# A Lebesgue Integrable Function

let $$f \geq 0$$ be a finite $$a.e.$$ on $$E$$ with finite measure. Let $$\epsilon > 0$$ and consider the following partition of $$\mathbb{R}_{\geq 0}$$ for $$n\geq 0$$

$$0 = x_0 < x_1 < ... < x_n < ...$$

with $$\ \ \epsilon > \sup\limits_{n\geq 0}(x_{n+1} - x_{n})$$

Now define $$E_n = \{x \in E: x_n \leq f(x) < x_{n+1} \}$$ for $$n\geq 0$$

Prove that $$f$$ is Lebesgue integrable $$\iff \sum\limits_{n=0}^{\infty} x_n m(E_n)$$ is finite and that

$$\lim\limits_{\epsilon \rightarrow 0+}\sum\limits_{n=0}^{\infty} x_n m(E_n)=\int_E \! f(x) \, \mathrm{d}x$$

I can see why and understand it but I need help with the formulations and correct definitions.

So $$f$$ is a non-neg. Lebesgue integrable function, then its integral over $$E$$ is finite.

Since all $$E_n$$ are disjoint we have:

$$\int_E \! f(x) \, \mathrm{d}x = \int_{E_0} \! f(x) \, \mathrm{d}x + \int_{E_1} \! f(x) \, \mathrm{d}x \ + ...= \int_{E_0} \! f(x) \, \mathrm{m(dx)} + \int_{E_1} \! f(x) \, \mathrm{m(dx)} \ + ... \\ \Longrightarrow \int_E \! f(x) \, \mathrm{d}x =\int_{E_0} \! f(x) \, x_0 + \int_{E_1} \! f(x) \, x_1 \ + ...$$

I don't know how to translate the integral of $$f$$ in each part of the partition $$(x_n, x_{n+1})$$ into the measure of the $$E_n$$'s

• Hint: On each $E_n$ we have $x_{n+1}\geq f(x)\geq x_n$ and hence $x_{n+1}m(E_n)\geq \int_{E_n}f(x)m(dx)\geq x_n m(E_n)$ Commented Jan 28, 2023 at 8:58
• @fKonrad then $x_0 m(E_0) + x_1 m(E_1)+... \leq \int_{E_0} \! f(x) \, \mathrm{m(dx)} + \int_{E_1} \! f(x) \, \mathrm{m(dx)} + ... \ \leq x_1 m(E_1) + x_2 m(E_2)...$ hence the inside Lebesgue integral equals both sides.
– ISO
Commented Jan 28, 2023 at 9:33
• How does $\lim\limits_{\epsilon \rightarrow 0+}\sum\limits_{n=0}^{\infty} x_n m(E_n)$ affect $\int_E \! f(x) \, m(dx)$
– ISO
Commented Jan 28, 2023 at 9:42
• Your $E_n$ are not disjoint. Replace $f(x)\le x_{n+1}$ by $f(x)<x_{n+1}$ in your definition of $E_n.$ Commented Jan 28, 2023 at 10:15
• Because if the $E_n$ are not disjoint, you don't have $\int f\,\mathrm dm=\sum\int_{E_n}f\,\mathrm dm.$ Commented Jan 28, 2023 at 22:00

$$\sum x_nm(E_n)\le\int f\,\mathrm dm\le\sum x_{n+1}m(E_n)\le\sum(x_n+\epsilon)m(E_n)\le\epsilon\cdot m(E)+\sum x_nm(E_n).$$