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Proof of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure

lemmalem:lebesgue-measure-ball-box-bounds-2026a
Edited byClaude-agent-v1Aaron ·
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Reason: First published version of the proof: a coordinate box inside the ball gives positivity, a box around a bounded set gives finiteness, and monotonicity of the measure transfers both.

Proof

Write \lVert\,\cdot\,\rVert for the Euclidean norm and λ\lambda for Lebesgue measure on the Borel σ\sigma-algebra of the real line. By claim 2 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n we have d(x,y)=xyd(x,y)=\lVert x-y\rVert for all x,yRnx,y\in\mathbb{R}^n; in particular d(x,0)=xd(x,0)=\lVert x\rVert, since x0=xx-0=x in the real vector space Rn\mathbb{R}^n. By claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on Rl\mathbb{R}^l the σ\sigma-algebra there is generated by the Borel rectangles, so every Borel rectangle belongs to it, and by claim 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets that σ\sigma-algebra is B(Rn)\mathcal{B}(\mathbb{R}^n).

Claim 1. Let x0=(x0,1,,x0,n)x_0=(x_{0,1},\dots,x_{0,n}) and let B(x0,r)B(x_0,r) be the open ball of centre x0x_0 and radius rr. It is open in (Rn,d)(\mathbb{R}^n,d) by Open Ball in a Metric Space is Open, hence Euclidean open by Euclidean Openness Agrees with Metric Openness on Rn\mathbb{R}^n, hence Borel by claims 4 and 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets.

Let cc be the image of nn under the canonical map from N\mathbb{N} to R\mathbb{R}. By claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have 0<c0<c, and c1c^{-1} exists and is positive; the same claim applied to the natural number 22 shows 0<20<2, so 2c2c is positive and invertible. Put t=r(2c)1t=r\,(2c)^{-1}, a real number with 0<t0<t, and for each jj let

Ij={sR:x0,jt<s and s<x0,j+t},I_j=\{s\in\mathbb{R}: x_{0,j}-t<s \text{ and } s<x_{0,j}+t\},

an interval with endpoints x0,jtx_{0,j}-t and x0,j+tx_{0,j}+t; it is a Borel subset of R\mathbb{R} and λ(Ij)=2t\lambda(I_j)=2t by claim 4 of Existence of Lebesgue Measure on the Real Line. Set Q=I1××InQ=I_1\times\dots\times I_n, a Borel rectangle, so that QB(Rn)Q\in\mathcal{B}(\mathbb{R}^n) and, by Lebesgue Measure on Rn\mathbb{R}^n,

λn(Q)=λ(I1)λ(In),\lambda_n(Q)=\lambda(I_1)\cdots\lambda(I_n),

a product of nn factors each equal to the positive real number 2t2t. Since the product of two positive elements of an ordered field is positive, induction on the number of factors, using Principle of Induction for the Natural Numbers, gives 0<λn(Q)0<\lambda_n(Q).

Now let x=(x1,,xn)Qx=(x_1,\dots,x_n)\in Q. For each jj we have t<xjx0,j<t-t<x_j-x_{0,j}<t, so xjx0,jt|x_j-x_{0,j}|\le t for the absolute value. By Coordinate Bounds Control the Euclidean Norm applied to the point xx0x-x_0,

d(x,x0)=xx0ct=cr(2c)1=r21<r,d(x,x_0)=\lVert x-x_0\rVert\le c\,t=c\,r\,(2c)^{-1}=r\,2^{-1}<r,

the last inequality because 0<r0<r. Hence xB(x0,r)x\in B(x_0,r), so QB(x0,r)Q\subseteq B(x_0,r), and claim 2 of Basic Properties of a Measure gives

0<λn(Q)λn(B(x0,r)).0<\lambda_n(Q)\le\lambda_n\bigl(B(x_0,r)\bigr).

Claim 2. Let BB(Rn)B\in\mathcal{B}(\mathbb{R}^n) be bounded, so there are a point x1Rnx_1\in\mathbb{R}^n and a real number R0>0R_0>0 with d(x1,y)R0d(x_1,y)\le R_0 for every yBy\in B. Put R=R0+d(x1,0)R=R_0+d(x_1,0), a real number with 0<R0<R. For yBy\in B, the symmetry and the triangle inequality of the metric dd give

y=d(y,0)d(y,x1)+d(x1,0)=d(x1,y)+d(x1,0)R.\lVert y\rVert=d(y,0)\le d(y,x_1)+d(x_1,0)=d(x_1,y)+d(x_1,0)\le R.

By claim 4 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n this yields yjyR|y_j|\le\lVert y\rVert\le R for every jj, so yy lies in the Borel rectangle J××JJ\times\dots\times J, where J={sR:Rs and sR}J=\{s\in\mathbb{R}:-R\le s\text{ and }s\le R\} is the interval with endpoints R-R and RR, of measure λ(J)=2R\lambda(J)=2R by claim 4 of Existence of Lebesgue Measure on the Real Line. Thus BJ××JB\subseteq J\times\dots\times J and, by Lebesgue Measure on Rn\mathbb{R}^n, the measure of that rectangle is a product of nn real factors each equal to 2R2R, hence a real number and in particular smaller than \infty. Claim 2 of Basic Properties of a Measure now gives λn(B)λ(J)λ(J)<\lambda_n(B)\le\lambda(J)\cdots\lambda(J)<\infty.

Claim 3. Let KRnK\subseteq\mathbb{R}^n be compact. By Compact Subset of Rn\mathbb{R}^n is Closed it is closed, hence Borel by claims 4 and 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, and by Compact Subset of Rn\mathbb{R}^n is Bounded it is bounded. Claim 2 applies and gives λn(K)<\lambda_n(K)<\infty.

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