Bounded Measurable Function is Uniform Limit of Simple Functions

Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f: X \to \overline{\R}$ be a bounded $\Sigma$-measurable function.

Then there exists a sequence $\left({f_n}\right)_{n \in \N} \in \mathcal E \left({\Sigma}\right)$ of simple functions, such that:


 * $\forall \epsilon > 0: \exists n \in N: \forall \in X: \left\vert{f \left({x}\right) - f_n \left({x}\right)}\right\vert < \epsilon$

That is, such that $f = \displaystyle \lim_{n \to \infty} f_n$ uniformly.

The sequence $\left({f_n}\right)_{n \in \N}$ may furthermore be taken to satisfy:


 * $\forall n \in \N: \left\vert{f_n}\right\vert \le \left\vert{f}\right\vert$

where $\left\vert{f}\right\vert$ denotes the absolute value of $f$.

Also see

 * Measurable Function Pointwise Limit of Simple Functions, a similar result when $f$ is not bounded