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Rudin Remark 11.23

  1. May 25, 2008 #1
    [SOLVED] Rudin Remark 11.23

    1. The problem statement, all variables and given/known data
    Prove: If f is measurable and bounded on E, and if [itex]\mu(E) < + \infty[/itex], then f is Lebesgue integrable on E.

    2. Relevant equations
    Here is how Rudin defines the Lebesgue integral. Everyone probably already knows this but anyway:

    (51) [tex]s(x) = \sum_{i=1}^n c_i K_{E_i} (x) \mbox{ (for x \in X, c_i >0)} [/tex]

    (where K is the indicator function) is measurable, and suppose E is a measurable set. We define

    (52) [tex] I_E (s) = \sum_{i=1}^n c_i \mu(E \cap E_i)[/tex]

    If f is measurab;e and nonnegative, we define the Lebesgue integral of f over the set E as

    (53) [tex]\int_E f d\mu = \sup I_E (s)[/tex]

    where the sup is taken over all measurable simple functions s such that 0 \leq s \leq f.

    Let f be measurable, and consider the two integrals

    (55) [tex]\int_E f^+ d\mu \mbox{ and} \int_E f^- d\mu [/tex]

    where [itex]f^+ = max(f,0)[/itex] and [itex]f^-= min(f,0)[/itex]. If at least one of the two integrals is finite, we define

    (56) [tex]\int_E f d\mu = \int_E f^+ d\mu - \int_E f^- d\mu[/tex]

    If both integrals on the RHS are finite, we say the f is Lebesgue integrable.

    3. The attempt at a solution
    Apparently this should follow directly from the definition, although I am having trouble figuring out why. Assume B is a bound for f. Then all of the c_i have to less then or equal to B (at least if E_i is nonempty). Then we someone need to get a bound for the I_E. So, we probably want to invoke the countable additivity of the set function mu but we then need to know that all of the [itex]E_i \cap E[/itex] are disjoint and we do not know that, right?
    Last edited: May 25, 2008
  2. jcsd
  3. May 25, 2008 #2


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    Try to invoke the monotonicity of the integral.
  4. May 26, 2008 #3
    The definition of a simple function is that the Ei are disjoint. If they're not, you can split them into new Ei that are disjoint. For instance, if
    E1 = (0,1/2), c1 = 1
    E2 = (1/4, 3/4), c2 = 2
    E3 = (1/2, 1), c3 = 1.
    you can easily separate the Ei and ci so that the new Ei are disjoint.
  5. May 26, 2008 #4
    Also, the reason that this follows directly from the definition is that it is the definition. What are you trying to prove?
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