$\| f \|_{L^1} > 0 $ is stronger than simply $f \neq 0$. But how much?

Question

My question is: for a non-negative function $f: \mathbb{R} \rightarrow [0,+\infty)$

• $ f \neq 0$ means that for some $x \in\mathbb{R}, f(x)>0$
• $\| f \|_{L^p} > 0 $ should imply something stronger. Is it true that $\| f \|_{L^p} > 0 $ if and only if there exists some nonzero-measure measurable set $E$ such that $|f|>0$ on $E$?

(I am sure this is a duplicate but I wasn't able to find the same question)

Answer 1

Yes; if $\|f\|_p>0$ then $f>0$ on some positive measure set, since otherwise $f=0$ a.e. and hence $\|f\|_p = 0$. Conversely, if $\|f\|_p = 0$ then $f=0$ almost everywhere.