note-math

affine-combination_(tag)

Affine combination

\begin{matrix} \sum_{0 . . 𝑁} 𝑡_{𝑖} \cdot 𝑥_{𝑖} \\ 𝑡_{0}, \dots, 𝑡_{𝑁} \in ℝ, \sum_{0 . . 𝑁} 𝑡_{𝑖} = 1 \end{matrix}

is a well-defined affine point, or rather the coordinate definition does not depend on the choice of origin. Let the coordinates of $𝑥_{𝑖}$ be $𝑣_{𝑖}$ . Change the origin $𝑣_{𝑖^{'}} = 𝑣_{𝑖} + Δ$

\sum 𝑡_{𝑖} (𝑣_{𝑖} + Δ) = (\sum 𝑡_{𝑖} 𝑣_{𝑖}) + (\sum 𝑡_{𝑖}) Δ = (\sum 𝑡_{𝑖} 𝑣_{𝑖}) + Δ

Regarding intuition, the simplest example is the proportional point of a straight line between two points

Can be iteratively or decomposed e.g. triangle $𝑡_{1} 𝑥_{1} + 𝑡_{2} 𝑥_{2} + 𝑡_{3} 𝑥_{3} ⟷ 𝑠_{1} (𝑡_{1} 𝑥_{1} + 𝑡_{2} 𝑥_{2}) + 𝑠_{2} 𝑥_{3}$ . And the decomposition operation is commutative. And it can be decomposed into multiple $\geq 1$ order

affine-coordinate_(tag) $𝑡_{𝑖}$ can be considered as a coordinate based on the point $𝑥_{𝑖}$ . Affine coordinates. alias Barycentric coordinates barycentric-coordinate_(tag)

affine-independent_(tag)

Affine independence := $𝑥_{𝑘}$ cannot be expressed as $\sum_{𝑖 = 0 . . 𝑁 ∖ 𝑘} 𝑡_{𝑖} 𝑥_{𝑖}$

Affine independence corresponds to the linear independence of $𝑥_{𝑖} - 𝑥_{0}$ after selecting one point e.g. $𝑥_{0}$ as the origin

If it is affine independent, then the vertices correspond to $𝑡_{𝑘} = 1 \land 𝑡_{1, \dots, 𝑁 ∖ 𝑘} = 0$

An $𝑛$ -dimensional affine space has at most $𝑛 + 1$ affine independent points

For the coordinates of $𝑛 + 1$ affine independent points of an $𝑛$ -dimensional affine space $𝑉$ , $𝑡_{0}, \dots, 𝑡_{𝑛}$ have a one-to-one correspondence with the affine points of $𝑉$

If $𝑡_{0}, \dots, 𝑡_{𝑁} \in ℝ, \sum_{0 . . 𝑁} 𝑡_{𝑖} = 0$ , although the coordinate $\sum 𝑡_{𝑖} 𝑣_{𝑖}$ will not change due to changing the origin, it is not an affine point

affine-map-point-ver_(tag) alias simplicial-map_(tag) Let $𝑦_{1}, \dots, 𝑦_{𝑛}$ be points in another affine space. The affine mapping is determined by $𝑓 (𝑥_{𝑖}) = 𝑦_{𝑖}$ , and the situation of other points can be obtained by generating them through affine homomorphism

\sum 𝑡_{𝑖} 𝑥_{𝑖} ⇝ \sum 𝑡_{𝑖} 𝑦_{𝑖}

center-of-affine-point-set_(tag) The center point of $\sum_{0 . . 𝑁} 𝑡_{𝑖} = 1$ is $𝑡_{1} = \dots = 𝑡_{𝑛} = \frac{1}{𝑁}$

convex-hull_(tag) := extra $0 \leq 𝑡_{𝑖} \leq 1$

simplex_(tag) := convex hull formed by affinely independent points

parallelogram_(tag) Due to symmetry, the description of parallelepiped can be simplified from the convex hull of $2 𝑛$ points to the description of $𝑛$ points, after selecting the origin

