prove limit with integral is equal supremum Let $f$ be continuous non-negative function at interval $[a,b]$, it has there supremum equal to $M$ prove that : $\displaystyle \lim_{n \to \infty} (\int _{a}^bf(x)^ndx)^\frac{1}{n}=M$
 A: Notice that the set $\left\{x\in [a,b]: \, |f(x)|>M\right\}$ has measure zero, So we have that $$\int_{a}^{b}{f(x)^{n} \, dx}\leq M^{n}\cdot (b-a)$$ hence $$\limsup_{n \rightarrow \infty}\,\left(\int_{a}^{b}{f(x)^{n} \, dx}\right)^{1/n}\leq M$$ Now for the reverse inequality let $0<\varepsilon<M$ be given and consider the set $$B_{\varepsilon}=\left\{x \in [a,b]: \, |f(x)| > M-\varepsilon\right\}$$ This set has nonzero Lebesgue measure and it is open, since f is continuous. Now using the fact that the open subintervals of $[a,b]$ forms a base for the usual Topology on $[a,b]$, we can find $(a_{\varepsilon},b_{\varepsilon})\subset [a,b]$ such that $(a_{\varepsilon},b_{\varepsilon}) \subset B_{\varepsilon}$. Now using this fact we have that $$\int_{a}^{b}{f(x)^{n} \, dx}\geq  \int_{a_{\varepsilon}}^{b_{\varepsilon}}{f(x)^{n} \, dx} \geq (M-\varepsilon)^{n}(b_{\varepsilon}-a_{\varepsilon})$$ Now taking the limit we obtain $$\liminf_{n \rightarrow \infty}\,\left(\int_{a}^{b}{f(x)^{n}}\right)^{1/n} \geq M-\varepsilon$$ Now since $\varepsilon>0$ was arbitrary, we finally conclude that $$\lim_{n \rightarrow \infty}\, \left(\int_{a}^{b}{f(x)^{n} \, dx}\right)^{1/n}= M$$
