# Exercise 4.E - The Elements of Integration and Lebesgue Measure by Bartle

$$4.E.$$ Let $$f,g\in M^{+},$$ let $$\omega\in M^{+}$$ be a simple function such that $$\omega\leq f+g$$ and let $$\phi_{n}(x)=\sup\{(m/n)\omega(x): 0\leq m\leq n, (m/n)\omega(x)\leq f(x)\}$$. Also let $$\psi_{n}(x)=\sup\{(1-\frac{1}{n})\omega(x)-\phi_{n}(x),0\}$$. Show that $$(1-\frac{1}{n})\omega\leq\psi_{n}+\phi_{n}$$, $$\phi_{n}\leq f$$, $$\psi_{n}\leq g$$.

I was able to prove the first two inequalities, it's just use the definition of $$\phi_n$$ and $$\psi_n$$ and observe that $$f$$ is an upper bound for the set $$\{(m/n)\omega(x): 0\leq m\leq n, (m/n)\omega(x)\leq f(x)\}$$, but I'm stuck in prove that $$\psi_n \leq g$$. Can anyone give me a hint in order to prove this inequality?

$$\textbf{P.S.: read the comments of mojobask's answer.}$$

This holds when $$\omega \gt f$$ : $$\omega/n \ge f - \phi_n \\ \omega/n + \phi_n + \psi_n \ge f + \psi_n \\ (1/n + (1-1/n)) \omega \ge f + \psi_n \\ w \ge f + \psi_n \\ f + g \ge w \ge f + \psi_n$$ Otherwise $$\psi_n = 0$$ and it holds trivially.
• I tried separate in two cases: $\left( 1 - \frac{1}{n} \right) \omega < f(x)$ and $\left( 1 - \frac{1}{n} \right) \omega \geq f(x)$, but I couldn't realize why is valid the first and third inequalities that you put are valid when I tried solve this problem. Can you explain with more details this inequalities? – Math enthusiast Sep 22 '18 at 14:34
• By definition $\phi_n$ is less than a fraction of $\omega$ away from $f$ (whenever $\omega \ge f$ as $m$ may not exceed $n$). The third is what you get by adding $\phi_n$ through the supremum in $\psi_n$. – mojobask Sep 22 '18 at 14:43
• I still do not see how the third line follows from the second: in fact, since, $\phi_n+\psi_n\ge (1-1/n)\omega$, the inequality goes in the wrong direction. – Matematleta Sep 22 '18 at 15:16
• But under the condition $\omega \gt f$ it does hold that $\phi_n + \psi_n = (1-1/n)\omega$ , though not for $\omega = f$ which is my mistake. – mojobask Sep 22 '18 at 15:22
• I think that I understood the third inequality, it's because if $\left( 1 - \frac{1}{n} \right) \omega > f$, then $$\left( 1 - \frac{1}{n} \right) \omega > f \geq \phi_n \Longrightarrow \left( 1 - \frac{1}{n} \right) \omega - \phi_n > 0 \Longrightarrow \psi_n = \left( 1 - \frac{1}{n} \right) \omega - \phi_n \Longrightarrow \phi_n + \psi_n = \left( 1 - \frac{1}{n} \right) \omega$$ This justify the third inequality. – Math enthusiast Sep 22 '18 at 15:47