𝑡_{1} 𝑣_{1} + \dots + 𝑡_{𝑛} 𝑣_{𝑛}, 0 \leq 𝑡_{𝑖} \leq 1

parallelogram-simplex-correspond_(tag)

A parallelepiped can be $⨆$ decomposed into $𝑛!$ simplexes that are equivalent under translation and reflection

The $𝑛$ permutations of points $𝑣_{1}, \dots, 𝑣_{𝑛}$

\begin{matrix} 𝑡_{𝑖 (1)} 𝑣_{𝑖 (1)} + \dots + 𝑡_{𝑖 (𝑛)} 𝑣_{𝑖 (𝑛)} \\ 0 \leq 𝑡_{𝑖 (𝑛)} \leq \dots \leq 𝑡_{𝑖 (1)} \leq 1 \end{matrix}

Corresponding simplex

\begin{matrix} 𝑠_{0} 0 + 𝑠_{1} 𝑣_{𝑖 (1)} + (𝑠_{2} 𝑣_{𝑖 (1)} + 𝑣_{𝑖 (2)}) + \dots + 𝑠_{𝑛} (𝑣_{𝑖 (1)} + \dots + 𝑣_{𝑖 (𝑛)}) \\ \sum_{𝑖 = 0 . . 𝑛} 𝑠_{𝑖} = 1, 0 \leq 𝑠_{𝑖} \leq 1 \end{matrix}

with

\begin{matrix} 𝑣_{𝑖 (𝑛)} & = & 𝑠_{𝑛} \\ 𝑣_{𝑖 (𝑛 - 1)} & = & 𝑠_{𝑛} + 𝑠_{𝑛 - 1} \\ ⋮ \\ 𝑣_{𝑖 (1)} & = & 𝑠_{𝑛} + 𝑠_{𝑛 - 1} + \dots + 𝑠_{1} \end{matrix}

Conversely, a simplex also gives many parallelepipeds with it as one of the $𝑛!$ simplex blocks

The structural strength given by these two things is about the same

volume-of-parallelogram_(tag) Volume assumption for $ℝ^{𝑛}$

Translation invariant
Reflection invariant (unsigned volume)
Finite $⨆$ -> finite volume $\sum$
If $𝑣_{1}, \dots, 𝑣_{𝑛}$ are not linearly independent, then in the lower-dimensional subspace, so the $𝑛$ -order volume is defined as zero

volume-of-simplex_(tag) is $\frac{1}{𝑛!}$ of volume-of-parallelogram

shear-transformation_(tag) After decomposing the parallelepiped into simplexes, cut and translate to form a new parallelepiped with the same volume. Called shear transformation. e.g. $𝑡_{1} (𝑣_{1} + 𝑣_{2}) + 𝑡_{2} 𝑣_{2} + \dots + 𝑡_{𝑛} 𝑣_{𝑛}$

(image from p.587 of ref-3)

Shear transformation volume invariance is algebraically e.g. $(𝑣_{1} + 𝑣_{2}) \land 𝑣_{2} \land \dots \land 𝑣_{𝑛} = 𝑣_{1} \land 𝑣_{2} \land \dots \land 𝑣_{𝑛}$ or $det (\begin{matrix} 1 & 1 \\ 1 \\ ⋱ \\ 1 \end{matrix}) = 1$

Scaling of edges $ℕ, ℤ, ℚ, ℝ$ . e.g. $\forall 𝑎 \in ℝ, Vol (𝑎 𝑣_{1}, 𝑣_{2}, \dots, 𝑣_{𝑛}) = 𝑎 Vol (𝑣_{1}, 𝑣_{2}, \dots, 𝑣_{𝑛})$

The stretching and shearing of parallelepipeds corresponds to the decomposition of $GL (𝑛, ℝ)$ into elementary linear transformations, which is also used in Gaussian elimination (although they can be used for $𝑚 \times 𝑛$ matrices)

volume-determinant_(tag) The volume change of parallelepiped $𝑣_{1}, \dots, 𝑣_{𝑛}$ is $𝐴 \in GL (𝑛, ℝ), Vol (𝐴 𝑣_{1}, \dots, 𝐴_{𝑛} 𝑣_{𝑛}) = det 𝐴 Vol (𝑣_{1}, \dots, 𝑣_{𝑛})$

