As I am studying real analysis a certain pattern keeps reoccurring - an example of this is from Measure Theory (2nd edition) by Cohn, Donald L., namely the proof of Lemma 1.5.3. The proof uses the following result:
Proposition 1.4.1: Let $ \lambda $ be the Lebesgue measure on $ \mathbb R^d $ and let $ A \subset \mathbb R^d $ be Lebesgue measurable. Then $ \lambda(A) = \sup \{ \lambda(K) \mid K \text{ is compact and } K \subseteq A \} $.
Then in the proof of Lemma 1.5.3 we do the following:
For each positive integer $ n $, use proposition 1.4.1 to choose a compact set $ K_n $ such that $ K_n \subseteq A $ and $ \lambda(A) - 1/n < \lambda(K_n) $.
I do not see why this is necessarily true. Intuitively, I feel like if $ n $ is large enough, so that $ 1/n $ is very small, then no such $ K_n $ can exist?
I am quite certain that I have seen this pattern elsewhere, i.e., that something is defined as a supremum/infimum and we can arbitrarily close to it. Does this hold in general?
Edit: To clarify, consider the following situation. We have some map $ f: 2^X \rightarrow \mathbb R $ and we define $ \tilde f: 2^X \rightarrow \mathbb R $ by $ \tilde f(A) := \sup \{ f(B) \mid B \subset A \} $. Is it true that: Given some $ \varepsilon > 0 $ there exists some $ B \subset A $ such that $ \tilde f(A) - \varepsilon < f(B) $?
This is the "approximation property" of a supremum (or infimum). It is very fundamental in analysis and can be considered an alternative definition of supremum.
If you have a nonempty bounded set $S\subset\Bbb R$, $\alpha:=\sup S$ exists in $\Bbb R$. What we would like to show is that for all $\epsilon>0$ - e.g., for all $1/n$ where $n$ is a large positive integer! - there will exist some $s\in S$, $\alpha-\epsilon<s\le\alpha$.
Well, suppose this is false. Then there is some $\epsilon>0$ for which there are no such $s$. Since for all $s\in S$ we know $s\le\alpha$ (supremum is an upper bound) the claim that $\alpha-\epsilon<s$ is false means that in fact we have $s\le\alpha-\epsilon$. This is true for any $s$, so $\alpha-\epsilon$ is also an upper bound of $S$.
But this is a contradiction since $\alpha-\epsilon<\alpha$ strictly - the supremum is the least upper bound.
Because of this contradiction, our claim is correct.
Example: Let $S$ be the set $\{\lambda(K):K\subseteq A\text{ is compact }\}$. Here, $\alpha=\lambda(A)$. For any $n\in\Bbb N$, consider $\epsilon:=1/n>0$. There must exist, by our theorem, some $s\in S$ with $\lambda(A)-1/n<s$. That is, there must exist some $K\subseteq A$ compact with $\lambda(A)-1/n<\lambda(K)$.
Yes, the statement in your edit, about $f:2^X\to\Bbb R$, is correct - if you assume all the suprema exist. By the way, if you consider suprema to always exist in the extended real valued sense then the theorem is basically still correct. The slightly more general claim is that for all $x<\alpha$, there is some $s\in(x,\alpha]$. In particular if $S$ is unbounded from above and $\sup S=+\infty$, the theorem says for any real number $x$ at all there exists $s>x$, $s\in S$. This is of course correct since $S$ is unbounded.
Very important exercise: state and prove a corresponding result about infima.