What does "take over" mean in the "the inf being taken over all countable coverings of E by open elementary sets"?

Question

I'm studying real analysis with Rudin Principles of Mathematical Analysis textbook.

I'm confused about the expression "the inf/sup being taken over ~~" in several definitions.

In principles of mathematical analysis Page 304, outer measure is defined as follows.

Definition $11.7$: Let $\mu$ be additive, regular, nonnegative, and finite on $\mathcal{E}$. Consider countable coverings of any set $E \subset \mathbb{R}^n$ by open elementary sets $A_n$: $$E \subseteq \bigcup_{n=1}^\infty A_n.$$ Define $$\mu^\ast(E) = \inf \sum_{n=1}^\infty \mu(A_n),$$ the infimum being taken over all countable coverings of $E$ by open elementary sets. $\mu^\ast(E)$ is called the outer measure of $E$, corresponding to $\mu$.

What does the "take over" mean in the expression "the infimum being taken over all countable coverings of E."?

I'm not a native English speaker. So I'm not sure what this expression exactly means.

I thought the expression is actually restricting the sets $A_n$.

Suppose E is a closed interval [0,1].

Then, $A_1 $ and $A_2$ can't be [0,2/3] and [1/3,1] because they are overlapped and the result of summation $\sum_{n=1}^\infty$ can't be infimum.

So I thought the expression "the inf being taken over all countable coverings of E by open elementary sets" is actually restricting possible $A_n$s.

Is my understanding correct?

Or could you explain the expression more detail?

Thanks!

Answer 1

Here is a simpler example, the formula for evaluating a finite geometric sum: $$\sum_{n=0}^N k^n = \frac{k^{n+1}-1}{k-1} $$ In English one can read this as

The sum, taken over all $n$ from $1$ to $N$, of $k^n$, equals $\frac{k^{N+1}-1}{k-1}$.

Or here is a slightly different way to read the same equation, in somewhat more colloquial English:

The ratio $\frac{k^{N+1}-1}{k-1}$ is equal the sum of all powers $k^n$ taken over all integers $k$ from $0$ to $N$.

In this example, the phrase "the sum, taken over ..." is telling you there is a set of numbers that is being summed. What then follows is a description of exactly which set is being summed, namely the all of the numbers $k^n$ as $n$ varies between $0$ and $N$.

In more general usage, the phrase "the OPERATION, taken over..." is telling you that there is some mathematical operation which takes as its input certain kinds of sets, and the language "taken over" is then telling you exactly which set is acting as the input for the operation.

There are a lot of variations to this.

Second example, some definite integrals: $$\int_{x=a}^{x=b} \, x^n \, dx = \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1} $$

The integral of $x^n$ taken over the interval from $x=a$ to $x=b$ equals...

In this example the operation is a definite integral of $x^n$, and this operation takes as its input any interval of real numbers $a \le x \le b$.

Third example, the supremum operator:

$$\sup \{x > 0 \mid x^2 < 2\} = \sqrt{2} $$

The supremum of $x$ taken over the set of all all positive $x$ such that $x^2 < 2$ is $\sqrt{2}$.

In this example the operation is the supremum, which takes as its input a certain subset of the real numbers.

Finally, in your example, the operation (which is rather more complicated than the previous examples) is an infimum-of-sums, that is, the infimum of sums of the form $\sum_{n=1}^{\infty} \mu(A_n)$. The input to this operation is a certain set of infinite sequences $\{A_n\}_{n=1}^\infty$, and the wording is telling you exactly which set of infinite sequences is used as the input for this operation, namely: the set of all sequences of the form $\{A_n\}_{n=1}^{\infty}$ such that each $A_n$ is an open elementary subset of $\mathbb R$ and such that $E \subseteq \bigcup_{n=1}^\infty A_n$.

You have brought up the example $A_1=[0,2/3]$ and $A_2=[2/3,1]$. One can certainly think of this examples as an infinite sequence by setting $A_3=A_4=...=\emptyset$. And this example certainly does cover the set $E=[0,1]$. But what rules this example out is that $A_1,A_2$ are certainly not open elementary subsets (although you have not said what that means, from context I am quite sure that those two sets are not "open elementary").

One last think to keep in mind: there is no assertion that one particular one of these sums is the infimum of all such sums, just as there is no assertion that one particular one of the set $\{x > 0 \mid x^2 < 2\}$ is the supremum of that set. You really do have to get into your mind the entire SET of sequences of the form $\{A_n\}_{n=1}^\infty$ such that [now repeating myself] each $A_n$ is an open elementary subset and $E \subseteq \bigcup_{n=1}^\infty A_n$, and the infimum of the entire SET of sums $\sum_{n=1^\infty}\mu(A_n)$.

Answer 2

It means:

$$\mu^\ast(E)\\=\inf\left\{\sum_{n=1}^\infty\mu(A_n):(A_n)_{n\in\Bbb N}\text{ is an elementary open cover of $E$}\right\}$$

The "taking" over means this: we are looking at $\inf\sum_n\mu(A_n)$, but what the hell is $A_n$? We have an inf of some quantities but we don't know where the quantities come from. So, "taken over the elementary open covers" (or words to that effect) means this (otherwise undefined!) symbol "$A_n$" refers to an element of a countable cover of $E$ by elementary open sets.

As you can see, the "fully correct" $\inf$ notation is long winded and takes up space; it's arguably nicer to describe the inf by adding some natural language and keeping the notation short.