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Homework Help: Proving a Function is Riemann Integrable

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f, g : [a, b] [tex]\rightarrow[/tex] R be integrable on [a, b]. Then, prove that h(x) = max{f(x), g(x)} for
    x [tex]\in[/tex] [a, b] is integrable.
    1

    2. Relevant equations

    Definition of integrability: for each epsilon greater than zero there exists a partition P so that U(f,P)-L(f,P)<epsilon

    3. The attempt at a solution


    Ok i have absolutley no clue how to do this one. The following graph is how i think the function would look : http://i31.photobucket.com/albums/c373/SNOOTCHIEBOOCHEE/Graph2.jpg

    Sorry about the crude drawing, but the very light blue would be h(x)

    But i honest to god cant see a way to make U(f,P)-L(f,P)<epsilon a true statement

    Thanks in advance
     
  2. jcsd
  3. Jan 21, 2008 #2

    morphism

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    Hint: Prove that if f is integrable then so is |f|.

    Why does this help?
     
  4. Jan 21, 2008 #3
    I know that that is a theorem in the book, but i dont see how its applicable here.


    i also know that |Integral(f)| < Integral(|f|) comes from that statement.
     
    Last edited: Jan 21, 2008
  5. Jan 21, 2008 #4

    morphism

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    Try writing max(f,g) in terms of absolute values. To get a feel for this, try to find such an expression for max(f,0).
     
  6. Jan 21, 2008 #5
    how do you mean in absolute values?

    like max(f,g)<max(|f|,|g|)?

    edit: guess not

    im trying to figure out this for h(x)= max(f,0)

    so we know h(x) = f if f is positive and 0 else.

    but i cant figure this out for the other thing.
     
  7. Jan 21, 2008 #6

    morphism

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    max(f,0) = (|f| + f)/2
     
  8. Jan 21, 2008 #7
    I see where this argument is going. Basically you are gunna describe h(x) as a sum of absolute values of functions we already know are integrable, therefore the result is integrable.

    Now to figure out max(f,g)...
     
  9. Jan 21, 2008 #8
    sorry for the tripple post. but i got a little bit farther


    max(f,g)= [(g+|g|)+ (f+|f|)] /2 - min(f,g)

    but then i gotta figure out how to describe the min function...
     
  10. Jan 21, 2008 #9

    morphism

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    Try to think about f-g and f+g.
     
  11. Jan 21, 2008 #10
    ok ya im not getting this

    [(g+|g|)+ (f+|f|)] /2 = f+g if f and g are the same sign
     
  12. Jan 21, 2008 #11

    morphism

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    I really don't know how to give you any more hints that don't involve giving you the answer! Maybe try thinking about why (|f|+f)/2 gives us max(f,0). Try to sleep on it.
     
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