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Let $(X,\mathcal A, \mu)$ be a measure space and let $E_n=\{x\in X:\frac{1}{n}\leq |f(x)|\leq n\}$ for all $n\in \mathbb N$. If $f\in L^p$, then it can be shown that $E_n$ has finite measure for each $n\in \mathbb N$. How to show that $\lim\limits_{n\to \infty}(\int\limits_{E_n}|f|^pd\mu)^{1/p}=\|f\|_p$?

One way is obvious. $\|f\|_p=(\int\limits_X|f|^pd\mu)^{1/p}\geq (\int\limits_{E_n}|f|^pd\mu)^{1/p}$ for all $n\in \mathbb N$ and so $\lim\limits_{n\to \infty}(\int\limits_{E_n}|f|^pd\mu)^{1/p}\leq\|f\|_p$. But I got stuck to show the reverse inequality. Any help will be appreciated in this regard.

2 Answers 2

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$\newcommand{\abs}[1]{\lvert{#1}\rvert} \newcommand{\dd}{\mathrm{d}}$I think that since $E_n\subseteq E_{n+1}$ for all $n\in\mathbb{N}$ and $\abs{f}^p$ is obviously non-negative, you have $$ \int_{E_n}\abs{f}^p\,\dd\mu\le\int_{E_{n+1}}\abs{f}^p\,\dd\mu $$ from which $$ \biggl(\int_{E_n}\abs{f}^p\,\dd\mu\biggr)^\frac1{p}\le\biggl(\int_{E_{n+1}}\abs{f}^p\,\dd\mu\biggr)^\frac1{p} $$ follows; this shows that the sequence $\bigl(\int_{E_n}\abs{f}^p\,\dd\mu\bigr)^\frac1{p}$ is monotonically increasing. It is also bounded from above, as you proved, so it converges, and the limit is the upper bound, that is $\lVert f\rVert_p$.

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There is also a slick proof:

We have $E_n \uparrow X$, therefore $\nu(E_n) \uparrow \nu(X)$ by the Continuity of measures, where $\nu(A) := \int_A |f|^p\,d\mu$. Then you apply $\sqrt[p]{}$ to both sides and you are done.