Choose a basis $𝑒_{1}, \dots, 𝑒_{𝑛}$ of $ℝ^{𝑛}$ , the volume of the parallelepiped generated by it is $1$ , and the volume of other parallelepipeds $𝐴 𝑒_{1}, \dots, 𝐴 𝑒_{𝑛}$ is $det 𝐴$

This is the oriented volume. $𝑣_{1} \land 𝑣_{2} \land \dots \land 𝑣_{𝑛} = - 𝑣_{2} \land 𝑣_{1} \land \dots \land 𝑣_{𝑛}$ The set of parallelepipeds remains the same, so the absolute volume remains the same, but the directions of $𝑣_{1}, 𝑣_{2}, \dots, 𝑣_{𝑛}$ and $𝑣_{2}, 𝑣_{1}, \dots, 𝑣_{𝑛}$ are opposite

Oriented volume = Unoriented volume + Direction factor

$𝑣_{1}, \dots, 𝑣_{𝑛}$ linearly dependent ==> in a lower-dimensional subspace ==> zero volume. At this time, $𝐴 \in GL$ can be extended to $𝐴 \in Lin$ , and zero volume algebraically corresponds to $𝐴 \notin GL ⟺ det (𝐴) = 0$

For $ℝ^{𝑛}$ 's $𝑘$ -th order parallelepiped and simplex

Map the parallelepiped to $ℝ^{𝑛}$ 's $𝑘$ -th order alternating tensor ${(ℝ^{𝑛})}^{\land 𝑘}$ 's decomposable element $𝑣_{1} \land \dots \land 𝑣_{𝑘}$

try-to-define-volume-of-low-dim_(tag) How to define low-dimensional volume? Consider two methods. Similar to linear form vs quadratic form. The first is like defining $(\begin{matrix} 𝑣_{1} \\ 𝑣_{2} \end{matrix})$ as $𝑣_{1} + 𝑣_{2}$ or $| 𝑣_{1} + 𝑣_{2} |$ , the second is similar to defining ${(𝑣_{1}^{2} + 𝑣_{2}^{2})}^{\frac{1}{2}}$ or $| 𝑣_{1}^{2} - 𝑣_{2}^{2} |^{\frac{1}{2}}$

A basis of $ℝ^{𝑛}$ gives a basis of the alternating tensor space $𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}}, 𝑖_{1} < \dots < 𝑖_{𝑘}$

Use it to define volume: For each $𝑘$ , a special alternating $𝑘$ multilinear function or $𝑘$ form of $ℝ^{𝑛}$ ${Vol}_{𝑛, 𝑘}$ , defined as ${Vol}_{𝑛, 𝑘} (𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}}) = 1$ , forall $𝑖_{1} < \dots < 𝑖_{𝑘}$

So for a general parallelepiped $𝑣_{1} \land \dots \land 𝑣_{𝑘} = (𝑣_{1}^{𝑖_{1}} 𝑒_{𝑖_{1}}) \land \dots \land (𝑣_{𝑘}^{𝑖_{𝑘}} 𝑒_{𝑖_{𝑘}}) = \sum_{𝑖_{1} < \dots < 𝑖_{𝑘}} det (\begin{matrix} 𝑣_{1}^{𝑖_{1}} & \dots & 𝑣_{𝑘}^{𝑖_{1}} \\ ⋮ & ⋮ \\ 𝑣_{1}^{𝑖_{𝑘}} & \dots & 𝑣_{𝑘}^{𝑖_{𝑘}} \end{matrix}) 𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}}$ the volume is

\begin{matrix} Vol (𝑣_{1} \land \dots \land 𝑣_{𝑘}) & ≔ & \sum_{𝑖_{1} < \dots < 𝑖_{𝑘}} det (\begin{matrix} 𝑣_{1}^{𝑖_{1}} & \dots & 𝑣_{𝑘}^{𝑖_{1}} \\ ⋮ & ⋮ \\ 𝑣_{1}^{𝑖_{𝑘}} & \dots & 𝑣_{𝑘}^{𝑖_{𝑘}} \end{matrix}) \\ or & ≔ & | \sum_{𝑖_{1} < \dots < 𝑖_{𝑘}} det (\begin{matrix} 𝑣_{1}^{𝑖_{1}} & \dots & 𝑣_{𝑘}^{𝑖_{1}} \\ ⋮ & ⋮ \\ 𝑣_{1}^{𝑖_{𝑘}} & \dots & 𝑣_{𝑘}^{𝑖_{𝑘}} \end{matrix}) | \end{matrix}

