrability of $g$.