# Simple function

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

${\displaystyle \sum _{k=1}^{n}a_{k}\mu (A_{k}),}$

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.