The volume of a nonzero decomposable alternating tensor can be zero, $𝐴 = (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) \in GL (2, ℝ)$ such that $Vol (2, 1) (𝐴 𝑒_{1}) = 1 - 1 = 0$ . The shear transformation of order $𝑛$ does not hold for order $𝑘$

Question A special basis is selected, so what other bases have the same result? or what is the linear subgroup that keeps the volume unchanged?

$SL (𝑛, ℝ)$ does not preserve $𝑘 < 𝑛$ dimensional volume. e.g. $(\begin{matrix} \frac{1}{2} \\ 2 \end{matrix})$ or $(\begin{matrix} - 1 \\ - 1 \end{matrix})$ does not preserve $1$ dimensional volume

matrix $𝐴$ that preserves all order volumes $𝐴 = (\begin{matrix} 𝑎_{1}^{1} & \dots & 𝑎_{𝑛}^{1} \\ ⋮ & ⋮ \\ 𝑎_{1}^{𝑛} & \dots & 𝑎_{𝑛}^{𝑛} \end{matrix}) \in GL (𝑛, ℝ)$ satisfies, for $𝑘 = 1, \dots, 𝑛$ for $𝑖_{1} < \dots < 𝑖_{𝑘}$ , ${Vol}_{𝑛, 𝑘} (𝐴 𝑒_{𝑖_{1}} \land \dots \land 𝐴 𝑒_{𝑖_{𝑘}}) = 1$ , or $\sum_{𝑗_{1} < \dots < 𝑗_{𝑘}} det (\begin{matrix} 𝑎_{𝑖_{1}}^{𝑗_{1}} & \dots & 𝑎_{𝑖_{𝑘}}^{𝑗_{𝑖}} \\ ⋮ & ⋮ \\ 𝑎_{𝑖_{1}}^{𝑗_{𝑘}} & \dots & 𝑎_{𝑖_{𝑘}}^{𝑗_{𝑘}} \end{matrix}) = 1$

Example ${Vol}_{𝑛, 1} (𝐴 𝑒_{𝑖}) = 𝑎_{𝑖}^{1} + \dots + 𝑎_{𝑖}^{𝑛}$ (sum of elements in the $𝑖$ th column). The $𝑛 - 1$ and $1$ cases are similar, i.e. $𝑎_{𝑗}^{𝑖}$ corresponds to the remaining sub式

(The cofactor is used in the $1, 𝑛 - 1$ alternating tensor decomposition representation of $det$ , which can be generalized to the $𝑘, 𝑛 - 𝑘$ alternating tensor decomposition representation or Laplace expansion of $det$ )

Example ${Vol}_{2, 1} (𝐴 𝑒_{𝑖}) = 𝑎_{𝑖}^{1} + 𝑎_{𝑖}^{2}$

matrix $𝐴$ preserves all $ℝ^{2}$ volumes $𝐴 = (\begin{matrix} 𝑎_{1}^{1} & 𝑎_{2}^{1} \\ 𝑎_{1}^{2} & 𝑎_{2}^{2} \end{matrix})$ satisfies

\begin{matrix} 𝑎_{1}^{1} 𝑎_{2}^{2} - 𝑎_{1}^{2} 𝑎_{2}^{1} & = & 1 \\ 𝑎_{1}^{1} + 𝑎_{1}^{2} & = & 1 \\ 𝑎_{2}^{1} + 𝑎_{2}^{2} & = & 1 \end{matrix}

