maths.free › Measure Theory › The integral › Simple function
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 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.
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 \(f\colon X \to \mathbb{C}\) of the form
\(f(x)=\sum_{k=1}^n a_k {\mathbf 1}_{A_k}(x),\)
where \({\mathbf 1}_A\) 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 \(\mathbb{C}\).
Integration of simple functions
If a measure \(\mu\) is defined on the space \((X, \Sigma)\), the integral of a simple function \(f\colon X \to \mathbb R\) with respect to \(\mu\) is defined to be
\(\int_X f d \mu = \sum_{k=1}^na_k\mu(A_k),\)
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 \(f\colon X \to\mathbb{R}^{+}\) 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 \(\mathbb{R}^{+}\) is the restriction of the Borel σ-algebra \(\mathfrak{B}(\mathbb{R})\) to \(\mathbb{R}^{+}\). The proof proceeds as follows. Let \(f\) be a non-negative measurable function defined over the measure space \((X, \Sigma,\mu)\). For each \(n\in\mathbb N\), subdivide the co-domain of \(f\) into \(2^{2n}+1\) intervals, \(2^{2n}\) of which have length \(2^{-n}\). That is, for each \(n\), define
\(I_{n,k}=\left[\frac{k-1}{2^n},\frac{k}{2^n}\right)\) for \(k=1,2,\ldots,2^{2n}\), and \(I_{n,2^{2n}+1}=[2^n,\infty)\),
which are disjoint and cover the non-negative real line (\(\mathbb{R}^{+} \subseteq \cup_{k}I_{n,k}, \forall n \in \mathbb{N}\)).
Now define the sets
\(A_{n,k}=f^{-1}(I_{n,k}) \,\) for \(k=1,2,\ldots,2^{2n}+1,\)
which are measurable (\(A_{n,k}\in \Sigma\)) because \(f\) is assumed to be measurable.
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.
Ugbu a ị Enweghị kalkulata na-ewepụ nke a, ma ọ bụ na ọ nwere ike ijikwa ya. Jiri otu n'okpuru, mọọbụ pịa nke gị.
Aịkọn ndị ahụ ejirila ebe a
Tinye ọbụla simbụl maka nkọwa zuru ezu, inyogo, na ihe ọbụla ọbụla na-egosi.
Ajụjụ ndị mmadụ jụrụ
What is wrong with the Riemann integral?
It fails on functions that oscillate too much, and it does not interact well with limits: the limit of integrable functions need not be integrable. Lebesgue's integral fixes both.
Oge nke ihuakwụkwọ a bụ nke a gbanwere site na Wikipedia (CC BY-SA 4.0). E kọwaara ya na ebe a; nsogbu bụ nke anyị.
Oge Measure Theory
σ-algebras and measuresThe Lebesgue integral and convergence theorems