I'm studying Measure Theory by myself and I would appreciate some guidance about my proof.
My textbook constructs an outer measure as following:
$$m_*(E)=\inf\sum_{k=1}^{\infty}|Q_k|$$
where the infimum is taken over all coverings of $E
$ with closed cubes.
In an exercise it asks to prove that the above definion, only with rectangles instead of cubes, will give rise to the same measure, i.e. I need to prove that:
$$m_*(E)=\inf\sum_{k=1}^{\infty}|Q_k|=\inf\sum_{k=1}^{\infty}|R_k|=m_*^R(E)$$
where on the left the infimum is taken over covering with cubes, and on the right is taken over covering with rectangles.
The fact that $m_*(E)\geq m_*^R(E)$ is obious.
STEP 1: When the infimum is taken only over "rational" rectangles, i.e. rectangles in which the ratio between all edges is rational, then each of these rectangles can be devided into a finitely many cubes and it follows that the infimum must be equal .
STEP 2: This is the step that I'm not sure about. What is left to prove that coverings with rational rectangles give rise to the same measure as covering with general rectangles. Let $\{R_j\}$ be a covering of $E$ with rectangles and let $\{R_j^*\}$ be the subseries of not rational rectangles in $\{R_j\}$. Then each $R_j^*$ can be extended to a rational rectangle $R_j^{**}$ so that $|R_j^{**}|=|R_j^{*}|+\epsilon/2^j$. We remain with a covering of $E$ and thus:
$$\inf \sum_{j=1}^{\infty}|R_j|=\inf\left(\sum_{rational\ rect.}|R_j|+\sum_{irrational\ rect.}|R_j^{*}|\right) = \inf\left(\sum_{rational\ rect.}|R_j|-\sum_{j=1^\ }^\infty\frac{\epsilon}{2^j}\right)$$
$$=\inf\left(\sum_{rational\ rect.}|R_j|\right)-\epsilon$$
And since $\epsilon$ is arbitrary the claim is proven.
Is it correct?
Thanks!
Best Answer
You have the right idea, but the notation could be improved in your second step. If $(R_j)_j$ is an enumeration of a countable covering of $E$ with rectangles, I wouldn't bother separating the rational ones from the ones that are not: Fix $\varepsilon > 0$. For every $j$, $R_j$ is contained in a slightly bigger rational rectangle $R^{**}_j$, say with $|R^{**}_j| \leq |R_j| + \varepsilon/2^j$. Thus $$\sum_j |R_j| \geq \sum_j |R^{**}_j| - \varepsilon \geq \inf \sum_j |Q_j| - \varepsilon,$$ where the infimum on the right runs over all rational rectangles covering $E$. Taking the infimum over all $(R_j)_j$ and then letting $\varepsilon \to 0$, we get the result together with step 1.