a coordinate representation of the solution

\begin{matrix} 𝑥 & \in & ℝ \\ 𝑎_{1}^{1} & = & 1 - 𝑥 \\ 𝑎_{1}^{2} & = & 𝑥 \\ 𝑎_{2}^{1} & = & - 𝑥 \\ 𝑎_{2}^{2} & = & 1 + 𝑥 \\ 𝐴 & = & (\begin{matrix} 1 - 𝑥 & - 𝑥 \\ 𝑥 & 1 + 𝑥 \end{matrix}) \end{matrix}

is an affine line of $gl (2, ℝ)$ passing through $𝟙 = (\begin{matrix} 1 \\ 1 \end{matrix})$ . $SO (2)$ or $SO (1, 1)$ is not its subset

Select a non-degenerate quadratic form

#link(<tensor-induced-quadratic-form>)[Derive] the quadratic form of the alternating space ${⟨ 𝑣_{1} \land \dots \land 𝑣_{𝑘} ⟩}^{2} = det ⟨ 𝑣_{𝑖}, 𝑣_{𝑗} ⟩$ . Undirected volume $| det ⟨ 𝑣_{𝑖}, 𝑣_{𝑗} ⟩ |^{\frac{1}{2}}$ or ${| det (\begin{matrix} ⟨ 𝑣_{1}, 𝑣_{1} ⟩ & \dots & ⟨ 𝑣_{1}, 𝑣_{𝑛} ⟩ \\ ⋮ & ⋮ \\ ⟨ 𝑣_{𝑛}, 𝑣_{1} ⟩ & \dots & ⟨ 𝑣_{𝑛}, 𝑣_{𝑛} ⟩ \end{matrix}) |}^{\frac{1}{2}}$ . According to the orthonormal basis and its coefficients $𝑣_{1} \land \dots \land 𝑣_{𝑘} = \sum_{𝑖_{1} < \dots < 𝑖_{𝑘}} det (\begin{matrix} 𝑣_{1}^{𝑖_{1}} & \dots & 𝑣_{𝑘}^{𝑖_{1}} \\ ⋮ & ⋮ \\ 𝑣_{1}^{𝑖_{𝑘}} & \dots & 𝑣_{𝑘}^{𝑖_{𝑘}} \end{matrix}) 𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}}$ , write it as a standard quadratic form

\begin{matrix} {⟨ 𝑣_{1} \land \dots \land 𝑣_{𝑘} ⟩}^{2} & = & \sum_{𝑖_{1} < \dots < 𝑖_{𝑘}} {(det (\begin{matrix} 𝑣_{1}^{𝑖_{1}} & \dots & 𝑣_{𝑘}^{𝑖_{1}} \\ ⋮ & ⋮ \\ 𝑣_{1}^{𝑖_{𝑘}} & \dots & 𝑣_{𝑘}^{𝑖_{𝑘}} \end{matrix}))}^{2} {⟨ 𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}} ⟩}^{2} \\ {Vol}_{𝑛, 𝑘} (𝑣_{1} \land \dots \land 𝑣_{𝑘}) & ≔ & {| \sum_{𝑖_{1} < \dots < 𝑖_{𝑘}} {(det (\begin{matrix} 𝑣_{1}^{𝑖_{1}} & \dots & 𝑣_{𝑘}^{𝑖_{1}} \\ ⋮ & ⋮ \\ 𝑣_{1}^{𝑖_{𝑘}} & \dots & 𝑣_{𝑘}^{𝑖_{𝑘}} \end{matrix}))}^{2} {⟨ 𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}} ⟩}^{2} |}^{\frac{1}{2}} \end{matrix}

${⟨ 𝑣_{1} \land \dots \land 𝑣_{𝑛} ⟩}^{2} = 0$ <==> volume is zero

In the non-Euclidean case, light-like will have an impact

Different signature volume definitions will be different for the same set of order $𝑘 < 𝑛$

The two volume definitions coincide for $𝑘 = 𝑛$

convex-hull-decomposition_(tag) convex hull optimal decomposition to simplex, the method is not unique. Troublesome combinatorial problem

