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Analysis proof showing discontinuous funtion is integrable?

  1. Mar 26, 2007 #1
    analysis proof...showing discontinuous function is integrable?

    1. The problem statement, all variables and given/known data
    if a function f : [a,b] is Riemann integrable and g :[a,b] is obtained by altering values of f at finite number of points, prove that g is Riemann integrable and that
    ∫ f = ∫ g (f and g integrated from a to b)




    2. Relevant equations



    3. The attempt at a solution

    g is bounded on [a,b] so for all E>0 let Q be a partition of [a, b] such that
    PcQ

    then L(P,f)<L(Q,g)<U(Q,g)<U(P,f) (inequalities should be less than or equal
    to...how to type that?)

    therefore U(Q,g)-L(Q,g)<E

    therefore g is Riemann integrable on [a,b]
     
    Last edited: Mar 26, 2007
  2. jcsd
  3. Mar 26, 2007 #2

    StatusX

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    What is P? How did you derive those inequalities?

    The important point is that the points are only going to affect a finite number of grids in a given partition, and by arranging these grids to be small enough, you can make their effect negligible.
     
  4. Mar 26, 2007 #3
    P is my partition of f, (f given as integrable) and the inequalities are given in a theorem. I think I am trying to do what you said. Trying to set up Q, my partition of g, as constants plus or minus a delta term and then deriving my U(Q,f) and L(Q,f). Does that make sense?
     
  5. Mar 26, 2007 #4

    StatusX

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    What theorem is specific enough that you can just write those inequalities down given the relation between f and g? And no, that didn't make sense (to me at least). Remember that there is no specific partition of f or g, you need to show that the result is the same over all partitions as their meshes go to zero.
     
    Last edited: Mar 26, 2007
  6. Mar 26, 2007 #5
    Alright. I see your point here, my inequality set up doesn't work yet. Thanks for your help thus far. I will go try something else. And I knew that I have to do this
    "The important point is that the points are only going to affect a finite number of grids in a given partition, and by arranging these grids to be small enough, you can make their effect negligible." but any further hints on how to do that.
     
  7. Mar 26, 2007 #6

    StatusX

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    Say the mesh (width of the largest grid) is e, and the places where g differs from f are x_1, x_2,..., x_n. Then what is the biggest the difference between the sums for f and g could be in terms of f(x_k), g(x_k), and e?
     
  8. Mar 27, 2007 #7

    matt grime

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    Why is g bounded?

    HINT: it is easy, and equivalent, to consider only a function g that is zero except at a finite number of points.
     
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