Prove the convergence of a sequence involving integrals I need to prove the following:
Assume $f:[a,b] \to R$ is continous, $f(x)\leq0$ for all $x \in [a,b]$, and $M=sup\{f(x):x \in [a,b]\}$.Show that:
$$\{[\int [f(x)]^{n}dx]^{1/n}\} \to M$$
This result is clear when $f(x)=0$ because $M=0$ and the sequence of only zeros(since the integral is cero for all $a,b$ and $n$) converge to zero, but the other case, I do not know how to attached it because is a little bit confusing for me.Can you help me to prove this .I appreciate all your anwers, thank you I really need your help.  
I forgot something, which is this m in the step (M−ϵ)m(E)1/n≤[∫E|f(x)|ndx]1/n≤[∫ba|f(x)|ndx]1/n≤M(b−a)1/n. –  @RRL?, How can I do it with out the Lebesgue measure ? thanks 
 A: For every $\epsilon > 0$ there exists $x \in [a,b]$ such that $M - \epsilon < |f(x)| \leq M.$
Let E = $\{x \in [a,b]: M - \epsilon < |f(x)| < M\} = |f|^{-1}[( M - \epsilon,M))].$
Since $f$ is continuous, E is an open set and there is an interval $(\alpha,\beta) \subset E \subset [a,b]$.
Note that
$$\int_\alpha^{\beta} |f(x)|^ndx \leq  \int_a^b |f(x)|^ndx$$
Hence, 
$$(M-\epsilon)\left( \beta-\alpha\right)^{1/n}\leq \left[\int_{\alpha}^{\beta} |f(x)|^{n}dx\right]^{1/n}\leq\left[\int_a^b |f(x)|^{n}dx\right]^{1/n}\leq M(b-a)^{1/n}.$$
Now use the squeeze property as $n \rightarrow \infty$.
Then
$$(M- \epsilon)\lim_{n \rightarrow \infty}\left(\beta-\alpha\right)^{1/n} \leq \lim_{n \rightarrow \infty}\left[\int_a^b |f(x)|^{n}dx\right]^{1/n}\leq M \lim_{n \rightarrow \infty}(b-a)^{1/n},$$
and for every $\epsilon > 0$,
$$M- \epsilon \leq \lim_{n \rightarrow \infty}\left[\int_a^b |f(x)|^{n}dx\right]^{1/n}\leq M $$
Hence,
$$\lim_{n \rightarrow \infty}\left[\int_a^b |f(x)|^{n}dx\right]^{1/n} =  M $$
A: Hint: the $L^n$-norm converges to $L^\infty$-norm as $n\to +\infty$, i.e.
$\left( \int_a^b f^n \right)^{1/n}=\|f\|_{L^n}\to \|f\|_{L^\infty}...$