Example

$ℝ^{2}$ 's $4$ points

$ℝ^{2}$ 's $5$ points. First select $2$ simplex, that is, select $3$ vertices

Find out which simplex combinations are decompositions of the convex hull

The intersection of convex hulls is a convex hull

Example

The reduced set of a simplex may not be a convex hull. But it can still be decomposed into simplex

Example

polyhedra_(tag) Polyhedron :=

n simplex finite union with

internally disjoint
transitively connected between two n simplex
the transitive boundary is n-1 simplex

The dimension of the transitive boundary is to give the polyhedron the best connectivity

low-dim-polyhedra_(tag) Low-dimensional sub-polyhedra. As a submanifold-like setting? i.e. Adjacent simplexes with $𝑘 - 1$ boundaries in $ℝ^{𝑘}$ dimension have only two -> piecewise embedded in $ℝ^{𝑛}$ . Otherwise, consider the example of a three-connected boundary

Countable generalization -> Countable polyhedron

polyhedra-measurable_(tag)

Polyhedron measurable set $𝐴$ . Approximate with a countable polyhedron $𝑃$ , #link(<symmetric-set-minus>)[symmetric difference] $𝐴 Δ 𝑃$ cover with countable simplexes as a measure estimate error

Sets $𝐴, 𝐵$ define distance (ref-12)

𝑑 (𝐴, 𝐵) ≔ \inf_{\begin{matrix} polyhedra 𝐶 \\ 𝐴 Δ 𝐵 \subset 𝐶 \end{matrix}} Vol (𝐶)

Measurable set $𝐴$ := $\inf_{polyhedra 𝑃} 𝑑 (𝐴, 𝑃) = 0$

Distance from set $𝐴$ to "origin" $\emptyset$ is $𝐴 Δ \emptyset = 𝐴$ and $𝑑 (𝐴) : = 𝑑 (𝐴, \emptyset) = \inf_{\begin{matrix} polyhedra 𝐶 \\ 𝐴 \subset 𝐶 \end{matrix}} Vol (𝐶)$

$𝑑 (𝐴 Δ 𝐵) = 𝑑 (𝐴, 𝐵)$

If $𝐴 \subset 𝐴^{'}$ then $𝑑 (𝐴) \leq 𝑑 (𝐴^{'})$

$𝑑 (𝐴 \cup 𝐴^{'}) \leq 𝑑 (𝐴) + 𝑑 (𝐴^{'})$ Proof by $(𝐴 \subset 𝑃) \land (𝐴^{'} \subset 𝑃^{'}) ⟹ (𝐴 \cup 𝐴^{'}) \subset (𝑃 \cup 𝑃^{'})$

Note that such measurable sets have good connectivity. In one dimension, there are only intervals, excluding the Smith–Volterra–Cantor set, etc. Operations such as the union of polyhedral measurable sets are also restricted.

Lebesgue-measurable_(tag) If transitive connectivity is not used, then the definition of a general measurable set is obtained. alias: Lebesgue measurable set. Non-measurable sets exist.

Lebesgue-measure_(tag)

The symmetric difference of sets satisfies

$𝐵 Δ 𝐵^{'} \subset (𝐴 Δ 𝐵) \cup (𝐴 Δ 𝐵^{'})$

Corresponding triangle inequality $𝑑 (𝐵, 𝐵^{'}) \leq 𝑑 (𝐴, 𝐵) + 𝑑 (𝐴, 𝐵^{'})$

Proof $𝐵 ∖ 𝐵^{'} \subset (𝐵 ∖ 𝐴) \cup (𝐴 ∖ 𝐵^{'})$

\begin{matrix} 𝑥 \in 𝐵 ∖ 𝐵^{'} & ⟺ & 𝑥 \in 𝐵 \land 𝑥 \notin 𝐵^{'} \\ ⟺ & (𝑥 \in 𝐵 \land 𝑥 \notin 𝐵^{'}) \land (𝑥 \notin 𝐴 \lor 𝑥 \in 𝐴) \\ ⟺ & (𝑥 \in 𝐵 \land 𝑥 \notin 𝐵^{'} \land 𝑥 \notin 𝐴) \lor (𝑥 \in 𝐵 \land 𝑥 \notin 𝐵^{'} \land 𝑥 \in 𝐴) \\ ⟹ & (𝑥 \in 𝐵 \land 𝑥 \notin 𝐴) \lor (𝑥 \in 𝐴 \land 𝑥 \notin 𝐵^{'}) \\ ⟺ & 𝑥 \in (𝐵 ∖ 𝐴) \cup (𝐴 ∖ 𝐵^{'}) \end{matrix}

