Using the Riesz Representation theorem to show a measure is unique.

437 Views Asked by At

I'm having problems with this homework question on Riesz Representation theorem and was wondering if anyone could give me some advice? Unfortunately, as an Economist taking a measure theory class I'm not as well prepared as I should be. So I'm finding this question rather challenging. Hopefully I'll improve my maths skills by taking this course.

$\textbf{Question}$

Let $X$ be a compact metric space and let $\mathcal{B}(X)$ denote the Borel $\sigma$-algebra on X. Also let $(M,\mathcal{M},\mu)$ be a measure space with $\mu(M)<\infty$. Now let $\phi: M \rightarrow X$ be a $\mathcal{M}−\mathcal{B}(X)$-measurable map and define $\Phi:C(X)\rightarrow \mathbb{R}$ by \begin{equation} \Phi(f)=\int f \circ \phi \,d\mu\;\;\text{for all}\;\;f\in C(X) \end{equation}

It follows from above and the Riesz Representation Theorem for positive functionals that there is a unique finite Borel measure $\mu_{\Phi}$ on $X$ such that \begin{equation} \Phi(f)=\int f \,d\mu_{\Phi}\;\;\text{for all}\;\;f\in C(X) \end{equation}

Define $\mu_{\phi}:\mathcal{B}(X)\rightarrow [0,\infty]$ by $\mu_{\phi}(B)=\mu(\phi^{-1}(B))$ for $B \in \mathcal{B}(X)$.

(iii) Show that $\mu_{\Phi}=\mu_{\phi}$.

[Hint: Recall that if $f:M\rightarrow [0,\infty]$ is a positive Borel function then there is an increasing sequence $(s_n)_n$ of positive and simple Borel functions $s_n:M\rightarrow [0,\infty]$ such that $s_n \nearrow f$ pointwise.]

I have a strategy that I'm confident will work. If I can show that $\Phi(f)$ is represented by the two integrals above, by the uniqueness part of the Riesz Representation theorem we can conclude that $\mu_{\Phi}=\mu_{\phi}$ (I think). From the hint, I know we need to first consider simple functions or characteristic functions and show they are the same on all Borel sets then use a convergence theorem to show that the simple functions converge to $f$. However, I don't know the mechanics of how to do this. It should be trivial (or some I'm told) to show this for characteristic functions and simple functions but I don't see how.

Many thanks for your help.

1

There are 1 best solutions below

1
On

Let $E$ be a measurable set, $\mu_{\Phi}(E)=\int 1_Edu_{\Phi}=\int 1_E\circ \phi d\mu=\int 1_{\phi^{-1}(B)}d\mu=\mu(\phi^{-1}(B))$.

Remark that $(1_E\circ\phi)(x)=1$ iff $1_E(\phi(x))=1$ this equivalent to saying that $\phi(x)\in E$ or equivalently $x\in \phi^{-1}(E)$ so $1_E\circ \phi =1_{\phi^{-1}(E)}$.