Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral of step functions

  1. Dec 5, 2011 #1
    In Apostol's book the definition of integral of step function is given as

    [itex]\int^{a}_{b}[/itex] s(x) dx = [itex]\sum^{n}_{k=1}[/itex] s[itex]_{k}[/itex] [itex]\times[/itex] (x[itex]_{k}[/itex] - x[itex]_{k-1}[/itex])

    and says that to compute the integral we multiply each constant value s[itex]_{k}[/itex] by the length of the k'th subinterval and add together all these products.
    I have understood till this point but he further says that the values of s at the subdivision points are immaterial since they do not appear on the right hand side of the equation.
    i.e. if s is constant on the open interval (a,b) say s(x) = c if a<x<b then

    [itex]\int^{a}_{b}[/itex] s(x) dx = c[itex]\sum^{n}_{k=1}[/itex] (x[itex]_{k}[/itex] - x[itex]_{k-1}[/itex]) = c(b-a) regardless of values s(a) and s(b). If c>0 and s(x) =c for all x in the closed interval [a,b], the ordinate set of s is a rectangle of base b-a and altitude c; the integral of s is c(b-a) the area of the rectangle.

    I am not able to understand the statement "changing the value of s at one or both endpoints a or b changes the ordinate set but does not alter the integral of s or the area of its ordinate set."

    If there is change in the ordinate set shouldn't there be a change in the area too? Also if the end points are ignored wouldn't the area change? Is it because we are computing the area of an open interval?
  2. jcsd
  3. Dec 5, 2011 #2


    User Avatar
    Science Advisor

    Changing the value at one point means changing the value on an interval of length 0. Therefore the contribution to the integral will be 0, irrespective of the value (as long as it is finite).
  4. Dec 6, 2011 #3
    So I can reason that in other words it is open interval is because the end points do not matter right?
  5. Dec 6, 2011 #4


    User Avatar
    Science Advisor

    As long as you are computing integrals, you are correct.
  6. Dec 7, 2011 #5
    Thanks a lot :)
  7. Dec 7, 2011 #6
    It is formally explained in a course including measure theory, however until then suffice to know that if you change your function at a single point or even say at every integer N in the domain your integral does not change.

    It is a quite natural thing. The integration variable dx is a sort of function that gives a weight to each set in your domain. For your means you can accept it as the usual length (or area or volume for higher dimensional cases). Then the set of all integers has length 0 for instance. In the same way in a two dimensional space, if you change your function at a line (which has 0 area hence zero measure) the integral does not change. In three dimensional case you can change your function at a hypersurface (which has zero volume) etc etc
  8. Feb 4, 2012 #7
    How can a set of all integers has length 0? Could you please elaborate on that?
  9. Feb 4, 2012 #8
    The measure of a disjoint union of countable sets is equal to the sum of the measures. If the measure of a point is zero then the measure of the set of all integers is the sum of the measures of a countable set of points and thus is zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook