11.5: Jordan Measurable Sets
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Volume and Jordan measurable sets
Given a bounded set S⊂Rn its characteristic function or indicator function is χS(x):={1 if x∈S0 if x∉S.
A bounded set S⊂Rn 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 x∈∂S, then for every δ>0, the sets S∩B(x,δ) (where χS is 1) and the sets (R∖S)∩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
- The closure ¯S is Jordan measurable.
- The interior S∘ is Jordan measurable.
- S∪T is Jordan measurable.
- S∩T is Jordan measurable.
- S∖T is Jordan measurable.
FIXME
If S⊂Rn 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)≥∫RχS−ϵ=V(S)−ϵ.
Now let R′1,…,R′ℓ be all the subrectangles of P such that χS is identically one on each R′j. In other words, these are the subrectangles contained in S. The interiors of the subrectangles R′∘j are disjoint and V(R′∘j)=V(R′j). It is easy to see from definition that m∗(ℓ⋃j=1R′∘j)=ℓ∑j=1V(R′∘j).
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 S⊂Rn be a bounded Jordan measurable set. A bounded function f:S→R is said to be Riemann integrable on S if for a closed rectangle R such that S⊂R, the function ˜f:R→R defined by ˜f(x)={f(x) if x∈S,0 otherwise,
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:R→R for a closed rectangle R, and S⊂R is a Jordan measurable subset then ∫Sf=∫RfχS.
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Images of Jordan measurable subsets
Let us prove the following FIXME. We will only need this simple
Suppose S⊂Rn is a closed bounded Jordan measurable set, and S⊂U for an open set U⊂Rn. If g:U→Rn 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 ∂T⊂g(∂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 ∂T⊂T. Suppose y∈∂T, then there must exist an x∈S such that g(x)=y. The Jacobian of g is nonzero at x.
We now use the inverse function theorem . We find a neighbourhood V⊂U 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)=y∈W. As y∈∂T, there exists a sequence {yk} in W with limyk=y and yk∉T. 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 yk∉T=g(S), clearly xk∉S. Since x∈S, we conclude that x∈∂S. The claim is proved, ∂T⊂g(∂S).
Exercises
Prove .
Prove that a bounded convex set is Jordan measurable. Hint: induction on dimension.
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