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$4.E.$ 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?

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

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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.

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  • $\begingroup$ 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? $\endgroup$ – Math enthusiast Sep 22 '18 at 14:34
  • $\begingroup$ 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 $\psi_n$. $\endgroup$ – mojobask Sep 22 '18 at 14:43
  • $\begingroup$ 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. $\endgroup$ – Matematleta Sep 22 '18 at 15:16
  • $\begingroup$ 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. $\endgroup$ – mojobask Sep 22 '18 at 15:22
  • $\begingroup$ 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. $\endgroup$ – Math enthusiast Sep 22 '18 at 15:47

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