Prove that $\lim_{n\to \infty} [\int_E f^n (x) dx)]^{1/n} = M$ where $M = \sup_{x\in E} f (x)$

Question

In the book of Mathematical Analysis II, by Zorich at page 128, it is given that

Let $E$ be a Jordan-measurable set of nonzero measure, $f : E → \mathbb{R}^n$ a continuous nonnegative integrable function on $E$, and $M = \sup_{x\in E} f (x)$. Prove that

$$\lim_{n\to \infty} [\int_E f^n (x) dx)]^{1/n} = M$$

I have already proved that $$\lim_{n\to \infty} [\int_E f^n (x) dx)]^{1/n} \leq M$$ since $E$ is Jordan measurable and has nonzero measure, but I'm having trouble proving the reverse of that statement, so any help is appreciated.

Edit:

I'm new to this topic, and I even do not understand the premise of the asked question, so I would appreciate if you could post an answer that is formulated in terms of the statement that I have given in the question.

Answer 1

Take $\epsilon > 0$. Let $F = \{x : f(x) > M-\epsilon\}$. Then,

$[\int_E f^n(x)dx]^{1/n} \ge [\int_F f^n(x)dx]^{1/n} \ge [(M-\epsilon)^n \mu(F)]^{1/n} = (M-\epsilon) \mu(F)^{1/n}$.

Since $f$ is continuous, $\mu(F) > 0$. Since $f$ is integrable, $\mu(F) < \infty$. So, if we let $n \to \infty$, we see that we can get a lower bound of $M-\epsilon$ for all $\epsilon > 0$.