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

Question

$4.E.$ 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?

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

Answer 1

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.

Answer 2

You can argue by contradiction. Suppose on the contrary that $g(x)<\psi_n(x)$ for some $x$ and some $n$. Then $\psi_n(x)$ cannot be zero since $g \in M^{+}$. This implies that $\psi_n(x)>0$ or equivalently $(1-\frac{1}{n})\omega(x)=\frac{n-1}{n}\omega(x)>\phi_n(x)$. This contradicts the definition of $\phi_n(x)$.

Answer 3

Fix $x \in X$ and $n \in \mathbb{Z}^+$. If $\psi_n(x)=0$, the result is immediate because then $g(x)\ge 0=\psi_n(p)$. So we may assume $\psi_n(x)= \left(1-\frac{1}{n}\right)\omega(x)-\phi_n(x)>0$. Recall that $\phi_n(x)$ is the supremum of a finite set, so there is a nonnegative integer $m_0\le n$ such that $\phi_n(x)=\frac{m_0}{n}\omega(x)$.

Observe that $$ \psi_n(x)= \left(1-\frac{1}{n}\right)\omega(x)-\phi_n(x) = \left(1-\frac{1}{n}-\frac{m_0}{n}\right)\omega(x) >0$$ can only be possible if $m_0 \le n-2$. This fact provides us with two key inequalities: $$ \frac{m_0+1}{n}\omega(x)>f(x)$$ and $$ \frac{m_0}{n}\omega(x)\le \phi_n(x).$$

The result follows now by some algebraic manipulation. Firstly, $$\begin{align} g(x)&\ge \omega(x)-f(x)\\&>\omega(x)-\frac{m_0+1}{n}\omega(x)\\ &=\frac{n-m_0-1}{n}\omega(x), \end{align} $$

and secondly,

$$\begin{align} \psi_n(x) & = \left(1-\frac{1}{n}\right)\omega(x)-\phi_n(x) \\ & \le \left(1-\frac{1}{n}\right)\omega(x)-\frac{m_0}{n}\omega(x) \\ & = \frac{n-m_0-1}{n}\omega(x) \\ &\le g(x), \end{align} $$

which concludes the proof.