# what does "measurable" mean in axiomatic definition of area??

I was studying Apostol's Calculus second edition (volume 1) chapter THE CONCEPTS OF INTEGRAL CALCULUS, section 1.6 The concept of area as a set function where I ran into this:

AXIOMATIC DEFINITION OF AREA. We assume there exists a class $M$ of measurable sets in the plane and a set function $a$, whose domain is $M$ , with the following properties: ... (page 58)

I don't know the definition of "measurable sets in the plane".

• mathworld.wolfram.com/MeasurableSet.html (or just google it!) Sep 15, 2016 at 18:02
• i saw this link but i wasn't sure they're the same , what is the sigma-algebra for real plane?? @Glougloubarbaki Sep 15, 2016 at 18:04
• click on the hyperlink for the definition of a sigma algebra. in the case of the plane, the usual sigma algebra is given by countable intersections and unions of open or closed sets. basically, the idea is that it's not possible to assign a meaningful notion of area (or length, or volume) to ANY set, so we first define what kind of sets are measurable, ie for what kind of set we want to define a notion of area Sep 15, 2016 at 18:09
• thank you , can you recommend a book about these basic notions like area ? @Glougloubarbaki Sep 15, 2016 at 18:12
• If nothing is specified then the sigma algebra for some set is assumed to be the borel algebra, i.e. the sigma algebra induced from the topology. If no topology is specified then is assumed that the topology is the standard. For $\Bbb R^n$ the standard topology is induced from the product topology of the standard topology of $\Bbb R$ (the topology generated from the open intervals). Sep 15, 2016 at 18:36