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Let $(X,\mathfrak{X},\mu)$ be a measure space and let $f \in L^p(\mu)$. Show that \begin{equation} \mu(|f| \geq n)\rightarrow 0 \end{equation}

as $n \rightarrow \infty$.

Since $f \in L^p(\mu)$, we know that: \begin{equation*} ||f||_{p} = \left(\int_X |f|^p \,d\mu\right)^{1/p} < \infty \end{equation*}

It is clear that if $m<n$ then \begin{equation} \mu(|f| \geq m) \geq \mu(|f|\geq n) \end{equation}

as \begin{equation} |f(x)|\leq m \Rightarrow |f(x)|\leq n \text{ whenever } m<n \end{equation} for $x \in X$.

Furthermore, we know that: \begin{equation} (||f||_{p})^p = \int_X |f|^p \,d\mu < \infty \end{equation}

However, I am unsure of how to proceed from here. Since $p$ is fixed, is it sufficient to claim thus, that there is some upper bound $M$ such that $|f(x)|\leq M$ for all $x\in X$ and hence when $n>M$ we have that $\{x \in X : |f(x)| \leq n \} = \varnothing$ which being empty, has measure zero?

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3 Answers 3

up vote 2 down vote accepted

Not exactly like you write in the end, $f$ may not be bounded and those sets may not be empty, neither of measure $0$, but.. something like that..

Hint: Define the set $A_n := (|f|\ge n)$ and its complement, and split the integral of $|f|^p$, and use $$\int_{A_n} |f|^pd\mu \ge \mu(A_n)\cdot n^p$$

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Define $E_n = \{x \in X : |f(x)| \ge n\}$. Notice that $\mu(E_1) < \infty$ and $E_1 \supset E_2 \supset \cdots$. What elements are in $\bigcap_{n=1}^\infty E_n$? How is it related to $\lim_{n \to \infty} \mu(E_n)$?

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How do you know that $\mu(E_1)<\infty$? The elements in the intersection are those for which $f(x)=\infty$. We should have that $\bigcap_{n=1}^\infty E_n=\lim_{n \rightarrow \infty} \mu(E_n)$, right? –  Jack Rousseau Sep 30 '12 at 9:44
    
@JackRousseau If $\mu(E_1) = \infty$, $f$ cannot be in $L^p(\mu)$. –  Ayman Hourieh Sep 30 '12 at 9:47
    
it is because what I wrote in the hint, $A_n=E_n$, and $\infty > \int_X|f|^pd\mu\ge\int_{E_n}|f|^p\ge\int_{E_n}n^p=\mu(E_n)\cdot n^p$. –  Berci Sep 30 '12 at 10:43
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To answer your last question, it is false that $f(x)$ has an upper bound $M$, because, for example, take $f$ which almost everywhere equals $x^{-1/3}$ on the interval $[-1,1]$ and zero elsewhere. Then $f \in L^2$ but is not bounded.

Without giving you the whole answer, consider the following intuition: if $\mu(|f| \geq n)$ does not become small as $n$ increases, then $f$ is large too often, and so the integral of $|f|^p$ will not be finite.

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Thanks, this is a nice example. –  Jack Rousseau Sep 30 '12 at 9:46
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