# Simple function

Short description: Function that attains finitely many values

In the mathematical field of real analysis, a simple function is a real (or complex)-valued function over a subset of the real line, similar to a step function. Simple functions are sufficiently "nice" that using them makes mathematical reasoning, theory, and proof easier. For example, simple functions attain only a finite number of values. Some authors also require simple functions to be measurable; as used in practice, they invariably are.

A basic example of a simple function is the floor function over the half-open interval [1, 9), whose only values are {1, 2, 3, 4, 5, 6, 7, 8}. A more advanced example is the Dirichlet function over the real line, which takes the value 1 if x is rational and 0 otherwise. (Thus the "simple" of "simple function" has a technical meaning somewhat at odds with common language.) All step functions are simple.

Simple functions are used as a first stage in the development of theories of integration, such as the Lebesgue integral, because it is easy to define integration for a simple function and also it is straightforward to approximate more general functions by sequences of simple functions.

## Definition

Formally, a simple function is a finite linear combination of indicator functions of measurable sets. More precisely, let (X, Σ) be a measurable space. Let A1, ..., An ∈ Σ be a sequence of disjoint measurable sets, and let a1, ..., an be a sequence of real or complex numbers. A simple function is a function $\displaystyle{ f: X \to \mathbb{C} }$ of the form

$\displaystyle{ f(x)=\sum_{k=1}^n a_k {\mathbf 1}_{A_k}(x), }$

where $\displaystyle{ {\mathbf 1}_A }$ is the indicator function of the set A.

## Properties of simple functions

The sum, difference, and product of two simple functions are again simple functions, and multiplication by constant keeps a simple function simple; hence it follows that the collection of all simple functions on a given measurable space forms a commutative algebra over $\displaystyle{ \mathbb{C} }$.

## Integration of simple functions

If a measure μ is defined on the space (X,Σ), the integral of f with respect to μ is

$\displaystyle{ \sum_{k=1}^na_k\mu(A_k), }$

if all summands are finite.

## Relation to Lebesgue integration

The above integral of simple functions can be extended to a more general class of functions, which is how the Lebesgue integral is defined. This extension is based on the following fact.

Theorem. Any non-negative measurable function $\displaystyle{ f\colon X \to\mathbb{R}^{+} }$ is the pointwise limit of a monotonic increasing sequence of non-negative simple functions.

It is implied in the statement that the sigma-algebra in the co-domain $\displaystyle{ \mathbb{R}^{+} }$ is the restriction of the Borel σ-algebra $\displaystyle{ \mathfrak{B}(\mathbb{R}) }$ to $\displaystyle{ \mathbb{R}^{+} }$. The proof proceeds as follows. Let $\displaystyle{ f }$ be a non-negative measurable function defined over the measure space $\displaystyle{ (X, \Sigma,\mu) }$. For each $\displaystyle{ n\in\mathbb N }$, subdivide the co-domain of $\displaystyle{ f }$ into $\displaystyle{ 2^{2n}+1 }$ intervals, $\displaystyle{ 2^{2n} }$ of which have length $\displaystyle{ 2^{-n} }$. That is, for each $\displaystyle{ n }$, define

$\displaystyle{ I_{n,k}=\left[\frac{k-1}{2^n},\frac{k}{2^n}\right) }$ for $\displaystyle{ k=1,2,\ldots,2^{2n} }$, and $\displaystyle{ I_{n,2^{2n}+1}=[2^n,\infty) }$,

which are disjoint and cover the non-negative real line ($\displaystyle{ \mathbb{R}^{+} \subseteq \cup_{k}I_{n,k}, \forall n \in \mathbb{N} }$).

Now define the sets

$\displaystyle{ A_{n,k}=f^{-1}(I_{n,k}) \, }$ for $\displaystyle{ k=1,2,\ldots,2^{2n}+1, }$

which are measurable ($\displaystyle{ A_{n,k}\in \Sigma }$) because $\displaystyle{ f }$ is assumed to be measurable.

Then the increasing sequence of simple functions

$\displaystyle{ f_n=\sum_{k=1}^{2^{2n}+1}\frac{k-1}{2^n}{\mathbf 1}_{A_{n,k}} }$

converges pointwise to $\displaystyle{ f }$ as $\displaystyle{ n\to\infty }$. Note that, when $\displaystyle{ f }$ is bounded, the convergence is uniform.