Simple function

<h2 id="definition">Definition</h2>
Formally, a simple function is a finite <a href="/facts/Linear_combination/CVjL6kuN">linear combination</a> of <a href="/facts/Indicator_function/QEuh04NM">indicator functions</a> of <a href="/facts/Measurable_set/yCq7zlI4">measurable sets</a>. More precisely, let (X, Σ) be a <a href="/facts/Sigma-algebra/8LJINLNc">measurable space</a>. Let A1, ..., An ∈ Σ be a <a href="/facts/Sequence/gV2S4uqp">sequence</a> of disjoint measurable sets, and let a1, ..., an be a sequence of <a href="/facts/Real_number/R02gw5Pb">real</a> or <a href="/facts/Complex_number/2w7mqI7e">complex numbers</a>. A simple function is a function 
 
 
 
 f
 :
 X
 →
 
 C
 
 
 
 {\displaystyle f\colon X\to \mathbb {C} }
 
 of the form

f
        (
        x
        )
        =
        
          ∑
          
            k
            =
            1
          
          
            n
          
        
        
          a
          
            k
          
        
        
          
            
              1
            
          
          
            
              A
              
                k
              
            
          
        
        (
        x
        )
        ,
      
    
    {\displaystyle f(x)=\sum _{k=1}^{n}a_{k}{\mathbf {1} }_{A_{k}}(x),}

where 
 
 
 
 
 
 
 1
 
 
 
 A
 
 
 
 
 {\displaystyle {\mathbf {1} }_{A}}
 
 is the <a href="/facts/Indicator_function/QEuh04NM">indicator function</a> of the set A.

<h2 id="properties-of-simple-functions">Properties of simple functions</h2>
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 <a href="/facts/Algebra_over_a_field/8T8ulWJb">commutative algebra</a> over 
 
 
 
 
 C
 
 
 
 {\displaystyle \mathbb {C} }
 
.

<h2 id="integration-of-simple-functions">Integration of simple functions</h2>
If a <a href="/facts/Measure_(mathematics)/yCq7zlI4">measure</a> 
 
 
 
 μ
 
 
 {\displaystyle \mu }
 
 is defined on the space 
 
 
 
 (
 X
 ,
 Σ
 )
 
 
 {\displaystyle (X,\Sigma )}
 
, the <a href="/facts/Lebesgue_integral/f4AgvC6x">integral</a> of a simple function 
 
 
 
 f
 :
 X
 →
 
 R
 
 
 
 {\displaystyle f\colon X\to \mathbb {R} }
 
 with respect to 
 
 
 
 μ
 
 
 {\displaystyle \mu }
 
 is defined to be

∫
          
            X
          
        
        f
        d
        μ
        =
        
          ∑
          
            k
            =
            1
          
          
            n
          
        
        
          a
          
            k
          
        
        μ
        (
        
          A
          
            k
          
        
        )
        ,
      
    
    {\displaystyle \int _{X}fd\mu =\sum _{k=1}^{n}a_{k}\mu (A_{k}),}

if all summands are finite.

<h2 id="relation-to-lebesgue-integration">Relation to Lebesgue integration</h2>
The above integral of simple functions can be extended to a more general class of functions, which is how the <a href="/facts/Lebesgue_integral/f4AgvC6x">Lebesgue integral</a> is defined. This extension is based on the following fact.

Theorem. Any non-negative <a href="/facts/Measurable_function/NgIHnb1b">measurable function</a> 
 
 
 
 f
 :
 X
 →
 
 
 R
 
 
 +
 
 
 
 
 {\displaystyle f\colon X\to \mathbb {R} ^{+}}
 
 is the <a href="/facts/Pointwise/dt4Mrs6w">pointwise</a> 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 
 
 
 
 
 
 R
 
 
 +
 
 
 
 
 {\displaystyle \mathbb {R} ^{+}}
 
 is the restriction of the <a href="/facts/Borel_%CF%83-algebra/jvSsfpkP">Borel σ-algebra</a> 
 
 
 
 
 
 B
 
 
 (
 
 R
 
 )
 
 
 {\displaystyle {\mathfrak {B}}(\mathbb {R} )}
 
 to 
 
 
 
 
 
 R
 
 
 +
 
 
 
