Measurable Function is Pointwise Limit of Simple Functions

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

Let $f: X \to \overline{\R}$ be a $\Sigma$-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:


 * $\forall x \in X: f \left({x}\right) = \displaystyle \lim_{n \to \infty} f_n \left({x}\right)$

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

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 $f$.

Proof
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:

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 Power 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 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.

Also see

 * Bounded Measurable Function Uniform Limit of Simple Functions, a strengthening when $f$ is bounded