Let $f$ be a bounded measurable function on $E$. Show that there are sequences of simple functions converging uniformly on $E$.

Let $f$ be a bounded measurable function on $E$. Show that there are sequences of simple functions on $E$, $\{\phi_n\}$ and $\{\psi_n\}$, such that $\{\phi_n\}$ is increasing and $\{\psi_n\}$ is decreasing and each of these sequences converges uniformly on $E$.

Let $f:E \rightarrow \mathbb{R}$ be measurable. There exist simple measurable functions $\phi_n$ on $E$ such that $$(a) \ \phi_1 \leq \phi_2 \leq \cdots \leq f.$$ $$(b) \ \forall x\in E, \text{ we have that } \phi_n(x) \rightarrow f(x), \text{ as }n \rightarrow \infty$$ Proof:

Notice that for each positive integer $n$ and each real number $t$ corresponds a unique integer $k = k_n(t)$ that satisfies $k2^{-n} \leq t < (k+1)2^{-n}$. Define: $$ s_n(t) = \left\{\begin{matrix} k_n(t)2^{-n} & 0 \leq t < n\\ n & n \leq t \leq \infty \end{matrix}\right. $$ Note that each $s_n$ is a borel function on $[0,\infty]$. In other words, for any open set $V$, $s_n^{-1}(V)$ is a borel set. Now, $$k2^{-n} \leq t < (k+1)2^{-n} \Rightarrow t - 2^{-n} < s_n(t) \leq t \text{ if } 0 \leq t\leq n,$$ thus, $0 \leq s_1 \leq s_2 \leq \cdots \leq t,$ and $s_n(t) \rightarrow t$ as $n \rightarrow \infty,$ for every $t\in [0,\infty]$. It follows that the function $\phi_n = s_n\circ f$ satisfy (a) and (b); since $f$ is measurable and $s_n$ is a borel function, then $\phi_n$ is also measurable. To obtain a decreasing function, let $\psi_n = -s_n(-f)$, thus $\phi_n$ and $\psi_n$ are steps functions, $ \phi_n \leq f\leq \psi_n$ and $\psi_n-\phi_n \leq 2^{-n}$ for every integer $n$.

My question is how I can obtain uniform convergence? I know that from what i proved, I didn't needed to assume that $f$ is bounded.

The following theorem might be useful, namely Dini's Theorem(obtained from wikipedia).

If $Y$ is a compact topological space, and $\{f_n\}_{n\in \mathbb{N}}$ is a monotonically increasing sequence (meaning $f_n(x) \leq f_{n+1}(x)$ for all $n$ and $x$) of continuous real-valued functions on $X$ which converges pointwise to a continuous function $f$, then the convergence is uniform. The same conclusion holds if $\{f_n\}_{n\in \mathbb{N}}$ is monotonically decreasing instead of increasing.

Is there a way to use this theorem under the assumption that $f$ is bounded to obtain uniform convergence?