 
 {\displaystyle \mathbb {R} ^{+}}
 
. The proof proceeds as follows. Let 
 
 
 
 f
 
 
 {\displaystyle f}
 
 be a non-negative measurable function defined over the measure space 
 
 
 
 (
 X
 ,
 Σ
 ,
 μ
 )
 
 
 {\displaystyle (X,\Sigma ,\mu )}
 
. For each 
 
 
 
 n
 ∈
 
 N
 
 
 
 {\displaystyle n\in \mathbb {N} }
 
, subdivide the co-domain of 
 
 
 
 f
 
 
 {\displaystyle f}
 
 into 
 
 
 
 
 2
 
 2
 n
 
 
 +
 1
 
 
 {\displaystyle 2^{2n}+1}
 
 intervals, 
 
 
 
 
 2
 
 2
 n
 
 
 
 
 {\displaystyle 2^{2n}}
 
 of which have length 
 
 
 
 
 2
 
 −
 n
 
 
 
 
 {\displaystyle 2^{-n}}
 
. That is, for each 
 
 
 
 n
 
 
 {\displaystyle n}
 
, define

I
 
 n
 ,
 k
 
 
 =
 
 [
 
 
 
 
 k
 −
 1
 
 
 2
 
 n
 
 
 
 
 ,
 
 
 k
 
 2
 
 n
 
 
 
 
 
 )
 
 
 
 {\displaystyle I_{n,k}=\left[{\frac {k-1}{2^{n}}},{\frac {k}{2^{n}}}\right)}
 
 for 
 
 
 
 k
 =
 1
 ,
 2
 ,
 …
 ,
 
 2
 
 2
 n
 
 
 
 
 {\displaystyle k=1,2,\ldots ,2^{2n}}
 
, and 
 
 
 
 
 I
 
 n
 ,
 
 2
 
 2
 n
 
 
 +
 1
 
 
 =
 [
 
 2
 
 n
 
 
 ,
 ∞
 )
 
 
 {\displaystyle I_{n,2^{2n}+1}=[2^{n},\infty )}
 
,
which are disjoint and cover the non-negative real line (
 
 
 
 
 
 R
 
 
 +
 
 
 ⊆
 
 ∪
 
 k
 
 
 
 I
 
 n
 ,
 k
 
 
 ,
 ∀
 n
 ∈
 
 N
 
 
 
 {\displaystyle \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
          
        
        )
        
      
    
    {\displaystyle A_{n,k}=f^{-1}(I_{n,k})\,}
  
 for 
  
    
      
        k
        =
        1
        ,
        2
        ,
        …
        ,
        
          2
          
            2
            n
          
        
        +
        1
        ,
      
    
    {\displaystyle k=1,2,\ldots ,2^{2n}+1,}

which are measurable (
 
 
 
 
 A
 
 n
 ,
 k
 
 
 ∈
 Σ
 
 
 {\displaystyle A_{n,k}\in \Sigma }
 
) because 
 
 
 
 f
 
 
 {\displaystyle f}
 
 is assumed to be measurable.
Then the increasing sequence of simple functions

f
          
            n
          
        
        =
        
          ∑
          
            k
            =
            1
          
          
            
              2
              
                2
                n
              
            
            +
            1
          
        
        
          
            
              k
              −
              1
            
            
              2
              
                n
              
            
          
        
        
          
            
              1
            
          
          
            
              A
              
                n
                ,
                k
              
            
          
        
      
    
    {\displaystyle f_{n}=\sum _{k=1}^{2^{2n}+1}{\frac {k-1}{2^{n}}}{\mathbf {1} }_{A_{n,k}}}

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

<h2 id="see-also">See also</h2>
<a href="/facts/Bochner_measurable_function/HxXFeAgt">Bochner measurable function</a>

<ul><li>J. F. C. Kingman, S. J. Taylor. Introduction to Measure and Probability, 1966, Cambridge.</li>
<li><a href="/facts/Serge_Lang/I7MW2rUu">S. Lang</a>. Real and Functional Analysis, 1993, Springer-Verlag.</li>
<li><a href="/facts/Walter_Rudin/9mxRAECY">W. Rudin</a>. Real and Complex Analysis, 1987, McGraw-Hill.</li>
<li>H. L. Royden. Real Analysis, 1968, Collier Macmillan.</li></ul>

Simple function open-in-new

Simple function