# $\sigma$-algebra on the space of all probabilty measures of a measurable space

I am trying to understand the arguments in a book I am reading.

Consider the probability space $\left( X, \mathcal{B} \right)$ and let $\mathcal{P}$ be the set of probability measures on it. Let $\mathcal{C}$ be the $\sigma$-algebra generated by the sets $A_{B, t} = \left\{ P \in \mathcal{P}: P \left( B \right) \leqslant t \right\}$ where $B \in \mathcal{B}$ and $t \in \left[ 0, 1 \right]$.

• How does one prove that $\mathcal{C}$ is the smallest $\sigma$-algebra making the functions $g_B$ from $\mathcal{P}$ into $\mathbb{R}$ defined by $g_B \left( P \right) = P \left( B \right)$ measurable?
• I am looking also for specific counterexamples of functions not measurable in this context to help me understand it more.
• It is also indicated in the book that there is some link to the topology of pointwise convergence. What is that link?
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If $X$ is a compact metric space (might work with something more general) one can endow $\mathcal{P}$ with the weak*-topology (sometimes called the topology of weak convergence or the narrow topology). The Borel sets in that topology coince with this $\sigma$-algebra. – Michael Greinecker Nov 17 '12 at 14:18
Hi Michael. Thanks a lot for the help. I will try to prove that the sets $A_{B,t}$ constitute a base for the weak*-topology and see if I am successful. – Learner Nov 19 '12 at 0:43

One definition of a measurable real-valued function on a measure space $(X,\mathcal{A},\mu)$ is that the pre-image of every interval $(-\infty,\alpha)$ is in $\mathcal{A}$. This characterisation of measurability comes from the fact that the Borel sigma algebra on $\mathbb{R}$ can be generated by such intervals.
For each $g_B$ to be measurable in this sense we certainly require that each $A_{B,t}$ is measurable in $\mathcal{P}$. Hence the sigma algebra generated by these $A_{B,t}$ is the smallest sigma algebra with this property.
I am unsure as to what you are asking here, $P$ as I used it was a probability measure rather than a set. – Sean Gomes Nov 19 '12 at 13:19
Also note that $\mathcal{P}=A_{B,1}$ for any $B\in\mathcal{B}$. – Sean Gomes Nov 19 '12 at 13:21