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I'm taking an introductory course on measure theory as an economist and I'm finding it quite confusing. I have a homework question where I must prove that a putative measure is a Borel measure. I'm slightly unsure of what I need to show to prove this.

I've seen a definition as follows:

$\textbf{Definition 1}$

$\mu$ is a function such that $\mu:\mathcal{B}(X)\rightarrow [0,\infty]$ that satisfies

  1. The empty set has measure zero, $\mu(\emptyset)=0$
  2. Countable additivity, Any countable collection $E_1,E_2,\ldots$ of pairwise disjoint measurable sets the following holds \begin{equation} \mu(\bigcup_i E_i)=\sum_i \mu(E_i) \end{equation} in which $i$ counts over the whole collection.

But when I speak to my tutor he asks that I check if the measure is positive and countably subaddative and defined over all Borel sets.

I'm sure this question is rather basic but i'm stuck until I actually know what i need to show. The problem is that we don't define the Borel measure explicitly in our lecture notes.

The question I have might put my query into context.

$\mathbf{Question}$ 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}

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

i) Show that $\mu_{\phi}$ is well-defined (i.e., show that if $B \in \mathcal{B}(X)$ then $\phi^{-1}(B)\in \mathcal{M}$) and that $\mu_{\phi}$ is a Borel measure on $X$.

Thanks very much for your help.

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  • $\begingroup$ Your definition of borel measure implies positive, countably subadditive and defined over all borel sets. Make sure you understand why. What did you try doing to answer the question? Do you know about the properties of preimage? $\endgroup$ – user159517 Jan 9 '16 at 14:45
  • $\begingroup$ Thanks for your comment. The definition is not the one used in the lecture notes as it wasn't really defined. It's just an example of what I've seen. I know I need to use the properties of the preimage to prove it but not sure which ones. At the moment I'm still unsure of what I actually need to show. I'm not too sure about the properties of the preimage. Cheers. $\endgroup$ – mark Jan 9 '16 at 14:59
  • $\begingroup$ Show the three points in your definition, 0) $\mu_{\phi}$ is a function from $\mathcal{B}(X) \to [0,\infty)$, (well-definedness). This follows immediately from the hint that was given in the question 1) The empty set has measure zero 2) $\mu_{\phi}$ is countably additive. For this you will have to use some of the properties of the preimage. Try writing out what you have to prove for a start and then see if you can do it, if you cant, post your progress here $\endgroup$ – user159517 Jan 9 '16 at 15:04
  • $\begingroup$ I've had a look and I might be making progress. $\endgroup$ – mark Jan 9 '16 at 15:44
  • $\begingroup$ I've had a look and I might be making progress. For well-definedness. Since $\phi^{-1}:X\rightarrow M$ and $\mu$ is a measure on $\mathcal{M}$ then $\mathcal{B}(X) \rightarrow [0,\infty]$. Then for the countable additivity. $\mu(\phi^{-1}(\cup_i B_i))=\mu(\cup_i\phi^{-1}(B_i))=\sum_i \mu(\phi^{-1}(B_i))$ since if $B_i$ are disjoint sets then so are $\phi^{-1}(B_i)$. But of course I could be completely wrong. $\endgroup$ – mark Jan 9 '16 at 15:52

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