# Exercise 1.12 (2) from Revuz and Yor. Finding a Riesz Representation form for a bounded measure on $[0,1]$.

by nomadicmathematician   Last Updated October 09, 2019 17:20 PM

We denote by $$H$$ the subspace of $$C([0,1])$$ of functions $$h$$ such that $$h(0)=0$$, $$h$$ is absolutely continuous and its derivative $$h'$$ (which exists a.e.) satisfies $$\int_0^1 h'(s)^2 ds < \infty.$$

Then $$H$$ is a Hilbert space for the scalar product

$$(g,h) = \int_0^1 g'(s)h'(s)ds.$$

2) For any bounded measure $$\mu$$ on $$[0,1]$$, show that there exists an element $$h$$ in $$H$$ such that for every $$f \in H$$ $$\int_0^1 f(x)d\mu(x) = (f,h),$$ and that $$h'(s) = \mu((s,1])$$.

[Hint: The canonical injection of $$H$$ into $$C([0,1])$$ is continuous; use Riesz's theorem.]

Following the hint, we know that since $$\int_0^1 f(x)d\mu(x)$$ as a functional of $$f$$ is bounded linear, so it can be represented in the form $$(f,h)$$ for some $$h \in H$$. However I am stuck at showing that $$h'(s)=\mu((s,1])$$. $$\mu((s,1]) = \int 1_{(s,1]} d\mu$$, but this function does not belong to $$H$$. I would greatly appreciate any help.

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We have \begin{align*} \int_{0}^{1}\left(\int_{0}^{s}\mu((t,1])dt\right)'(s)f'(s)ds&=\int_{0}^{1}\mu((s,1])f'(s)ds\\ &=\int_{0}^{1}\int_{0}^{1}\chi_{s so by the uniqess of Riesz, $$\displaystyle\int_{0}^{s}\mu((t,1])dt=h(s)$$ and hence $$\mu((s,1])=h'(s)$$.

user284331
October 09, 2019 16:55 PM

We fix an $$s<1$$.

Let $$(t_\epsilon)$$ be the family of test functions (parametrized by $$\epsilon >0$$, possibly only with values $$\epsilon =1/n$$, $$n>0$$ natural,) which is piecewise linear, given by the values:

• zero on the interval $$[0,s]$$,
• one on the interval $$[s+\epsilon, 1]$$,
• linear on the remaining interval $$[s,s+\epsilon]$$, where it has the slope $$1/\epsilon$$, the jump of one, divided by the width of the interval. Then we have a.e. in $$s$$ the equality:

\begin{aligned} \mu(\ (s,1]\ ) &= \lim_{\epsilon\searrow 0} \mu(t_\epsilon) \\ &= \lim_{\epsilon\searrow 0} \int_0^1t_\epsilon'(u)\; h'(u)\; du \\ &= \lim_{\epsilon\searrow 0} \int_s^{s+\epsilon}\frac 1\epsilon\; h'(u)\; du \\ &= \lim_{\epsilon\searrow 0} \frac 1\epsilon (h(s+\epsilon)-h(s) \\ &=h'(s)\ . \end{aligned}

dan_fulea
October 09, 2019 16:58 PM

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