Measurable Function is Pointwise Limit of Simple Functions

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

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

Then there exists a sequence $\sequence {f_n}_{n \mathop \in \N} \in \map \EE \Sigma$ of simple functions, such that:


 * $\forall x \in X: \map f x = \ds \lim_{n \mathop \to \infty} \map {f_n} x$

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

The sequence $\sequence {f_n}_{n \mathop \in \N}$ may furthermore be taken to satisfy:


 * $\forall n \in \N: \size {f_n} \le \size f$

where $\size f$ 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_k}^n := \begin{cases}

\set {k 2^{-n} \le f < \paren {k + 1} 2^{-n} } & : k \ne n 2^n \\ \set {f \ge n} & : k = n 2^n \end{cases}$

where for example $\set {f \ge n}$ is short for $\set {x \in X: \map f x \ge n}$.

It is immediate that the ${A_k}^n$ are pairwise disjoint, and that:


 * $\ds \bigcup_{k \mathop = 0}^{n 2^n} {A_k}^n = X$

Subsequently, define $f_n: X \to \overline \R$ by:


 * $\map {f_n} x := \ds \sum_{k \mathop = 0}^{n 2^n} k 2^{-n} \map {\chi_{ {A_k}^n} } x$

where $\chi_{ {A_k}^n}$ is the characteristic function of ${A_k}^n$.

Now if $\map f x < n$, then we have for some $k < n 2^{-n}$:


 * $x \in {A_k}^n$

so that:

since $x \in {A_k}^n$ $k 2^{-n} \le \map f x < \paren {k + 1} 2^{-n}$.

In particular, since $\map {f_n} x \le n$ 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:


 * ${A_{n 2^n} }^n = \set {f \ge n}$


 * ${A_k}^n = \set {f \ge k 2^{-n} } \cap \set {f < \paren {k + 1} 2^{-n} }$

are all $\Sigma$-measurable sets.

Hence, by definition, all $f_n$ are $\Sigma$-simple functions.

It remains to show that $\ds \lim_{n \mathop \to \infty} f_n = f$ pointwise.

Let $x \in X$ be arbitrary.

If $\map f x = +\infty$, then for all $n \in \N$, $x \in {A_{n 2^n} }^n$, so that:


 * $\map {f_n} x = n$

Now clearly, $\ds \lim_{n \mathop \to \infty} n = +\infty$, showing convergence for these $x$.

If $\map f x < +\infty$, then for some $n \in \N$, $\map f x < n$.

By the reasoning above, we then have for all $m \ge n$:


 * $\size {\map f x - \map {f_m} x} < 2^{-m}$

which by Sequence of Powers of Number less than One implies that:
 * $\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$

Thus $\ds \lim_{n \mathop \to \infty} f_n = f$ 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 $x \in X$:


 * $\ds \lim_{n \mathop \to \infty} \map { {f_n}^+} x - \map { {f_n}^-} x = \map {f^+} x - \map {f^-} x = \map f x$

Furthermore, we have for all $n \in \N$ and $x \in X$:

Hence the result.

Also see

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