= Solution
Let $M=\|f\|_\infty$, the <essential supremum> of $|f|$. If $M=0$, $f=0$ <almost everywhere> and every <Lp norm> is zero. Suppose $M>0$. Since $|f|\leq M$ <almost everywhere> and $m((0,1))=1$,
$$
\|f\|_p\leq M\qquad(1\leq p<\infty).
$$
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. Hence
$$
\|f\|_p^p\geq\int_{A_a}|f|^p\,dx\geq a^pm(A_a),\qquad
\|f\|_p\geq a\,m(A_a)^{1/p}.
$$
For fixed $a$, $m(A_a)^{1/p}\to1$. Thus $\liminf_{p\to\infty}\|f\|_p\geq a$. Letting $a\uparrow M$ and combining with the upper bound proves
$$
\boxed{\lim_{p\to\infty}\|f\|_p=\|f\|_\infty.}
$$
The positive-measure superlevel sets explain why the <Lp norms converge to the supremum norm> even when the essential maximum is never attained.
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