I'm taking an introductory measure theory course as an economics student and I've quickly found that I'm not as prepared as I should be. Anyway, I'll keep persevering. I've been given the following homework problem that I just can't figure out how to tackle. Any advice / prod in the right direction would be greatly appreciated! Can I ask that any explanations be as basic as possible. Many thanks.


Let $X$ be a compact metric space and let $\mathcal{B}(X)$ denote the Borel $\sigma$-algebra on X. Also let $(M,M,\mu)$ be a measure space with $\mu(M)<\infty$. Now let $\phi: M \rightarrow X$ be a $M−\mathcal{B}(X)$-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}

i) Show that $\Phi$ is well defined. (i.e. show that if $f \in C(X)$ then $f \circ \phi$ is integrable with respect to $\mu$) and that $\Phi$ is a positive functional on C(X).

The thought I had was that since $f$ is continuous then it is Borel measurable. Thus the composition is Borel measurable. Unfortunately I don't really know what to do next. Thanks for your help.


As consecuence of $f$ is in $C(X)$ and $X$ is compact, we conclude $f$ is a bounded function. i.e there is $K>0$ such that for all $y\in X$ we have $|f(y)|\le K$. In farticular $|f(\phi(x))|\le K$ for all $x\in M$.

Then, using $\mu(M)<\infty$ we get:

$$\left|\int_M f\circ\phi d\mu\right| \le\int_M |f\circ\phi| d\mu\le\int_MKd\mu=K\mu(M)<\infty$$.

Then, $f\circ\mu$ is integrable.

On the other hand, do you have clear that $\Phi$ is a positive functional?

  • $\begingroup$ Thanks a lot! That makes more sense. Cheers $\endgroup$ – mark Jan 3 '16 at 10:42

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