1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?