# analytic set

For a continuous map  of topological spaces  it is known that the preimages  of open sets are open, preimages of closed (http://planetmath.org/ClosedSet) sets are closed and preimages of Borel sets are themselves Borel measurable. The situation is more difficult for direct images  (http://planetmath.org/DirectImage). That is if $f\colon X\rightarrow Y$ is continuous  then it does not follow that $S\subseteq X$ being open/closed/measurable implies the same property for $f(S)$. In fact, $f(X)=\operatorname{Image}(f)$ need not even be measurable. One of the few things that can be said, however, is that $f(S)$ is compact  whenever $S$ is compact. Analytic sets are defined in order to be stable under direct images, and their theory relies on the stability of compact sets. This is a fruitful concept because, as it turns out, all measurable sets  are analytic  and all analytic sets are universally measurable.

A subset $S$ of a Polish space  $X$ is said to be analytic (or, a Suslin set) if it is the image of a continuous map $f\colon Z\rightarrow X$ from another Polish space $Z$ — see Cohn. It is then clear that $g(S)$ will again be analytic for any continuous map $g\colon X\rightarrow Y$ between Polish spaces. Indeed, $g(S)$ will be the image of $g\circ f$.

###### Definition.

Let $(X,\mathcal{F})$ be a paved space. Then a set $A\subseteq X$ is said to be $\mathcal{F}$-analytic, or analytic with respect to $\mathcal{F}$, if there exists a compact paved space $(K,\mathcal{K})$ and an $S\in(\mathcal{F}\times\mathcal{K})_{\sigma\delta}$ such that

 $A=\pi(S).$

Here, $\mathcal{F}\times\mathcal{K}$ is the product paving and $\pi\colon X\times K\rightarrow X$ is the projection map $\pi(x,y)=x$.

Writing this out explicitly, there are doubly indexed sequences of sets $A_{m,n}\in\mathcal{F}$ and $K_{m,n}\in\mathcal{K}$ such that

 $S=\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}A_{m,n}\times K_{m,n}.$

and,

 $A=\left\{x\in X\colon(x,y)\in S\text{ for some }y\in K\right\}.$

Although this allows for $\mathcal{K}$ to be any compact paving it can be shown that it makes no difference  if it is just taken to be the collection of compact subsets of, for example, the real numbers.

For a measurable space $(X,\mathcal{F})$, a subset of $X$ is simply said to be analytic if it is $\mathcal{F}$-analytic, and a subset of a topological space is said to be analytic if it is analytic with respect to the Borel $\sigma$-algebra (http://planetmath.org/BorelSigmaAlgebra).

## References

• 1 K. Bichteler, Stochastic integration with jumps. Encyclopedia of Mathematics and its Applications, 89. Cambridge University Press, 2002.
• 2 Donald L. Cohn, Measure theory. Birkhäuser, 1980.
• 3 Claude Dellacherie, Paul-André Meyer, Probabilities and potential. North-Holland Mathematics Studies, 29. North-Holland Publishing Co., 1978.
• 4 Sheng-we He, Jia-gang Wang, Jia-an Yan,Semimartingale theory and stochastic calculus. Kexue Chubanshe (Science Press), CRC Press, 1992.
• 5 M.M. Rao, Measure theory and integration. Second edition. Monographs and Textbooks in Pure and Applied Mathematics, 265. Marcel Dekker Inc., 2004.
 Title analytic set Canonical name AnalyticSet1 Date of creation 2013-03-22 18:44:52 Last modified on 2013-03-22 18:44:52 Owner gel (22282) Last modified by gel (22282) Numerical id 9 Author gel (22282) Entry type Definition Classification msc 28A05 Synonym Suslin set Related topic PavedSpace Related topic BorelSigmaAlgebra Related topic UniversallyMeasurable Defines Suslin set Defines analytic with respect to