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I am trying to understand a proof of a variant of the Riesz representation theorem.

Consider a linear functional $l$ on the space of continuous functions on $[0,1]$. Assume further that $l(f)\ge 0$ when $f\ge0$ on $[0,1]$.

We are going to define a function $F(u)$ on $[0,1]$ that will be used to construct a Lebesgue-Stieltjes measure. Define a collection of auxiliary functions $f_\epsilon(x,u)$ so they are $1$ on $[0,u]$, $0$ on $[u+\epsilon, 1]$, and linearly interpolated between those two intervals. The graph from left to right of such a function is a horizontal line with $y=1$, a downward sloping line, and then a horizontal line along 0.

Define $$F(u)=\lim_{\epsilon\rightarrow 0}\ l(f_\epsilon(x,u)).$$

We see that $F$ is increasing, and as shown in the answers, right-continuous. Why do we have

$$l(f)=\int_0^1 f(x)\ dF(x)$$

where the above is a Lebesgue integral.

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This may help to understand it intuitively, $dF(x)=dl(\epsilon(x))=l(\delta(x))$. – caozhu Jun 16 '12 at 8:57
up vote 1 down vote accepted

If you look at the Riemann sums for $\int_0^1 f(x)dF(x)$, they are like $$\tag{1} \sum_j f(t_{j+1})\,(F(t_{j+1})-F(t_j))=\; \lim_{\varepsilon\to0}\; l\left(\sum_jf(t_{j+1})(f_\varepsilon(x,t_{j+1})-f_\varepsilon(x,t_j))\right) $$

If you look carefully at the functions $x\mapsto \sum_jf(t_{j+1})(f_\varepsilon(x,t_{j+1})-f_\varepsilon(x,t_j))$, you'll notice they are piecewise linear functions that take the values $f(t_j)$ at $t_j$. So it is not hard to show that if the partitions for your Riemann sums are good (say, equally spaced) then these functions converge uniformly to $f$ (remember that $f$ is uniformly continuous).

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Although I guess this works, I was looking for an argument that used Lebesgue integration. – Potato Jun 16 '12 at 21:54
If you have convergence of Riemann sums, you have a Lebesgue integral. And the sum to the left of my equality is nothing but the Lebesgue integral of a step function; and the Lebesgue integral is the limit of the integrals of the approximating simple functions. – Martin Argerami Jun 16 '12 at 22:06
I'm aware of that. It just caught me a little off guard after I was learning all of this measure theory, I guess. – Potato Jun 16 '12 at 22:09

Right continuity of $F$ can be derived as follows. If $C = l(\mathbf{1})$ then $l(f) \leq C \cdot \max(f)$ for all continuous $f$ on $[0,1]$. In particular if $f_n$ is a sequence of non-negative functions such that $\max(f_n) \to 0$ then $l(f_n) \to 0$. Use this on $f_{\varepsilon}(x,u + \varepsilon') - f_{\varepsilon}(x, u)$ to conclude that

$$\lim_{\varepsilon' \downarrow 0}l(f_{\varepsilon}(x, u + \varepsilon')) = l(f_{\varepsilon}(x,u)). $$

Then for all $\delta > 0$ there exist $\varepsilon, \varepsilon' > 0$ such that

$$F(u) \leq F(u + \varepsilon') \leq l(f_{\varepsilon}(x, u + \varepsilon')) \leq l(f_{\varepsilon}(x, u)) + \frac{\delta}{2} \leq F(u) + \delta. $$

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Thank you. Why do we also have the integral expression in the problem? – Potato Jun 16 '12 at 21:21

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