Results 1 to 4 of 4

Math Help - Integration

  1. #1
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347

    Integration

    Using the definition of integration  \int_a^b f(t)dt = sup\{\int_{- \infty}^{\infty} g(t)dt where g is a step function supported on [a,b],  g(x) \le f(x) \forall x \in [a,b] \} (f continuous on [a,b] )

    show that for  \lambda \ge 0,  \int_a^b \lambda f(t)dt = \lambda \int_a^b f(t)dt .

    I would really appreciate any help with this. It's easy to show that if lambda is 0 you get the result but I am having trouble proving the rest. I just get tangled up in algebra, though I think I am trying to prove sup A =  \lambda sup B, where A =  \{\int_{- \infty}^{\infty} g(t)dt where g is a step function supported on [a,b],  g(x) \le \lambda f(x) \forall x \in [a,b] \} and B =  \{\int_{- \infty}^{\infty} g(t)dt where g is a step function supported on [a,b],  g(x) \le f(x) \forall x \in [a,b] \} Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,696
    Thanks
    1467
    sup\{\lambda F(x)\}= \lambda sup\{ F(x)\}

    (and, of course, \int \lambda F(x)dx= \lambda \int F(x)dx)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    provide \lambda \geq 0:
    for any step function g which supported on [a,b], g(x) \le f(x) \forall x \in [a,b], \lambda g(x) is a step function supported on [a,b], and \lambda g(x) \le \lambda f(x) \forall x \in [a,b].
    taking the superem over all step function g, we get that the LHS is no less than the RHS.
    and similarly the RHS is no less than the LHS. QED
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347
    thank you, worked it out
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum