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5.1 Integrals of step functions  (Page 2/3)

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Let [ a , b ] be a fixed closed bounded interval in R , and let H ( [ a , b ] ) denote the set of all step functions on [ a , b ] .

  1. Using Part (c) of [link] , prove that the set H ( [ a , b ] ) is a vector space of functions; i.e., it is closed under addition and scalar multiplication.
  2. Show that H ( [ a , b ] ) is closed under multiplication; i.e., if h 1 , h 2 ∈ H ( [ a , b ] ) , then h 1 h 2 ∈ H ( [ a , b ] ) .
  3. Show that H ( [ a , b ] ) is closed under taking maximum and minimum and that it contains all the real-valued constant functions.
  4. We call a function χ an indicator function if it equals 1 on an interval ( c , d ) and is 0 outside [ c , d ] . To be precise, we will denote this indicator function by χ ( c , d ) . Prove that every indicator function is a step function, and show also that every step function h is a linear combination of indicator functions:
    h = ∑ j = 1 n a j χ ( c j , d j ) .
  5. Define a function k on [ 0 , 1 ] by setting k ( x ) = 0 if x is a rational number and k ( x ) = 1 if x is an irrational number. Prove that the range of k is a finite set, but that k is not a step function.

Our first theorem in this chapter is a fundamental consistency result about the “area under the graph” of a step function.Of course, the graph of a step function looks like a collection of horizontal line segments, and the region under this graph is just a collection of rectangles.Actually, in this remark, we are implicitly thinking that the values { a i } of the step function are positive. If some of these values are negative, thenwe must re-think what we mean by the area under the graph. We first introduce the following bit of notation.

Let h be a step function on the closed interval [ a , b ] . Suppose P = { x 0 < x 1 < ... < x n } is a partition of [ a , b ] such that h ( x ) = a i on the interval ( x i - 1 , x i ) . Define the weighted average of h relative to P to be the number S P ( h ) defined by

S P ( h ) = ∑ i = 1 n a i ( x i - x i - 1 ) .

REMARK Notice the similarity between the formula for a weighted average and the formula for a Riemann sum.Note also that if the interval is a single point, i.e., a = b , then the only partition P of the interval consists of the single point x 0 = a , and every weighted average S P ( h ) = 0 .

The next theorem is not a surprise, although its proof takes some careful thinking.It is simply the assertion that the weighted averages are independent of the choice of partition.

Let h be a step function on the closed interval [ a , b ] . Suppose P = { x 0 < x 1 < ... < x n } is a partition of [ a , b ] such that h ( x ) = a i on the interval ( x i - 1 , x i ) , and suppose Q = { y 0 < y 1 < ... < y m } is another partition of [ a , b ] such that h ( x ) = b j on the interval ( y j - 1 , y j ) . Then the weighted average of h relative to P is the same as the weighted average of h relative to Q . That is, S P ( h ) = S Q ( h ) .

Suppose first that the partition Q is obtained from the partition P by adding one additional point. Then m = n + 1 , and there exists an i 0 between 1 and n - 1 such that

  1. for 0 ≤ i ≤ i 0 we have y i = x i .
  2.     x i 0 < y i 0 + 1 < x i 0 + 1 .
  3. For i 0 < i ≤ n we have x i = y i + 1 .

In other words, y i 0 + 1 is the only point of Q that is not a point of P , and y i 0 + 1 lies strictly between x i 0 and x i 0 + 1 .

Because h is constant on the interval ( x i 0 , x i 0 + 1 ) = ( y i 0 , y i 0 + 2 ) , it follows that

  1. For 1 ≤ i ≤ i 0 , a i = b i .
  2.     b i 0 + 1 = b i 0 + 2 = a i 0 + 1 .
  3. For i 0 + 1 ≤ i ≤ n , a i = b i + 1 .

So,

S P ( h ) = ∑ i = 1 n a i ( x i - x i - 1 ) = ∑ i = 1 i 0 a i ( x i - x i - 1 ) + a i 0 + 1 ( x i 0 + 1 - x i 0 ) + ∑ i = i 0 + 2 n a i ( x i - x i - 1 ) = ∑ i = 1 i 0 b i ( y i - y i - 1 ) + a i 0 + 1 ( y i 0 + 2 - y i 0 ) + ∑ i = i 0 + 2 n b i + 1 ( y i + 1 - y i ) = ∑ i = 1 i 0 b i ( y i - y i - 1 ) + a i 0 + 1 ( y i 0 + 2 - y i 0 + 1 + y i 0 + 1 - y i 0 ) + ∑ i = i 0 + 3 n + 1 b i ( y i - y i - 1 ) = ∑ i = 1 i 0 b i ( y i - y i - 1 ) + b i 0 + 1 ( y i 0 + 1 - y i 0 ) + b i 0 + 2 ( y i 0 + 2 - y i 0 + 1 ) + ∑ i = i 0 + 3 m b i ( y i - y i - 1 ) = ∑ i = 1 m b i ( y i - y i - 1 ) = S Q ( h ) ,
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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