Proposition
Let $f$ be a bounded measurable function on $E$.
Show that there are $\{\phi_n(x)\}$ and $\{\psi_n(x)\}$ – sequences of simple functions on $E$ such that $\{\phi_n(x)\}$ is increasing and $\{\psi_n(x)\}$ is decreasing and each of these converge to $f$ uniformly on $E$.
By the simple approximation lemma I know that both of the mentioned sequences exists such that $\phi_n(x) \leq f \leq \psi_n(x)$ and $\psi_n(x)-\phi_n(x) < \frac{1}{n}$ for each $n \in \mathbb{N}$.
Using uniform continuity: for some $\varepsilon > 0$ there exists an $n \geq N$ such that $|\phi_n(x)-f| < \frac{1}{n}$. Since simple functions take on a finite number of values, it is reasonable to take a $\max$. So I let $\phi(x)=\max\{\phi_n(x)\}$.
I guess I'm not sure how to pull together the lemma and the definition of simple function.
Best Answer
For every nonnegative integer $n$, define $A_n:\mathbb R\to\mathbb R$ by $A_n(t)=k2^{-n}$ for every $t$ such that $k\leqslant2^nt\lt k+1$, for some integer $k$ (in other words, $2^nA_n(t)$ is the integer part of $2^nt$).
Let $\phi_n=A_n(f)$ and $\psi_n=-A_n(-f)$. Then $\phi_n$ and $\psi_n$ are step functions, $\phi_n\leqslant f\leqslant \psi_n$, and $\psi_n-\phi_n\leqslant2^{-n}$ for every $n$. Furthermore, the sequence $(\phi_n)_n$ is nondecreasing and the sequence $(\psi_n)_n$ is nonincreasing.