Help Stuck in proof of Riesz Representation Theorem

[Tommy Jensen]

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$^{-2}$) $\leq$ g(t) + 2$\epsilon$ ?

The inequality G($\psi$$_{n}$) $\leq$ g(t) + 2$\epsilon$ from the previous line does not seem to help much, since
G($\psi$$_{n}$) $\leq$ G($\varphi$$_{t + n^{-2}, n}$)
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

 

[mathman]

g(t + n⁻²) ≤ g(t) + 2ϵ

This looks like a result of g being continuous. Since I don't have the book, I can't comment on the rest.

 

[Tommy Jensen]

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 $\in$ [a,b) and a sufficiently large integer n. Let $\varphi$$_{t,n}$ be the function that maps x to 1 if a ≤ x ≤ t, maps x to 0 if t+$\frac{1}{n}$ ≤ x ≤ b, and maps x to 1 - n(x-t) if t < x < t+$\frac{1}{n}$. Then $\varphi$$_{t,n}$ $\in$ C(J), and G($\varphi$$_{t,n}$) is non-negative and non-increasing as a function of n.

Let g(t) = lim$_{n → ∞}$ G($\varphi$$_{t,n}$). 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 > $\frac{1}{ε}$||G||, and g(t) ≤ G($\varphi$$_{t,n}$) ≤ g(t) + ε.
Then ||$\varphi$$_{t+\frac{1}{n^2},n}$ - $\varphi$$_{t,n}$|| = $\varphi$$_{t+\frac{1}{n^2},n}(t+\frac{1}{n})$ = $\frac{1}{n}$ implies ||G|| ≥ |G(n($\varphi$$_{t+\frac{1}{n^2},n}$ - $\varphi$$_{t,n}$))| = n(G($\varphi$$_{t+\frac{1}{n^2},n}$) - G($\varphi$$_{t,n}$)), hence
G($\varphi$$_{t+\frac{1}{n^2},n}$) ≤ G($\varphi$$_{t,n}$) + $\frac{1}{n}$||G|| ≤ g(t) + 2ε.
The right continuity of g at t follows by noting that
g(t) ≤ g(t+$\frac{1}{n^2}$) = lim$_{k → ∞}$ G($\varphi$$_{t+\frac{1}{n^2},k}$) ≤ G($\varphi$$_{t+\frac{1}{n^2},n}$) ≤ g(t) + 2ε.