Let $f$ be a real valued measurable function on $[0,1]$ and let $h(x,y)= f(x)f(y)$ be integrable on $[0,1] \times [0,1]$ . Prove that $f$ is integrable on $[0,1]$.
Using Fubini the $x$-section $h_x(y)$ is integrable for a.e. $x \in [0,1]$ , Pick $x_0$ from that set, so we get that $f(x_0)f(y)$ is integrable and thus $f$ is integrable on $[0,1]$. The problem is if $f(x_0)=0$, then my argument fails.
There is set $E$ of measure $0$ such that $h_x(y)$ is integrable if $x \notin E$. Suppose there exists $x_0 \notin E$ such that $f(x_0) \neq 0$. Then we are done (by the Fubini argument). Otherwise $f=0$ on $E^{c}$. This means $f=0$ almost everywhere so it is intergable.