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Rudin Remark 11.23

  • Thread starter ehrenfest
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  • #1
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[SOLVED] Rudin Remark 11.23

Homework Statement


Prove: If f is measurable and bounded on E, and if [itex]\mu(E) < + \infty[/itex], then f is Lebesgue integrable on E.


Homework Equations


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

Suppose
(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.

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?
 
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Answers and Replies

  • #2
morphism
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Try to invoke the monotonicity of the integral.
 
  • #3
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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.
 
  • #4
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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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