Are all compact convex sets in R^2 Borel-measurable?

Question

Emphasis on convex. If not, counterexample please. Thanks.

Thanks for the quick reply. It's not clear to me, though, that the inner and outer areas converge to the same value. Can you clarify the argument on this point?

Hmm maybe this works.
If you have an inscribed polygon I and an enclosing polygon E, if they are different pick a point in S-I to create a larger polygon equal to the convex hull of I and the new point. By convexity of S, this larger polygon is still inscribed in S, let it be the new inscribed polygon. Similarly construct a new enclosing polygon of smaller size than the previous enclosing polygon.
I think it follows that the extrema of these two polygons must contact the boundary of S infinitely often, but not sure what else can be said.

Clarification: by extrema I mean any maximal inscribed polygons and any minimal enclosing polygons.

Ignore what I said about extrema, it's nonsense :(

@kaksi: Pourquoi, ne comprenez-vous pas la théorie de mesure ? LOL

@Steiner: Which theorem are you quoting? Or is there a short proof? I don't recall that closed=>measurable in R^n, but then grad school was a long time ago.

@Steiner (continued): All Cartesian products of closed intervals of real numbers are measurable, of course, but that seems far weaker than the statement that all closed sets in R^2 are measurable.

@Steiner: You are right that the sets are Borel sets. My original intention was to find out if such sets always have the same inner and outer measure. Does S is a Borel set=>S has identical inner and outer measure?

Answer 1

Every compact set of R^2 is closed, and therefore, is Borel (as well as Lebesgue) measurable, right? Convex or not.

EDIT

The Borel sigma-algebra is the sigma-algebra generated by the open sets of R^n, so it includes all open sets and all closed sets, because closed sets are complements of open sets. The Lebesgue sigma-algebra includes all Borel sets, this can be proved.

Both sigma-algebras include all the F-sigma and G-delta and countable unions and intersection s of F-sigmas and G-deltas, etc.

Every Borel set is, by its very definition, Borel measurable. And also Lebesgue measurable (not by definition).

EDIT2

Every open set in R^n is a countable union of open cells(cartesian products of intervals of the real line). Such cells are Borel and Lebesgue measurable. Therefore, every open set of R^n is Borel and Lebesgue measurable.

Since closed sets are the complement of open sets, it then follows closed sets are Borel and lebesgue measurable.

And since any compact set in R^n is closed, compact sets are Borel and Lebesgue measurable.

But I didn't know that convex sets in R^2 are measurable. I guess this is true in R^n, too.

EDIT3

Yes, Borel sets have the samer inner ans outer measure. And if S is a Borel set and A is any set in R^n, then m*(A) = m*(A Inter S) + m*(A' inter S), where m* is the outer Lebesgue measure and A' is the complement of A.

Answer 2

occasion huge sort seventy 5 (Irrational Slope Topology) of “Counterexamples in Topology” by utilising Steen and Seebach gives you a counterexample. I summarize it right here. define X = {(x,y): x,y?? and y?0} (rational pairs in the better a million/2 airplane). restore some non-0 irrational huge sort ?. The topology ? on X is generated by utilising neighborhoods N(?,(x?,y?)) = {(x?,y?)} ? {(x,0)?X : |x - (x? - y?/?)|

Answer 3

Yes.

You can approximate the set both from within and without by sequences of convex polygons. The areas from both sequences will converge (from above for the outers, below for the inners) on the measure of the set.