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Help! Stuck in proof of Riesz Representation Theorem

  1. Oct 2, 2011 #1
    I am lecturing out of R.Bartle, The Elements of Integration and Lebesgue Measure, for the first time. In the most recent lecture I got stuck not being able to argue an inequality on page 107. I cannot post the text here, sorry. But if anyone has the book, can you also explain how to derive the inequality in line 6:

    g(t + n[itex]^{-2}[/itex]) [itex]\leq[/itex] g(t) + 2[itex]\epsilon[/itex] ?

    The inequality G([itex]\psi[/itex][itex]_{n}[/itex]) [itex]\leq[/itex] g(t) + 2[itex]\epsilon[/itex] from the previous line does not seem to help much, since
    G([itex]\psi[/itex][itex]_{n}[/itex]) [itex]\leq[/itex] G([itex]\varphi[/itex][itex]_{t + n^{-2}, n}[/itex])
    follows from G being positive, and this is the opposite inequality of the desired.
    Or what am I missing here? Thanks for any enlightenment!

    Kind regards, Tommy
     
  2. jcsd
  3. Oct 2, 2011 #2

    mathman

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    This looks like a result of g being continuous. Since I don't have the book, I can't comment on the rest.
     
  4. Oct 3, 2011 #3
    Thanks for the reply!

    Actually, in the part of the proof which I am asking about, the aim is the prove that the function g is continuous.

    There appears to be a mistake in the proof though. So for those who do have the book, here is a fix for later reference.

    Recall that J = [a,b] and C(J) is the set of real-valued continuous functions on J. A linear, bounded and positive functional G on C(J) is assumed given. Fix any t [itex]\in[/itex] [a,b) and a sufficiently large integer n. Let [itex]\varphi[/itex][itex]_{t,n}[/itex] be the function that maps x to 1 if a ≤ x ≤ t, maps x to 0 if t+[itex]\frac{1}{n}[/itex] ≤ x ≤ b, and maps x to 1 - n(x-t) if t < x < t+[itex]\frac{1}{n}[/itex]. Then [itex]\varphi[/itex][itex]_{t,n}[/itex] [itex]\in[/itex] C(J), and G([itex]\varphi[/itex][itex]_{t,n}[/itex]) is non-negative and non-increasing as a function of n.

    Let g(t) = lim[itex]_{n → ∞}[/itex] G([itex]\varphi[/itex][itex]_{t,n}[/itex]). Let g(s)=0 if s < a, and g(s)=G(1) if s ≥ b. Then g is monotone increasing. The aim is to show that g is everywhere continuous from the right, then to extend the Borel-Stieltjes measure generated by g to a measure defined on the Borel algebra, using the Hahn Extension Theorem.

    I cannot make the proof in the book work to prove continuity of g from the right. But the fix is easier than in the suggested proof.

    Let ε > 0 and assume that n is large enough to satisfy n > 2, n > [itex]\frac{1}{ε}[/itex]||G||, and g(t) ≤ G([itex]\varphi[/itex][itex]_{t,n}[/itex]) ≤ g(t) + ε.
    Then ||[itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex] - [itex]\varphi[/itex][itex]_{t,n}[/itex]|| = [itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}(t+\frac{1}{n})[/itex] = [itex]\frac{1}{n}[/itex] implies ||G|| ≥ |G(n([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex] - [itex]\varphi[/itex][itex]_{t,n}[/itex]))| = n(G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex]) - G([itex]\varphi[/itex][itex]_{t,n}[/itex])), hence
    G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex]) ≤ G([itex]\varphi[/itex][itex]_{t,n}[/itex]) + [itex]\frac{1}{n}[/itex]||G|| ≤ g(t) + 2ε.
    The right continuity of g at t follows by noting that
    g(t) ≤ g(t+[itex]\frac{1}{n^2}[/itex]) = lim[itex]_{k → ∞}[/itex] G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},k}[/itex]) ≤ G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex]) ≤ g(t) + 2ε.
     
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