# Arbitrary $\epsilon$ in Analysis Proof

I searched the Math stack, but I was unsuccessful in obtaining an adequate response. Please feel free to share a link if this is a repeat AND there is a good response.

I am encountering picking or fixing an $$\epsilon$$ for the following proof in the Exterior Measure section by Stein and Shakarchi (2009).

We are trying to prove $$|Q|\leq\sum^\infty_j|Q_j|.$$

The authors uses the fact that $$Q$$ is a compact set and chooses an open cube arbitrarily close to the original closed cube we started with, in conjunction with the Heine-Borel theorem and claim:

$$|Q|\leq\sum^N_j(1+\epsilon)|Q_j|\leq\sum^\infty_j(1+\epsilon)|Q_j|.$$

Then conclude, $$\epsilon$$ was chosen arbitrary, so it shows what we want to prove.

Why is this true? Is this because in Analysis, $$a_1=a_2$$ means $$|a_1-a_2|<\epsilon$$ for some arbitrarily small $$\epsilon$$?

More importantly, why do we use this trick? The pragmatic reason I see here is that we are starting with an arbitrary covering $$(Q_j)$$ that covers a closed cube $$Q$$. To get around it, we choose an aribtarily close-looking cube $$(S_j)$$ that each contains $$Q_j^s$$, so in order to do this the authors establish:

$$|S_j|\leq(1+\epsilon)|Q_j|$$

It seems to be without the epsilon such construction would not be possible. So is this the real reason why we say fix an arbitrary $$\epsilon$$ and then eventually make that go to zero for the inequality to ultimately hold?

Reference: $$\textit{Real Analysis: Measure Theory, Integration, and Hilbert Spaces}$$. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

Answering the "why do we use this trick?" question:

The authors want to use an open covering so they can reduce the countable covering to a finite one (by compactness of $$E$$). So, they are enclosing each $$Q_j$$ in an open cube $$S_j$$.

If $$Q_j$$ is not open, and $$S_j$$ is an open cube containing $$Q_j$$, then $$S_j$$ cannot have the same volume as $$Q_j$$; it must be strictly larger.

For example, in one dimension, the "cube" (interval) $$[a,b]$$ has length $$b-a$$, but any open interval containing $$[a,b]$$ has length strictly greater than $$b-a$$. However, we can find an open interval with length as close to $$b-a$$ as desired, for example, we can make it have length $$(b-a)(1+\epsilon)$$ by choosing the interval $$(a - \delta, b + \delta)$$, where $$\delta = (b-a)\epsilon/2$$.

Similarly, in $$n$$ dimensions, you can do something quite similar with a cube of volume $$V$$: you can find an open cube containing the original cube, with volume $$V(1+\epsilon)$$, or with volume $$V + \epsilon$$, or any other function of $$V$$ and $$\epsilon$$ that is slightly larger than $$V$$ and converges to $$V$$ as $$\epsilon \to 0$$.

Btw, I'm not sure why the authors aren't using open $$Q_j$$'s to begin with. Most authors define outer measure in terms of coverings by countable sequences of open intervals / cubes in part because they can be reduced to a finite subcover when covering a compact set.

• Hey, thanks for the response. Did you mean though $S_j$ being open cube rather than $Q_j$? Because $Q_j$ is a closed cube, and then we are finding an open cube is arbitrarily close to that. Please correct if that is what you meant. Thx. Commented Sep 9, 2019 at 18:55
• @FrankSwanton Yeah, that's what I meant when I said "if the $Q_j$'s are not open, we can't necessarily replace them with open cubes of the same volume." To put this more precisely, if $Q_j$ is not open, then any open cube $S_j$ containing $Q_j$ cannot have the same volume as $Q_j$. The volume of $S_j$ must be larger. I will edit to clarify this.
– user169852
Commented Sep 9, 2019 at 18:59
• Brilliant. Thank you. Commented Sep 9, 2019 at 19:29
• @FrankSwanton In this case, can we say $V=V+\mathcal{E}$? I mean you said $V$ converges to $V+\mathcal{E}$ as $\mathcal{E}$ goes to 0, but is this convergence means equal? Commented Dec 12, 2019 at 3:47

Is this because in Analysis, $$a_1=a_2$$ means $$|a_1-a_2|<\epsilon$$ for some arbitrarily small $$\epsilon$$?

What's in fact true is that $$|a_1-a_2|<\epsilon$$ for all $$\epsilon>0$$ means $$|a_1-a_2|=0,$$ which of course implies $$a_1=a_2.$$ The converse is even easier to see.

More importantly, why do we use this trick?

I don't see a trick here. I simply see it as an application of the triangle inequality.