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. $7.$
Best Answer
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 $$ \sum_{n\ge 1}m(\{f>n\})\le\int_{0}^{\infty}m(\{f>x\})\,dx\le \sum_{n\ge 0}m(\{f>n\}). $$