# Simple function

In the mathematical field of real analysis, a simple function is a real-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.) Note also that 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 a 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 measurable sets, and let a1, ..., an be a sequence of real or complex numbers. A simple function is a function $f:X\to \mathbb {C}$ of the form

$f(x)=\sum _{k=1}^{n}a_{k}{\mathbf {1} }_{A_{k}}(x),$ ## 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 $\mathbb {C}$ .

## Integration of simple functions

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

$\sum _{k=1}^{n}a_{k}\mu (A_{k}),$ if all summands are finite.

## Relation to Lebesgue integration

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

Now define the measurable sets

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

Then the increasing sequence of simple functions

$f_{n}=\sum _{k=1}^{2^{2n}+1}{\frac {k-1}{2^{n}}}{\mathbf {1} }_{A_{n,k}}$ converges pointwise to $f$ as $n\to \infty$ . Note that, when $f$ is bounded, the convergence is uniform. This approximation of $f$ by simple functions (which are easily integrable) allows us to define an integral $f$ itself; see the article on Lebesgue integration for more details.