Showing that $\left\{ T(f) \neq 0\right\}$ is $\sigma$-finite

Question

Let $(X,\mu)$, $(Y,\nu)$ be measure spaces and $$T: L^{p_0} + L^{p_1} \rightarrow L^{q_0} + L^{q_1}$$ be a linear operator for $p_0,p_1,q_0,q_1 \in [1,\infty]$. Further assume $T(f) \in L^{q_0}\cap L^{q_1}$ for all finitely simple functions $f$. Why does it hold, that the set $$\left\{ T(f) \neq 0\right\}$$ is $\sigma$-finite unless $q_0 = q_1 = \infty$? I encountered this statement in Folland's Real analysis Book, however I do not see how I can proove this.

Answer 1

If $g\in L^p$ for any $p<\infty$, then $A=\{x:g(x)\neq 0\}$ must be $\sigma$-finite. Indeed, in order for $\int |g(x)|^p dx$ to be finite, for any $n$ the set $A_n=\{x:|g(x)|^p>1/n\}$ must have finite measure, since $\int |g(x)|^p dx\geq \mu(A_n)/n$. Since $A=\bigcup_n A_n$, $A$ is $\sigma$-finite.