Help with understanding `If $A\subset B$ then $m^*(A) \leq m^*(B)$'

Question

I use the book of Marek and Ekkehard, most things are explained quite thoroughly, I just don't understand some things which all should be obvious. First I will clarify $m*(A) = \inf Z_{A} = \inf \left\{\sum_{n}^{}l(I_n): I_n \text{ are intervals}, A\subseteq\bigcup_n I_n \right\}$.

Now the proof of `If $A\subset B$ then $m^*(A) \leq m^*(B)$' is even in the book, although I just don't understand it fully. I understand that if the intervals $J_n$ cover $B$ then certainly covers A: $A\subset B\subset \bigcup_n J_n$. Which results in $Z_B \subset Z_A$ and that is what I don't understand.

So $Z_B = \left\{\sum_{m}^{}l(J_m): J_n \text{ are intervals}, B\subseteq\bigcup_m J_m \right\}$ and $Z_A$ defined as above how. But why is $Z_B\subset Z_A$ even though $A\subset B$? Since $b\in Z_B$ is also $b\in Z_A$ I guess, but that is weird right since $B$ is a bigger set than $A$ and I would think $\bigcup_n I_n \subseteq \bigcup_m J_m$. Where do my thoughts go wrong or rather could anybody tell me why I don't understand this?

Thank you for your help

Answer 1

As you said yourself in the question, if $A \subset B$, every sequence of intervals $I_1,I_2,\ldots$ that cover the set $B$ also covers the set $A$, since $$A \subset B \subset \bigcup_{i=1}^{\infty}I_i$$ This implies that $Z_B \subset Z_A$, which is not that unexpected — as the set gets smaller, more and more sequences (of intervals) begins to cover it (note that the set $Z_A$ consists of lengths of all the sequences that cover $A$). Moreover, $$\inf Z_A \leq \inf Z_B$$ since amongst additional lengths in $Z_A \setminus Z_B$ some could be smaller then $\inf Z_B$.

Answer 2

You could look into what $Z_B\subset Z_A$ means. The meaning is twofold:

First it's that if $x\in Z_B$ then $x\in Z_A$.

Now suppose that $x\in Z_B$ it will mean that we have intervals $J_n$ such that $B\subseteq \bigcup J_n$, but since $A\subset B$ we have therefore that $B\subseteq \bigcup J_n$. So the union is one of the measured set in the $Z_A$ and it's measure is the same as in the $Z_B$.

Second it's that $Z_A\ne Z_B$

We have to find some $x\in Z_A$ that is not in $Z_B$ and frankly I don't think that's possible. You can for example take $A=\{0\}$ and $B=\{0,1\}$ now for each covering of $A$ i think one could come up with an equally small cover of $B$ (by just halfing the sizes of the interval and then clone them to cover both $0$ and $1$).

It look's like the book is rather sloppy regarding $\subset$ vs $\subseteq$, I realize that there are books that only use $\subset$ to denote a subset that isn't necessarily proper, but then those usually don't use the $\subseteq$ symbol (as it would be redundant).

Answer 3

For each set $X$, you have a set of covers by intervals, I'll call it $C(X)$. Call the set of sums of lengths of these covers $M(X)$. If $A \subseteq B$ then $C(B) \subseteq C(A)$, because any cover of $B$ will also be a cover of $A$. Hence $M(B) \subseteq M(A)$. The infimum operation is monotone: if you have $X,Y \subseteq \mathbb{R}$ and $X \subseteq Y$ then $\inf X \geq \inf Y$. (That's because the infimum is the greatest lower bound). So $\inf M(B) \geq \inf M(A)$, which is your statement.