Lebesgue Measure on Euclidean Space — Examples

By [L1], every interval with endpoints $a\le b$ and every box with real side parameters has Lebesgue measure equal to its geometric length or volume, whatever choice of open and closed faces is made.

The rationals form a Lebesgue null subset of $\mathbb{R}$, so ${\lambda }_1(\mathbb{Q})=0$.

Since $[0,1]=(\mathbb{Q}\cap [0,1])\bigsqcup ([0,1]\setminus \mathbb{Q})$ and ${\lambda }_1([0,1])=1$ by step 1.1, step 1.2 and [L3] give ${\lambda }_1([0,1]\setminus \mathbb{Q})=1$.

Step 1.1 also covers the boundary cases: if $a=b$ then the interval has measure $0$, and the empty box has measure $0$.

By induction from [F1], the $n$-th stage removes exactly $2^n$ pairwise disjoint open intervals, each of length $3^{-n-1}$; the case $n=0$ is the single interval $(\frac{1}{3},\frac{2}{3})$.

By [L2], the union removed at stage $n$ has Lebesgue measure $2^n{3}^{-n-1}$, so the full removed set has measure ${\sum }_{n=0}^{\mathrm{\infty }}{2}^n{3}^{-n-1}=\frac{1}{3}{\sum }_{n=0}^{\mathrm{\infty }}(\frac{2}{3}{)}^n=1$ by [F2].

The complement of $C$ in $[0,1]$ is exactly the union of those removed intervals, so step 2.1 computes ${\lambda }_1([0,1]\setminus C)=1$; this agrees with [L1], which already gives ${\lambda }_1(C)=0$.

At stage $n$, the set $S_n$ is a disjoint union of $2^n$ closed intervals of common length ${\lambda }_n$, so ${\lambda }_1(S_n)=2^n{\lambda }_n$.

Put $u_n:=2^n{\lambda }_n$. Then $u_0=1$ and $u_{n+1}=u_n-2^{-n-2}$ by [F1], so an induction together with [F4] gives $u_n=\frac{1}{2}+2^{-n-1}$ for every $n\in \mathbb{N}$.

The sets $S_n$ decrease to $S$, and ${\lambda }_1(S_0)=1<+\mathrm{\infty }$, so [F2] yields ${\lambda }_1(S)={\inf }_n{\lambda }_1(S_n)={\lim }_n(\frac{1}{2}+2^{-n-1})=\frac{1}{2}$.

Since the rationals in $(0,1)$ are countably infinite, fix an enumeration $(q_k{)}_{k\in \mathbb{N}}$ of $\mathbb{Q}\cap (0,1)$.

Put $U:={\bigcup }_{k\in \mathbb{N}}((q_k-\epsilon 2^{-k-3},q_k+\epsilon 2^{-k-3})\cap (0,1))$. This set is open, and [L1], [F1] and [F3] give ${\lambda }_1(U)\le {\sum }_{k=0}^{\mathrm{\infty }}\epsilon 2^{-k-2}=\epsilon \mathrm{/}2<\epsilon$.

Every rational point of $(0,1)$ lies in $U$, so $U$ is dense in $(0,1)$.

The rational line $\mathbb{Q}$ is countable, hence null by [L2]. Applying [L1] to $\mathbb{Q}\subseteq \mathbb{R}$ gives a $G_{\delta }$ set $G\supseteq \mathbb{Q}$ with ${\lambda }_1(G)=0$.

Because $\mathbb{Q}$ is dense in $\mathbb{R}$ and $\mathbb{Q}\subseteq G$, the set $G$ is dense; and [F1] records that it is $G_{\delta }$.

