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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.

Our
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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 $\epsilon > 0$.

mathworker21
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  • Thanks for your answer. I have some questions. First of all, How does the fact that $f$ is continuous implies $\mu(F) > 0$ ? – Our Feb 02 '18 at 12:33
  • Secondly, how does $f$ is integrable imply that $\mu(F) < \infty.$ ? – Our Feb 02 '18 at 12:34
  • I was hoping you would fill in those details. For continuity, if you are greater than $M-\frac{\epsilon}{2}$ at some point, you are greater than $M-\epsilon$ in a neighborhood around that point. For integrability, if you are greater than $M-\epsilon$ on a set with infinite measure.... that's bad – mathworker21 Feb 02 '18 at 12:39
  • I'm sorry, I'm new to this topic, so I would appreciate if you could be more specific. – Our Feb 02 '18 at 12:40
  • More specific about what? If you are doing measure theory, you should already be comfortable with continuity. For the integrability argument, note $\infty > \int_F f^(x)dx \ge (M-\epsilon)^n \mu(F)$, so of course $\mu(F)$ can't be infinite. (Think intuitively - if you have a function which is bounded away from 0 for a set with infinite measure, then the integral will be infinite) – mathworker21 Feb 02 '18 at 12:44
  • I'm not doing measure theory, please see the book that I'm using. – Our Feb 02 '18 at 12:45
  • By the way, how do we know that $f$ is integrable on $F$, I mean we need to know that the boundary of $F$ is Jordan measure zero because $\int_F f dx = \int_{F \subset I} f\chi_F dx$, and $f\chi_F$ is discontinuous on $\partial F$. – Our Feb 03 '18 at 07:16