# Show Regularity of two dimensional lebesgue measure

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.