Lebesgue Measure: Existence and Uniqueness

Let $x \in \mathbb {R}$ and $k \geqslant 2$. Then $\lambda (\left\{ x\right\} )=\lambda_{k}\left(\left\{ x\right\} \times \mathbb {R}^{k-1}\right)=0$. Next, for any $j \in \left\{ 1, \ldots , k-1\right\}$ and $\boldsymbol {x} \in \mathbb {R}^{j}$ we have $\lambda_{k}\left(\left\{ \boldsymbol {x}\right\} \times \mathbb {R}^{k-j}\right)=0$.

Enumerate the rationals in $(0,1]$ by $\left\{ q_{1}, q_{2}, \ldots \right\}$. Given arbitrarily small $\varepsilon >0$, remove the interval $A_{n}=\left(q_{n}-\varepsilon / 2^{n+1}, q_{n}+\varepsilon / 2^{n+1}\right) \cap (0,1]$. Let $A=\bigcup_{n=1}^{\infty } A_{n}$. Then $\bar{\lambda }(A) \leqslant \varepsilon$, despite the fact that $A$ is an open dense subset of $(0,1]$. Also, we have $\bar{\lambda }((0,1] \backslash A) \geqslant 1-\varepsilon$, even though $(0,1] \backslash A$ is a nowhere dense subset of $(0,1]$.

There cannot exist a closed subset of $(0,1]$ whose interior is empty, yet has $\bar{\lambda }$-measure of one.

$\lambda_{k}$ is nonatomic: any $A \in \mathscr {B}^{k}$ with $\lambda_{k}(A)>0$ has a proper subset $B \in \mathscr {B}^{k}$ with $0<\lambda_{k}(B)<\lambda_{k}(A)$. This forces $\bar{\lambda }_{k}$ to be nonatomic as well.

Let $k \geqslant 2$. There exists an uncountable set $U \in \mathscr {B}^{k}$ with $\lambda_{k}(U)=0$.

$\left|\mathscr {L}^{k}\right|=2^{\mathrm{c}}$ and $\left|\mathscr {B}^{k}\right|=\mathrm{c}$ for each $k \in \mathbb {N}$.

Assume that $\mathscr {L}=2^{\mathbb {R}}$; in particular every one of the $2^{\mathrm{c}}$ subsets of $\left[0,1\right]$ is a Lebesgue set. Let $B=\left\{ \bar{\lambda }(A): A \subseteq [0,1], \bar{\lambda }(A) \notin A\right\}$. Consideration of the set $B$ (which is in $\mathscr {L}$ be assumption) leads to contradiction.

Let $k \geqslant 2$. Every line is in $\mathscr {B}^{k}$, has $\lambda_{k}$-measure zero, and hence has $\bar{\lambda }_{k}$-measure.

Let $\mathscr {B}_{(0,1]}=\sigma (\left\{ (a, b]:(a, b] \subseteq (0,1]\right\} )$.