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I want to show the regularity of the $n$-dimensional lebesgue measure. I figured, that if I can show it for the two-dimensional case, I should be able to generalize this result to the $n$-dimensional case. More precisely I want to show that \begin{gather*} m^n(A)=\inf\{m^n(V)\mid A\subset V, V \text{ open}\} \\ m^n(V)=\sup\{m^n(K)\mid K\subset V, K \text{ compact}\}. \end{gather*}

Prelims

It should be noted that

  • I know that the one dimensional lebesgue measure is regular. So I am hoping to be able to utilize this fact.

  • The way the $n$-dimensional lebesgue measure was formulated as the product measure of $n$ one-dimensional lebesgue measures. In particular, for the two dimensional case the formulation for $m^2$ I know is the following

Consider all measurable rectangles $A\times B\in \mathcal{B}\times\mathcal{B}$, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$. Consider the algebra of finite unions of disjoint measurable rectangles and define a premeasure $\mu_{0}$ over this algebra as follows.

$$ \mu_{0}\left(\bigcup_{i=1}^{k}(A_{i}\times B_{I})\right) =\sum_{i=1}^{n}m(A_{i})m(B_{i})$$

Then, using the caratheodory extension theorem, we get a measure on $\mathcal{B}\otimes \mathcal{B}=\sigma(\mathcal{B}\times \mathcal{B})$. The reason why I am saying all this is because I am aware that there are many formulations of the $n$-dimensional lebesgue measure, however I would like answers that appeal to this construction of the lebesgue measure (even though I am aware that all the formulations give the same thing, note I only have awareness, the only formulation I know is the one described above).

My attempt

My attempt for outer regularity: I have tried first showing outer regularity on measurable rectangles.

Consider a measurable rectangles $A\times B\in \mathcal{B}\times \mathcal{B}$. Then, since the lebesgue measure is outer regular, for $\epsilon>0$ there exist open sets in \mathbb{R} $V_{A}$ and $V_{B}$ such that $m(V_{A}-A)<\epsilon$ and $m(V_{B}-B)<\epsilon$.

Then, we have that $$ m^2(V_{A}\times V_{B}-A\times B )<\epsilon^2 + \epsilon m(A) + \epsilon m(B). $$ Provided $m(A)$ and $m(B)$ are finite, we can choose epsilon to be arbitrarily small and hence make $m^2(V_{A}\times V_{B}-A\times B)$ arbitrarily small. I am concerned if one of $m(A)$ or $m(B)$ is infinity and the other is zero. In such a case $m^2(A\times B)$ is defined to be zero and we cannot use the above argument to show outer regularity, what should I do?

Supposing that all worked though, I can see how this would work for general set in $\mathcal{B}\otimes \mathcal{B}$. Since every element of $\mathcal{B}\otimes \mathcal{B}$ is the countable union of measurable rectangles we can find open rectangles of the form $V_{i}\times W_{i}\in \mathcal{B}\times \mathcal{B}$ such that $$ m^2(V_{i}\times W_{i}-A_{i}\times B_{i})<\frac{\epsilon}{2^{i}}. $$

My attempt for inner regularity: Since there is a countable basis for the topology of $\mathbb{R}\times \mathbb{R}$ we know that any open set in $\mathbb{R}\times \mathbb{R}$ can be expressed as the countable union of elements of the form $V_{i}\times W_{i}$ where $V_{i}$ and $W_{i}$ are in the countable basis for the topology of $\mathbb{R}$.

Let $U\in \mathbb{R}\times \mathbb{R}$ be an open set and let $U=\bigcup_{i=1}^{\infty}V_{i}\times W_{i}$. For each $V_{i}\times W_{i}$ we can use a similar trick we used in the proof of outer regularity to find $K_{i}^1$ and $K_{i}^2$, which are compact sets, such that $$ m^2(V_{i}\times W_{i}-K_{i}^1\times K_{i}^2) <\frac{\epsilon}{2^{i}}, $$ which is all good, but the set $\bigcup_{i=1}^{\infty}K_{i}^1\times K_{i}^2$ is not necessarily compact anymore.

Any help on how to solve this question is greatly appreciated. Sorry my attempt and explanation are a bit long.

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