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 $\displaystyle \lim_{n \to \infty} f_n = f$ pointwise. Let $x \in X$ be arbitrary. If $f \left({x}\right) = +\infty$, then for all $n \in \N$, $x \in A^n_{n 2^n}$, so that: :$f_n \left({x}\right) = n$ Now clearly, $\displaystyle \lim_{n \to \infty} n = +\infty$, showing convergence for these $x$. If $f \left({x}\right) < +\infty$, then for some $n \in \N$, $f \left({x}\right) < n$. By the reasoning above, we then have for all $m \ge n$: :$\left\vert{f \left({x}\right) - f_m \left({x}\right)}\right\vert < 2^{-m}$ which by Sequence of Powers of Number less than One implies that: : $\displaystyle \lim_{n \to \infty} f_n \left({x}\right) = f \left({x}\right)$ Thus $\displaystyle \lim_{n \to \infty} f_n = f$ pointwise. 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 pointwise to $f^+$ and $f^-$, respectively. The Sum Rule for Real Sequences implies that for all $x \in X$: :$\displaystyle \lim_{n \to \infty} f^+_n \left({x}\right) - f^-_n \left({x}\right) = f^+ \left({x}\right) - f^- \left({x}\right) = f \left({x}\right)$ 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}}