 Solution

Let $M=\Vert f{\Vert }_{\infty }$, the essential supremum of $|f|$. If $M=0$, $f=0$ almost everywhere and every Lp norm is zero. Suppose $M>0$. Since $|f|\le M$ almost everywhere and $m((0,1))=1$,For any $a$ with $0<a<M$, the measurable set $A_a=\{x\in (0,1):|f(x)|>a\}$ has positive Lebesgue measure; otherwise $a$ would be an essential upper bound. HenceFor fixed $a$, $m(A_a{)}^{1/p}\to 1$. Thus ${\mathrm{liminf}}_{p\to \infty }\Vert f{\Vert }_p\ge a$. Letting $a\uparrow M$ and combining with the upper bound provesThe positive-measure superlevel sets explain why the Lp norms converge to the supremum norm even when the essential maximum is never attained.