First, let us prove the theorem when $f$ is a positive $\Sigma$-measurable function. Now for any $n \in \N$, define for $0 \le k \le n 2^n$: :$A^n_k := \begin{cases}  \left\{{ k 2^{-n} \le f < \left({k + 1}\right) 2^{-n} }\right\} & : k \ne n 2^n \\  \left\{{f \ge n}\right\} & : k = n 2^n \end{cases}$ where e.g. $\left\{{f \ge n}\right\}$ is short for $\left\{{x \in X: f \left({x}\right) \ge n}\right\}$. It is immediate that the $A^n_k$ are pairwise disjoint, and that: :$\displaystyle \bigcup_{k \mathop = 0}^{n 2^n} A^n_k = X$ Subsequently, define $f_n: X \to \overline{\R}$ by: :$f_n \left({x}\right) := \displaystyle \sum_{k \mathop = 0}^{n 2^n} k 2^{-n} \chi_{A^n_k} \left({x}\right)$ where $\chi_{A^n_k}$ is the characteristic function of $A^n_k$. Now if $f \left({x}\right) < n$, then we have for some $k < n 2^{-n}$: :$x \in A^n_k$ so that: {{begin-eqn}} {{eqn | l = \left\vert{f \left({x}\right) - f_n \left({x}\right)}\right\vert       | r = \left\vert{f \left({x}\right) - k 2^{-n} }\right\vert }} {{eqn | o = <       | r = 2^{-n} }} {{end-eqn}} since $x \in A^n_k$ {{iff}} $k 2^{-n} \le f \left({x}\right) < \left({k + 1}\right) 2^{-n}$. In particular, since $f_n \left({x}\right) \le n$ for all $x \in X$, we conclude that pointwise, $f_n \le f$, for all $n \in \N$. By Characterization of Measurable Functions and Sigma-Algebra Closed under Intersection, it follows that: :$A^n_{n 2^n} = \left\{{f \ge n}\right\}$ :$A^n_k = \left\{{f \ge k 2^{-n}}\right\} \cap \left\{{f < \left({k + 1}\right) 2^{-n}}\right\}$ are all $\Sigma$-measurable sets. Hence, by definition, all $f_n$ are $\Sigma$-simple functions. It remains to show that $\displaystyle \lim_{n \to \infty} f_n = f$ uniformly. Let $\epsilon > 0$ be arbitrary. Let $n \in \N$ be a bound for $f$ satisfying $2^{-n} < \epsilon$. By the reasoning above, we then have for all $m \ge n$, and all $x \in X$: :$\left\vert{f \left({x}\right) - f_m \left({x}\right)}\right\vert < 2^{-n}$ This establishes the result for positive measurable $f$. For arbitrary $f$, by Difference of Positive and Negative Parts, we have: :$f = f^+ - f^-$ where $f^+$ and $f^-$ are the positive and negative parts of $f$. By Function Measurable iff Positive and Negative Parts Measurable, $f^+$ and $f^-$ are positive measurable functions. Thus we find sequences $f^+_n$ and $f^-_n$ converging uniformly to $f^+$ and $f^-$, respectively. That is, given $\epsilon > 0$, there exist $n, n'$ such that: :$\forall m \ge n: \forall x \in X: \left\vert{f^+ \left({x}\right) - f^+_m \left({x}\right)}\right\vert < \epsilon$ :$\forall m \ge n': \forall x \in X: \left\vert{f^- \left({x}\right) - f^-_m \left({x}\right)}\right\vert < \epsilon$ By the Triangle Inequality, for all $m \in \N$: :$\left\vert{\left({f^+ \left({x}\right) - f^- \left({x}\right)}\right) - \left({f^+_m \left({x}\right) - f^-_m \left({x}\right)}\right)}\right\vert \le \left\vert{f^+ \left({x}\right) - f^+_m \left({x}\right)}\right\vert + \left\vert{f^- \left({x}\right) - f^-_m \left({x}\right)}\right\vert$ Hence, given $\epsilon > 0$, if $n, n'$ are sufficient for $\epsilon / 2$ for $f^+_m$ and $f^-_m$ respectively, it follows that for $m \ge \max \left\{{n, n'}\right\}$: :$\left\vert{f \left({x}\right) - f_m \left({x}\right)}\right\vert = \left\vert{\left({f^+ \left({x}\right) - f^- \left({x}\right)}\right) - \left({f^+_m \left({x}\right) - f^-_m \left({x}\right)}\right)}\right\vert < \epsilon$ where $f_m \left({x}\right) = f^+_m \left({x}\right) - f^-_m \left({x}\right)$. Thus $f_m$ converges to $f$ uniformly. Furthermore, we have for all $n \in \N$ and $x \in X$: :$\left\vert{f^+_n \left({x}\right) - f^-_n \left({x}\right)}\right\vert = f^+_n \left({x}\right) + f^-_n \left({x}\right) \le f^+ \left({x}\right) + f^- \left({x}\right) = \left\vert{f \left({x}\right)}\right\vert$  where the last equality follows from Sum of Positive and Negative Parts. Hence the result. {{qed}}