 Solution

Put $M=\Vert f{\Vert }_{\infty }>0$ and $M_0=\mu (E)$. There is a positive-measure set on which $|f|>M/2$, so $M_0>0$ and each $M_n$ is positive; boundedness and finite measure make each finite. For $n\ge 1$, apply Cauchy-Schwarz to $|f{|}^{(n-1)/2}$ and $|f{|}^{(n+1)/2}$ to getTherefore the positive ratios $r_n=M_{n+1}/M_n$ are nondecreasing. Telescoping gives $M_n/M_0=r_0\cdots r_{n-1}\le r_n^n$, while $|f{|}^{n+1}\le M|f{|}^n$ almost everywhere gives $r_n\le M$. HenceFor $0<\epsilon <M$, the set $A_{\epsilon }=\{|f|>M-\epsilon \}$ has positive measure by the definition of essential supremum. Thus $\Vert f{\Vert }_n\ge (M-\epsilon )\mu (A_{\epsilon }{)}^{1/n}$, whereas $\Vert f{\Vert }_n\le M\mu (E{)}^{1/n}$. Letting $n\to \infty$ and then $\epsilon \downarrow 0$ shows $\mu (E{)}^{-1/n}\Vert f{\Vert }_n\to M$. The preceding squeeze yields $M_{n+1}/M_n\to \Vert f{\Vert }_{\infty }$. This proves the ratio limit rather than assuming it from the moment-root limit.