Let $f : E \to \Bbb{R}$ be a bounded measurable function. Show that there are sequences of simple functions on $E$, $\{\varphi_n \}$ and $\{\phi_n \}$, such that one is increasing an the other is decreasing, and both of which converge uniformly to $f$ on $E$.
I realize this has been asked on this website, but all of the solutions offered seemed rather complicated. So I want to present my solution which I open up to criticism (please do, especially if you think it is wrong)
Since $f$ is bounded and measurable, we know by the simple approximation theorem that, for each $n \in \Bbb{N}$, there exist simple functions $\varphi_n$ and $\phi_n$ such that $\varphi_n \le f \le \phi_n$ and $0 \le \phi_n - \varphi_n < \frac{1}{n}$ on $E$. Define the functions $h_n := \max \{\varphi_1,...,\varphi_n\}$ and $g_n := \min \{\phi_1,...,\phi_n\}$. Clearly $h_n$ and $g_n$ are simple functions since their ranges are a finite union of finite sets, and $\{h_n\}$ and $\{g_n\}$ are increasing and decreasing, respectively, sequences of functions. Moreover, since $\varphi_n \le f$ and $f \le \phi_k$ for every $n,k \in \Bbb{N}$, we see that $\varphi_n \le \phi_k$ and therefore $h_n \le g_n$.
Hence, we have $h_n \le f \le g_n$ and $0 \le g_n - h_n < \frac{1}{n}$. From this we get $0 \le f - h_n \le g_n - h_n < \frac{1}{n}$ or $|f(x)-h_n(x)| < \frac{1}{n}$ for all $n$ and $x \in E$, implying that $h_n \to f$ uniformly. Similarly, from $h_n \le f \le g_n$ we get $-\frac{1}{n} < h_n - g_n \le f - g_n \le 0$ or $\vert g_n - f\vert < \frac{1}{n}$, which leads to the conclusion that $g_n \to f$ uniformly.
