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Let $f : \Bbb R \longrightarrow [0,\infty)$ be a Borel measurable function. Show that $$\displaystyle{\sum\limits_{n = 1}^{\infty} \text {m} \left (\left \{f \gt n \right \} \right ) \leq \int f\ \text {dm} \leq \sum\limits_{n = 1}^{\infty} \text {m} \left (\left \{f \geq n \right \} \right ) },$$ where $\text {m}$ denotes the Lebesgue measure.

Approximating $f$ by a non-negative sequence of increasing simple measurable functions I find that $$\begin{align*} \int f\ \text {dm} & = \lim\limits_{n \to \infty} \left [\sum\limits_{k = 0}^{n 2^n - 1} \dfrac {k} {2^n}\ \text {m} \left (\left \{\dfrac {k} {2^n} \leq f \lt \dfrac {k + 1} {2^n} \right \} \right ) + n\ \text {m} \left (\left \{f \geq n \right \} \right ) \right ]. \end{align*}$$

How do I proceed now? Any help will be highly appreciated.

Thanks in advance.

Source $:$ ISI (Indian Statistical Institute) PhD entrance test in Mathematics held in $20$th September this year, TEST CODE $:$ MTA (FORENOON SESSION), Question No. $7.$

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    $\begingroup$ If $0 \leq f <1$ the inequality fails. $\endgroup$
    – Kavi Rama Murthy
    Commented Nov 2, 2020 at 8:24
  • $\begingroup$ You are making mess of things by omitting summation over $n$. $\endgroup$
    – Kavi Rama Murthy
    Commented Nov 2, 2020 at 8:26
  • $\begingroup$ What does that prove? Does it prove what you are asked to prove? $\endgroup$
    – Kavi Rama Murthy
    Commented Nov 2, 2020 at 8:33
  • $\begingroup$ @KaviRamaMurthy yes. For example, I don't see how it holds for $f=1/2$. $\endgroup$
    – mathemather
    Commented Jul 14, 2021 at 11:36
  • $\begingroup$ @mathmather for $f \equiv \frac {1} {2}$ we have $0$ everywhere. Isn't it so? $\endgroup$
    – Anacardium
    Commented Jul 15, 2021 at 16:51

1 Answer 1

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Since $$ \int f\,dm=\int_{0}^{\infty}m(\{f>x\})\,dx=\sum_{n\ge 0}\int_{n}^{n+1}m(\{f>x\})\,dx $$ and $$ m(\{f>n+1\})\le\int_{n}^{n+1}m(\{f>x\})\,dx\le m(\{f>n\}), $$ one gets $$ \sum_{n\ge 1}m(\{f>n\})\le\int_{0}^{\infty}m(\{f>x\})\,dx\le \sum_{n\ge 0}m(\{f>n\}). $$

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  • $\begingroup$ I have checked your arguments for indicator functions. How do I extend it to non-negative simple measurable functions? To show those results for any non-negative measurable function we need to first approximate the function by an increasing sequence of simple measurable function and then apply monotone convergence theorem. But applying it how do I get the result? You should add some more things to it so that it becomes accesible. May I know which book do you follow? $\endgroup$
    – Anacardium
    Commented Nov 2, 2020 at 16:00
  • $\begingroup$ \begin{align} \int f\, dm &=\int\left(\int_0^{\infty} 1\{f>x\}\, dx\right)dm \\ &=\int_0^{\infty}\left(\int 1\{f>x\}\, dm\right)dx \\ &=\int_0^{\infty} m(\{f>x\})\, dx \end{align} $\endgroup$
    – user140541
    Commented Nov 2, 2020 at 18:02
  • $\begingroup$ What is $1 \left \{f \gt x \right \}\ $? Do you mean $\chi_{\left \{f \gt x \right \}}\ $? $\endgroup$
    – Anacardium
    Commented Nov 3, 2020 at 5:13
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    $\begingroup$ Indeed, it is a typo. Here is the correct result: $$ z=\int_0^z 1\, dx=\int_0^{\infty}\chi_{[0,z]}(x)\,dx=\int_0^{\infty}\chi_{[0,z)}(x)\,dx. $$ $\endgroup$
    – user140541
    Commented Nov 3, 2020 at 16:27
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    $\begingroup$ For $x\in [n, n+1]$, $\{f>n+1\}\subseteq \{f>x\}\subseteq \{f>n\}$. $\endgroup$
    – user140541
    Commented Nov 3, 2020 at 16:54

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