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Mathematics LibreTexts

11.5: Jordan Measurable Sets

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Volume and Jordan measurable sets

Given a bounded set SRn its characteristic function or indicator function is χS(x):={1 if xS0 if xS.

A bounded set S is said to be Jordan measurable if for some closed rectangle R such that SR, the function χS is in R(R). Take two closed rectangles R and R with SR and SR, then RR is a closed rectangle also containing S. By and , χSR(RR) and hence χSR(R), and furthermore RχS=RχS=RRχS.
We define the n-dimensional volume of the bounded Jordan measurable set S as V(S):=RχS,
where R is any closed rectangle containing S.

A bounded set SRn is Jordan measurable if and only if the boundary S is a measure zero set.

Suppose R is a closed rectangle such that S is contained in the interior of R. If xS, then for every δ>0, the sets SB(x,δ) (where χS is 1) and the sets (RS)B(x,δ) (where χS is 0) are both nonempty. So χS is not continuous at x. If x is either in the interior of S or in the complement of the closure ¯S, then χS is either identically 1 or identically 0 in a whole neighbourhood of x and hence χS is continuous at x. Therefore, the set of discontinuities of χS is precisely the boundary S. The proposition then follows.

The proof of the following proposition is left as an exercise.

[prop:jordanmeas] Suppose S and T are bounded Jordan measurable sets. Then

  1. The closure ¯S is Jordan measurable.
  2. The interior S is Jordan measurable.
  3. ST is Jordan measurable.
  4. ST is Jordan measurable.
  5. ST is Jordan measurable.

FIXME

If SRn is Jordan measurable then V(S)=m(S).

Given ϵ>0, let R be a closed rectangle that contains S. Let P be a partition of R such that U(P,χS)RχS+ϵ=V(S)+ϵandL(P,χS)RχSϵ=V(S)ϵ.

Let R1,,Rk be all the subrectangles of P such that χS is not identically zero on each Rj. That is, there is some point xRj such that xS. Let Oj be an open rectangle such that RjOj and V(Oj)<V(Rj)+\nicefracϵk. Notice that SjOj. Then U(P,χS)=kj=1V(Rk)>(kj=1V(Ok))ϵm(S)ϵ.
As U(P,χS)V(S)+ϵ, then m(S)ϵV(S)+ϵ, or in other words m(S)V(S).

Now let R1,,R be all the subrectangles of P such that χS is identically one on each Rj. In other words, these are the subrectangles contained in S. The interiors of the subrectangles Rj are disjoint and V(Rj)=V(Rj). It is easy to see from definition that m(j=1Rj)=j=1V(Rj).

Hence m(S)m(j=1Rj)m(j=1Rj)
Therefore m(S)V(S) as well.

Integration over Jordan measurable sets

In one variable there is really only one type of reasonable set to integrate over: an interval. In several variables we have many very simple sets we might want to integrate over and these cannot be described so easily.

Let SRn be a bounded Jordan measurable set. A bounded function f:SR is said to be Riemann integrable on S if for a closed rectangle R such that SR, the function ˜f:RR defined by ˜f(x)={f(x) if xS,0 otherwise,

is in R(R). In this case we write Sf:=R˜f.

When f is defined on a larger set and we wish to integrate over S, then we apply the definition to the restriction f|S. In particular note that if f:RR for a closed rectangle R, and SR is a Jordan measurable subset then Sf=RfχS.

FIXME

Images of Jordan measurable subsets

Let us prove the following FIXME. We will only need this simple

Suppose SRn is a closed bounded Jordan measurable set, and SU for an open set URn. If g:URn is a one-to-one continuously differentiable mapping such that Jg is never zero on S. Then g(S) is Jordan measurable.

Let T=g(S). We claim that the boundary T is contained in the set g(S). Suppose the claim is proved. As S is Jordan measurable, then S is measure zero. Then g(S) is measure zero by . As Tg(S), then T is Jordan measurable.

It is therefore left to prove the claim. First, S is closed and bounded and hence compact. By , T=g(S) is also compact and therefore closed. In particular TT. Suppose yT, then there must exist an xS such that g(x)=y. The Jacobian of g is nonzero at x.

We now use the inverse function theorem . We find a neighbourhood VU of x and an open set W such that the restriction f|V is a one-to-one and onto function from V to W with a continuously differentiable inverse. In particular g(x)=yW. As yT, there exists a sequence {yk} in W with limyk=y and ykT. As g|V is invertible and in particular has a continuous inverse, there exists a sequence {xk} in V such that g(xk)=yk and limxk=x. Since ykT=g(S), clearly xkS. Since xS, we conclude that xS. The claim is proved, Tg(S).

Exercises

Prove .

Prove that a bounded convex set is Jordan measurable. Hint: induction on dimension.

FIXME


This page titled 11.5: Jordan Measurable Sets is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jiří Lebl via source content that was edited to the style and standards of the LibreTexts platform.

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