research36 min studyLesson 837 in Mathematics

Sets of finite perimeter and reduced boundary

Choose an outward-normal convention and recover geometric boundary area from a distributional measure.

Before you begin

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The question for today

What you’ll understand

Define relative perimeter through the characteristic-function gradient.
State reduced-boundary structure with a consistent outward normal sign.
Compute perimeter and flux while separating topological boundaries from null-set-invariant geometry.

A topological boundary can change drastically when a null set is added or removed. Perimeter should instead reflect the interface seen by volume and distributional differentiation. Sets of finite perimeter accomplish this by differentiating their characteristic functions. Their reduced boundary carries a measure-theoretic normal and almost all the perimeter, even when the chosen set representative has extra cracks or a dense collection of isolated points.

Perimeter is the variation of a characteristic function

For a Lebesgue-measurable set E and an open domain Omega, define relative perimeter as the total variation of the distributional gradient of its characteristic function. Local finite perimeter means finiteness on every compactly contained subdomain. Finite perimeter in Omega means finite total variation there. If the characteristic function is also globally integrable, this is its BV seminorm; finite-volume or bounded-domain conditions supply that L1 requirement. A half-space shows why local BV and global L1 should not be conflated.

The characteristic function and its perimeter depend only on the set up to Lebesgue-null changes. We use the outward-normal convention $D\chi_E=-\nu_E|D\chi_E|$ . Some sources name the opposite polar vector the normal, so the sign must be declared before using Gauss–Green. For a smooth bounded region, the outward vector points out of the region and this convention matches the usual divergence theorem.

P(E;\Omega)=|D\chi_E|(\Omega),\qquad D\chi_E=-\nu_E\,|D\chi_E|.

Our ν_E is outward; it is the negative of the polar vector of the distributional gradient.↔ Scroll wide equations horizontally.

Reduced boundary selects a unit limiting normal

The reduced boundary consists of points in the support of the perimeter measure where its averaged gradient direction has a unit limiting vector. With our convention, the negative of that limit is the outward normal. De Giorgi's structure theorem says the reduced boundary is countably $(n-1)$ -rectifiable and its normalized Hausdorff measure is exactly the perimeter measure. This is a deep structural theorem, rather than a definition that assumes the desired surface area formula.

At a reduced-boundary point, blow-ups of the set approach a half-space in local measure, and its volume density is one half. Polygon vertices need not belong to the reduced boundary, but they contribute zero boundary-area measure. The theorem does not identify perimeter with the Hausdorff measure of every topological boundary representative. Nor is density one half by itself a general substitute for the limiting-normal condition.

|D\chi_E|=\mathcal H^{n-1}\!\llcorner\partial^*E,\qquad \int_E\operatorname{div}F\,dx=\int_{\partial^*E}F\cdot\nu_E\,d\mathcal H^{n-1}.

Gauss–Green is stated for compactly supported C¹ fields, or for fields smoothly localized around the bounded example under consideration.↔ Scroll wide equations horizontally.

Null-set modifications can create topological boundary without perimeter

Removing a line segment from the interior of a planar region changes no planar volume and no characteristic-function class. The new topological crack therefore does not automatically create two units of perimeter. Points on its interior still have density one for the region, not density one half, so they are not reduced-boundary points. This behavior is deliberate: distributional perimeter measures the volume interface rather than every topological omission.

Adding a dense countable subset outside a region creates an even larger discrepancy. The modified set can have closure equal to the whole plane while agreeing almost everywhere with the original region. Its topological boundary may contain large planar pieces, yet its perimeter and reduced-boundary measure remain unchanged. Consequently, normal traces, perimeter compactness, and variational problems must specify measure-theoretic conventions instead of relying on the visual boundary of an arbitrary representative.

\mathcal L^n(E\triangle F)=0\quad\Longrightarrow\quad D\chi_E=D\chi_F,\quad P(E;\Omega)=P(F;\Omega).

Null modifications preserve the distributional interface, even when they radically change the topological boundary.↔ Scroll wide equations horizontally.

Work it through

A rectangle checks both perimeter and the normal sign

Take $E=(0,3)\times(0,2)$ and the field $F(x,y)=(x,2y)$ , smoothly localized outside a neighborhood of its closure. Compute perimeter, the distributional-gradient directions on two edges, and both sides of Gauss–Green.

The area is six and perimeter is ten. On the bottom edge the outward normal is $(0,-1)$ , so the gradient measure points $(0,1)$ . On the right edge the outward normal is $(1,0)$ , so the gradient measure points $(-1,0)$ .
The field divergence is three, hence its integral over the rectangle is eighteen. The right edge contributes flux $3\cdot2=6$ , and the top edge contributes $4\cdot3=12$ . The left and bottom fluxes vanish.
The boundary flux is therefore eighteen, agreeing with the divergence integral under the declared sign. The four vertices have zero one-dimensional measure. Adding the dense countable set of rational-coordinate points changes the topological boundary but leaves this characteristic-function class and perimeter ten unchanged.

P(E)=2(3+2)=10,\qquad \int_E3\,dx=18=6+12.

The edge calculation independently verifies the normal convention used in Gauss–Green.↔ Scroll wide equations horizontally.

The result

Perimeter is ten and outward flux is eighteen. The distributional gradient points inward, as required by Dχ_E=−ν_E|Dχ_E|.

The living notebook

Explore the model

Interactive lab

Interactive lab · matrix

Affine area and edge perimeter change differently

Pause and Reset, then set scale a to 1.5 and shear b to 1. The transformed unit-square edge vectors are (1.5,0) and (1,1), so predict area 1.5 and perimeter 3+2√2. For a fresh comparison Reset and set a to 1.5 and b to 0; area remains 1.5 while perimeter is five. The scene gives the matrix geometry and determinant, not a general finite-perimeter solver. Parameter edits retain time; set both controls after Reset for each independent comparison. For each run, take eight Steps to reach t=4 seconds, when the drawn geometry equals the selected matrix; its readouts describe that selected matrix throughout.

Interpolated geometryOriginal unit squareSelected final transform

Area scale · det A: 1.4
Singular values · σmax, σmin: 1.597, 0.876
Condition number · κ₂(A): 1.823

Diagonal scale a1.4Shear b0.6

t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA determinant controls area, while boundary length depends on the transformed edge vectors.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

A planar-null crack does not change the characteristic-function class or its distributional gradient. The topological boundary acquires a crack, but that is different from reduced-boundary perimeter.

Your turn

Try it yourself

Remove the vertical segment x=1/2, 0<y<1, from the open unit square. Find its perimeter and the density of the modified set at an interior crack point. Is that crack point in the reduced boundary?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A line segment has zero planar measure. Compare volume fractions in small balls with the normal-bearing half-space behavior of a true interface.

Compare with the solution

The perimeter remains four. At an interior crack point, a small ball loses only a planar-null line, so the volume density is one. Such a point is not in the reduced boundary: it has no perimeter measure nearby from the crack and no half-space blow-up. The original outer edges retain the usual reduced-boundary contribution.

Keep the idea

Take it with you

Perimeter depends on a characteristic function up to null sets.
Our outward normal is the negative polar direction of Dχ_E.
The reduced boundary carries perimeter; an arbitrary topological boundary can contain much more.

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