Write $G={\bigcap }_{n\in \mathbb{N}}{V}_n$ with each $V_n$ open, as in [F1]. Since $G$ is dense and $G\subseteq V_n$, every $V_n$ is dense, so each $F_n:=\mathbb{R}\setminus V_n$ is closed with empty interior; hence $\mathbb{R}\setminus G={\bigcup }_{n\in \mathbb{N}}{F}_n$ is meager by [F2]. Now let $I$ be any bounded interval in $\mathbb{R}$. Step 1.1 gives ${\lambda }_1(I\cap G)=0$ because $I\cap G\subseteq G$, while [L3] makes $I$ Lebesgue measurable with finite measure equal to its length; therefore [L4] applied to the disjoint union $I=(I\cap G)\bigsqcup (I\setminus G)$ gives ${\lambda }_1(I\setminus G)={\lambda }_1(I)$, so the complement has full measure on every bounded interval.

Fix a real $\eta >0$. For each integer $m$, the interval $[m,m+1]$ is compact by [F3], so [L3] gives a real ${\delta }_m>0$ such that $x,y\in [m,m+1]$ and $\mathrm{\mid }x-y\mathrm{\mid }<{\delta }_m$ imply $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\eta 2^{-\mathrm{\mid }m\mathrm{\mid }-3}$. By [F4] choose a natural number $N_m\ge 1$ with $1\mathrm{/}{N}_m<{\delta }_m$, and put $t_j:=m+j\mathrm{/}{N}_m$ for $0\le j\le N_m$. Then for every $j$ with $1\le j\le N_m$ and every $x\in [t_{j-1},t_j]$ one has $\mathrm{\mid }x-t_{j-1}\mathrm{\mid }\le 1\mathrm{/}{N}_m<{\delta }_m$, hence $\mathrm{\mid }f(x)-f(t_{j-1})\mathrm{\mid }<\eta 2^{-\mathrm{\mid }m\mathrm{\mid }-3}$.

For each such subinterval, the corresponding graph piece lies in the closed box $[t_{j-1},t_j]\times [f(t_{j-1})-\eta 2^{-\mathrm{\mid }m\mathrm{\mid }-3},f(t_{j-1})+\eta 2^{-\mathrm{\mid }m\mathrm{\mid }-3}]$, whose width is $t_j-t_{j-1}$ and whose height is $\eta 2^{-\mathrm{\mid }m\mathrm{\mid }-2}$. Therefore [L2] gives a box cover of ${\mathrm{\Gamma }}_f\cap ([m,m+1]\times \mathbb{R})$ whose total area is ${\sum }_{j=1}^{N_m}(t_j-t_{j-1})\eta 2^{-\mathrm{\mid }m\mathrm{\mid }-2}=\eta 2^{-\mathrm{\mid }m\mathrm{\mid }-2}$.

Summing these covers over all integers $m$ and using [F1], the whole graph is covered by countably many closed boxes with total area at most $\eta {\sum }_{m\in \mathbb{Z}}{2}^{-\mathrm{\mid }m\mathrm{\mid }-2}=\eta (2^{-2}+2{\sum }_{n=1}^{\mathrm{\infty }}{2}^{-n-2})=3\eta \mathrm{/}4<\eta$, the geometric-series identity coming from [F5]. Since $\eta >0$ was arbitrary, [L1] gives that ${\mathrm{\Gamma }}_f$ is Lebesgue null.

The matrix of $T$ is $\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$, so [F1] gives $\det T=2\cdot 3-1\cdot 1=5$; therefore [L1] gives ${\lambda }_2(T[(0,1{]}^2])=5{\lambda }_2((0,1{]}^2)=5$.

The matrix of $U$ is the upper triangular matrix $\left(\begin{array}{ccc}2& 1& 0\\ 0& 3& -1\\ 0& 0& 4\end{array}\right)$, so [F2] gives $\det U=2\cdot 3\cdot 4=24$; hence [L1] gives ${\lambda }_3(U[(0,1{]}^3])=24{\lambda }_3((0,1{]}^3)=24$.

The matrix of $S$ is $\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)$, so [F1] gives $\det S=0$, and $S[(0,1{]}^2]=\{(t,t):0<t\le 1\}$ lies in the proper linear subspace $\{(u,v):u=v\}$; therefore [L1] and [L2] give that $S[(0,1{]}^2]$ is Lebesgue null.