The other side is similar

Triangle inequality

\begin{matrix} 𝑑 (𝐵, 𝐵^{'}) & = & 𝑑 (𝐵 Δ 𝐵^{'}) \\ \leq & 𝑑 ((𝐴 Δ 𝐵) \cup (𝐴 Δ 𝐵^{'})) \\ \leq & 𝑑 (𝐴, 𝐵) + 𝑑 (𝐴, 𝐵^{'}) \end{matrix}

For polyhedra $𝑃, 𝑃^{'}$ with finite volume and $𝑑 (𝐴, 𝑃), 𝑑 (𝐴, 𝑃^{'}) < 𝜀$

Unique limit

| Vol (𝑃) - Vol (𝑃^{'}) | = Vol (𝑃 Δ 𝑃^{'}) = 𝑑 (𝑃, 𝑃^{'}) \leq 𝑑 (𝐴, 𝑃) + 𝑑 (𝐴, 𝑃^{'}) < 2 𝜀

If we use the #link(<net>)[net] of a polyhedron approximating $𝐴$ , then there is a #link(<hom-limit>)[limit homomorphism] $Vol (𝐴) ≔ lim_{𝑑 (𝐴, 𝑃) \to 0} Vol (𝑃)$

Obtain the definition of finite measure. The definition of infinite measure comes from the countable approximation of finite measure, or $\sum_{𝑛 = 1 . . \infty} 𝜀_{𝑛} < 𝜀$ technique

try-to-define-low-dim-measure_(tag) Try to define $ℝ^{𝑛}$ 's $𝑘 < 𝑛$ dimensional measurable set. Since the codimension of the $𝑘$ region $\neq 0$ , it is obvious that we cannot use set difference and simplex covering as measure estimation errors to approximate a general " $𝑘$ dimensional set"

pathologic-example-measure-of-boundary_(tag)

Using the Euclidean metric structure, some low-dimensional measurable sets can be defined, but there are still pathological examples (temporarily ignore the details, wiki it yourself)

Painter's paradox. Measure is finite but the measure of the boundary is infinite. Unbounded region is used
Koch snowflake. The measure is finite, but the measure of the boundary is either undefinable or infinite. Uses a boundary that is nowhere differentiable.

An example of approximating $𝑛$ volume but not approximating the boundary volume

Schwarz Lantern
Infinite staircase approaching the hypotenuse of a triangle $\sqrt{2} = 2$ or circle ( $𝜋 = 4$ ) or as long as there is large normal oscillation, $\sqrt{2} = 𝜋 = \infty$

measure-theoretic-boundary_(tag)

Measure theory boundary. Dimension — some supremum $𝑑 < 𝑛$ — may not be a natural number but a real number

For measurable sets of polyhedra, intuitively, boundary = the maximum minus the minimum in the zero-measure set quotient of the measurable set $⋃ [𝐴] ∖ ⋂ [𝐴]$

For general measurable sets, intuitively, boundary =

{𝑥 \in ℝ^{𝑛} : \neg lim_{simp \to 𝑥} \frac{Vol (𝐴 \cap simp)}{Vol (simp)} = 0, 1}

where $simp \to 0$ means that the overall scaling of a simplex centered on any $𝑥$ goes to zero

or boundary = neither inside nor outside. Inside = limit $1$ , outside = limit $0$

Lebesgue differentiation theorem says that the measure of the boundary is zero

The interval division of the sides of a rectangle gives a rectangular product-style division
Connecting the center of a simplex to $𝑛$ points has $(\binom{𝑛 + 1}{𝑛}) = 𝑛$ ways to divide a simplex into $𝑛$ sub simplices
Or use the midpoint of all lower-dimensional simplices on the boundary