The measure theory notes (John K. Hunter) states:
Definition 2.1
A $n$-dimensional closed rectangle is a subset $R \subset \mathbb{R}^{n}$ such that
$$R = [a_{1}, b_{1}] \times [a_{2}, b_{2}] \times \cdots \times [a_{2}, b_{2}]$$
where $-\infty \lt a_{i} \le b_{i} \lt \infty$ for $i = 1, \ldots, n$.
Denote collection of all $n$-dimensional rectangles as $\mathcal{R}(\mathbb{R}^{n})$
Volume of $R$ is:
$$\mu(R) = (b_{1} – a_{1})(b_{2} – a_{2})\cdots(b_{n} – a_{n})$$
We also consider empty set as a rectangle with $\mu(\emptyset) = 0$.
Then, $R \mapsto \mu(R)$ defines a map $\mu : \mathcal{R}(\mathbb{R}^{n}) \rightarrow [0, \infty)$.
Definition 2.2
The outer Lesbesgue measure $\mu^{*}(E)$ of a subset $E \subset \mathbb{R}^{n}$ is
$$\mu^{*}(E) = \inf \left \lbrace \sum_{i = 1}^{\infty} \mu(R_{i}) : E \subset \bigcup_{i = 1}^{\infty} R_{i}, R_{i} \in \mathcal{R}(\mathbb{R}^{n}) \right \rbrace$$
where the infinum is taken over all countable collections of rectangles whose union contains $E$.
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Question
In proof of 2.4, the notes states:
Consider the case that $\mu^{*}(E_{i})$ is finite for all $i \in \mathbb{N}$.
For all $\varepsilon > 0$, there exists a countable covering $\lbrace R_{ij} : j \in {N} \rbrace$ of $E_{i} \subset \mathbb{R}^{n}$ by rectangles $R_{ij}$ such that
$$\sum_{j=1}^{\infty} \mu(R_{ij}) \le \mu^{*}(E_{i}) + \frac{\varepsilon}{2^{i}}, \qquad E_{i} \subset \bigcup_{j=1}^{\infty} R_{ij}$$
I can prove $\sum_{j=1}^{\infty} \mu(R_{ij}) \lt \mu^{*}(E_{i})$ by a theorem about infimum (but not $\le$). How to show the two sides can be equal?
I think the proof should show the two sides can be equal in some case because a later proof (proposition) shows the $\ge$ case. Together the two cases show $\mu^{*}(R) = \mu(R)$.
I see the same thing in the proof here (equation 2 and 3): http://mathonline.wikidot.com/countable-subadditivity-of-the-lebesgue-outer-measure
Attempt to justify
Consider the one dimensional case.
$$\mu^{*}(E_{i}) = \inf \left \lbrace \sum_{i = 1}^{\infty} \mu(R_{i}) : E_{i} \subset \bigcup_{i = 1}^{\infty} R_{i}, R_{i} \in \mathcal{R}(\mathbb{R}) \right \rbrace$$
Theorem 2.19 in introduction to real analysis notes (Lee Larson) states:
For $\mathbb{F}$ be an ordered field, $A \subset \mathbb{F}$, and $\alpha \in \mathbb{F}$, we have:
$\alpha$ is the infimum (or greatest lower bound) of $A$ iff:
- $(-\infty, \alpha) \cap A = \emptyset$, and
- $\forall \varepsilon' > 0: [\alpha, \alpha + \varepsilon') \cap A \ne \emptyset$
Here,
- the ordered field $\mathbb{F} = \mathbb{R}$,
- the subset $A \subset \mathbb{F}$ is $\left \lbrace \sum_{i = 1}^{\infty} \mu(R_{i}) : E_{i} \subset \bigcup_{i = 1}^{\infty} R_{i}, R_{i} \in \mathcal{R}(\mathbb{R}) \right \rbrace$,
- the infimum is $\alpha = \mu^{*}(E_{i})$, and
- $\varepsilon' = \varepsilon / 2^{i}$.
Therefore, $\forall \varepsilon' > 0: [\alpha, \alpha + \varepsilon') \cap A \ne \emptyset$.
Then,
$$\forall \varepsilon' > 0: [\alpha, \alpha + \varepsilon') \cap \left \lbrace \sum_{i = 1}^{\infty} \mu(R_{i}) : E_{i} \subset \bigcup_{i = 1}^{\infty} R_{i}, R_{i} \in \mathcal{R}(\mathbb{R}) \right \rbrace \ne \emptyset$$
This implies for all $\varepsilon' > 0 $, there exists a collection $B = \lbrace R_{ij} \in \mathcal{R}(\mathbb{R}) : j \in \mathbb{N} \rbrace$ such that:
- $E_{i} \subset \bigcup_{j=1}^{\infty} B_{j}$, and
- $ \alpha \le \sum_{j = 1}^{\infty} \mu(B_{j}) \lt \alpha + \varepsilon'$
where $B_{j}$ is the jth element of the collection $B$.
That means:
$$ \sum_{j = 1}^{\infty} \mu(R_{ij}) \lt \mu^{*}(E_{i}) + \varepsilon / 2^{i}, \qquad E_{i} \subset \bigcup_{j=1}^{\infty} R_{ij} $$
Unfortunately, it is a $\lt$ sign instead of $\le$ sign …
Best Answer
In your edit, you have a completely different statement, $$ \sum_{j=1}^{\infty} \mu(R_{i j})<\mu^{*}\left(E_{i}\right)+\varepsilon / 2^{i} $$ is not at all the same as $$\sum_{j=1}^{\infty} \mu\left(R_{i j}\right)<\mu^{*}\left(E_{i}\right) $$ which is what was originally in your question.
Here, the $\le$ and $<$ statements are equivalent. A similar situation you may have encountered - Conflicting definitions of continuity (strict or non-strict inequality)? If its true that for every $\epsilon$, then the $\le$ statement for $\epsilon/2$ will give you $$ \sum_{j=1}^{\infty} \mu\left(R_{i j}\right)\le \mu^{*}(E_i)+(\varepsilon/2) / 2^{i} < \mu^*(E_{i})+\varepsilon / 2^i. $$
If you followed the proof all the way to the end with $< \dots + \epsilon$, when using the fact that $\epsilon$ is arbitrary, you again get $\le$, so there is no contradiction.