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.