If $G\le (\mathbb{R},+)$ is measurable and proper, then the contrapositive of [L1] gives ${\lambda }_1(G)=0$.

The integers and rationals are at most countable subgroups of $(\mathbb{R},+)$, so [L2] gives them Lebesgue measure zero; they are proper because [F1] and [F2] show $\mathbb{Q}\mathrm{\ne }\mathbb{R}$, and $\mathbb{Z}\subseteq \mathbb{Q}$.

Step 1.1 uses measurability in an essential way: it says nothing about subgroups of $\mathbb{R}$ that are not Lebesgue measurable.

The witness set $E=\mathbb{Q}\cap [0,1]$ is countable, hence Lebesgue null by [L1].

Let finitely many bounded intervals cover $E$. Their union is closed in $[0,1]$, and because it contains the dense subset $E$ of $[0,1]$, [F3] makes it contain all of $[0,1]$.

Therefore [F2] gives total covering length at least $1$.

The single interval $[0,1]$ realises total length $1$, so [F1] gives Jordan outer content exactly $1$. Thus $E$ is Lebesgue null and still has Jordan outer content one.

What is true is [L1]: one inequality holds for every bounded set, and equality holds when the set is Jordan measurable.

The witness $E:=\mathbb{Q}\cap [0,1]$ has Lebesgue outer measure $0$ by [L2]; and if finitely many bounded intervals cover $E$, their union is closed and contains the dense subset $E$ of $[0,1]$, so it contains all of $[0,1]$, and [F2] forces total length at least $1$, while the single interval $[0,1]$ realises $1$, so [F1] gives Jordan outer content $1$.

So the false statement drops the Jordan measurability hypothesis from the true theorem [L1], and the witness of step 1.2 refutes it.

The Smith-Volterra-Cantor set is nowhere dense by [L1], so [F1] gives that it has empty interior.

The same source item [L1] says that $S$ is not null, so [L2] gives ${\lambda }_1(S)\mathrm{\ne }0$; since Lebesgue measure is nonnegative, this means ${\lambda }_1(S)>0$.

So a measurable set can have empty interior and still have positive Lebesgue measure; the Smith-Volterra-Cantor set refutes the statement.

By [F1] and [F2], counting measure restricted to the Borel sets of $\mathbb{R}$ is a measure. It is translation invariant because for every real $a$ the map $x\mapsto x+a$ is a bijection of $\mathbb{R}$, so a Borel set and its translate have the same finite cardinality or are both infinite.

It gives the singleton $\{0\}$ the value $1$ and the unit interval $[0,1]$ the value $+\mathrm{\infty }$, while [L2] gives Lebesgue measure $0$ for $\{0\}$ and $1$ for $[0,1]$; so no nonnegative scalar multiple of Lebesgue measure equals counting measure.

This does not contradict [L1], because [L1] fixes the value of the unit cube and thereby forces finiteness on bounded sets, which the false statement omits.

By [F4], the Axiom of Choice gives countable choice, so [L1] applies and the Cantor set has Lebesgue measure zero.

By [F1], the Cantor set is in bijection with $\{0,1{\}}^{\mathbb{N}}$, hence with $\mathcal{P}(\mathbb{N})$, so its power set has cardinality $2^{\mathfrak{c}}$.

Since [F4] gives countable choice and [L2] says Lebesgue measure is complete under that hypothesis, every subset of the Cantor set is Lebesgue measurable.

Assuming the Axiom of Choice, [F2] gives only $\mathfrak{c}$ Borel subsets of $\mathbb{R}$, while [F3] gives $2^{\mathfrak{c}}>\mathfrak{c}$; therefore not every Lebesgue measurable subset of $\mathbb{R}$ can be Borel.

The refutation is purely cardinal: it produces no particular measurable non-Borel set, only shows that one must exist.