We have to understand what each thing means geometrically. Then, we can move on to what it means analytically.
For example, we know geometrically what the length of an interval is : it should be the difference of the end points. That is, the length of $(a,b]$ (or of $[a,b]$ or of $[a,b)$ or any other combination) should be $b-a$, and if say $a = \infty$ or $b = -\infty$ then the length should be infinite. Thus , we know the length of a set which is an interval. If some set is not an interval, but rather a finite disjoint union of intervals(plurinterval), like $[1,2] \cup (3,6]$, then the length of such a set should be the sum of the lengths of the intervals. In the above case, it is just $(2-1)+(6-3) = 4$.
This is what $\lambda_0$ captures : it tells you what the length of $(a,b]$ is, and what the length of $(a,+\infty)$ is for each $a,b$. From here, we can find the length of any plurinterval by simply summing the lengths of the intervals which that plurinterval is made of.
However, we have a small issue with the definition of $\mathcal I$ that is given above : indeed, $\mathcal I$ does not contain $[0,1]$, for example, or any interval of the form $[a,b]$. Why?
In fact, if $[a,b] = \cup(a_i,b_i]$, then $a \in (a_j,b_j]$ for some $j$ in the union, so $a > a_j$. But then $(a_j,b_j] \subset [a,b]$, however the left endpoint $a_j$ is not contained in $[a,b]$ because $a_j < a$. Therefore, $[a,b]$ does not belong to $\mathcal I$.
You can similarly check that intervals of the form $[a,b)$ don't belong to $\mathcal I$.
Now, the problem with $T_1$ is that it does not preserve membership in $\mathcal I$, so $\lambda_0$ is not invariant under $T_1$ because it is very possible that $I \in \mathcal I$ but $T_1(I) \notin \mathcal I$, so that $\lambda_0(T_1(I))$ does not even make sense.
For example, consider $a = -1$, in which case $T_1x = -x$ for all $x$. Then, for example, we have $(1,2]\in \mathcal I$ but $T_1((-1,2]) = [-2,1) \notin \mathcal I$.
Therefore the author has cleverly ensured that he doe not use anywhere the invariance of $T_1$ on $\mathcal I$. It is, however, invariant on $\mathcal L$, and we shall see why.
Moving to $T_2$, this is a "stretch" map : given a set $S$, the set $T_2 S$ is the set obtained by "strecthing" $S$ by a factor $a$. Now, imagine chewing gum : it has some length at the start, and then when you stretch it or compress it, its length changes, right? So, $T_2$ is not going to preserve lengths! $\lambda_0$ is by no means going to be invariant under $T_2$. For a counterexample, take $(0,1]$ and $a=2$, then $T(0,1] = (0,2]$. The first has $\lambda_0$ - length $1$, the second has $\lambda_0$ - length $2$.
However, $T_2$ does ensure that the length is multiplied by a constant factor, namely $|a|$. For this, you need to see the proof of how $\lambda^*$ changes under $T_2$, which you say you have no doubts in above.
Now, $T_3$ is a translation : it is moving a set around without affecting it in any other way. For example, it is like taking your notebook out of your bag : you move the notebook's location, but are not changing its volume, right?
Now, we need to know what $T_3$ does on an interval. For example, what does it do to $(x,y]$? Indeed, $T_3z = z + b$ for all $z$, so $x \to x+z$, $y \to y+z$ under $T_3$, and so on. You can see from here that $T_3(x,y] = (x+z,y+z]$. Similarly, $T_3((x,+\infty)) = (x+b,\infty)$. On these intervals, you can check from the definition of $\lambda_0$ that $\lambda_0((x+z,y+z]) = \lambda_0((x,y])$, and similarly that $\lambda_0((x,+\infty)) = \lambda_0((x+z,\infty))$.
Now, for a plurinterval in $\mathcal I$, we write it as the union of disjoint $(a_i,b_i]$, then verify two things :
We need to check that $T_3(\cup (a_i,b_i]) \in \mathcal I$, but it is easy to see that $(a_i,b_i] \to (a_i+x,b_i+x]$, so $T_3(\cup(a_i,b_i]) = \cup(a_i+b,b_i+b]$ remains disjoint. Something similar if one of the endpoints is $+\infty$. So, if a set is in $\mathcal I$, then $T_3$ of that set is also in $\mathcal I$(1).
From above, it is not difficult to see that $\lambda_0$ is invariant under $T_3$, since the length of $\cup (a_i+b,b_i+b]$ is the sum of the lengths of $(a_i+b,b_i+b]$ because of disjointness, then the length of $(a_i+b,b_i+b]$ is obviously the same as that of $(a_i,b_i]$, even if $b_i = +\infty$.
Thus, $\lambda_0$ is preserved under $T_3$(2). This is question $2$ : note that $T_3^{-1}$ is the inverse translation i.e. if $T_3 x = x+b$, then $T_3^{-1} x = x-b$, so it is also a translation, but by something different. So it is also $T_3$ but for some different $b$. The same proves that $\lambda(I_k') = \lambda(T_3^{-1}(I_k'))$ for each $k$, so the inequality in green in obvious. This answers questions two and three.
For question 1 , we just need to reverse the argument. This is because for any $x$, we have :
$$
T_1(T_1x) = T_1(\frac{a}{|a|}x) = \frac{a^2}{|a|^2} x = x
$$
So, for any set $F$, we have $T_1(T_1 F ) = F$. Keep this in mind.
Further, we have $\lambda^*(T_1(F)) \leq \lambda^*(F)$ for any set $F$.
Now, apply the above inequality with $F = T_1(E)$, then we get:
$$
\lambda^*(T_1 E) = \lambda^*(F) \geq \lambda^*(T_1 F) = \lambda^*(T_1(T_1(E))) = \lambda^* (E)
$$
which is the other direction, and answers why $\lambda^*(T_1(E)) \geq \lambda^* E$, if you look at the extreme left and right hand side of what I have written in the above sequence of statements.
For question four, the point is if $Tx = ax+b$ is a linear transformation, then so is $T^{-1}$! In fact, it is easy to check that $T^{-1} y = \frac{1}{a}y - \frac ba$. So, the same theorem applies in this case, giving $\lambda^*(T_1^{-1}(E)) = \frac 1{|a|} \lambda^*(E)$.
Since this is a bountied question, you may ask me to elaborate until satisfaction.
EDIT :
Since I think an example clarifies things best, I will take an example. That is, I will take separate values of $a,b$ and show that for these $a,b$ the above argument works out. This will help you understand how the argument works for general $a,b$.
Let us go with the example $a=2,b=3$.
Then, we have to find what $T_1,T_2$ and $T_3$ are.
What is $T_1$? It is $T_1 x = \frac{2}{|2|}x$, but $2 = |2|$ so $T_1x = x$. That is, $T_1$ is the identity map.
It should be obvious that $T_1$ preserves the length of any set, simple because it doesn't even change the set. For example, $T_1([2,3]) = [2,3]$, so there is no change of length.
Now, what is $T_2$? $T_2 x = |a|x = 2x$. This amounts to stretching a set by a factor of $2$. For example, $T_2([5,8]) = [10,16]$. Note that $[5,8]$ has length $3$ while $[10,16]$ has length $6$, so we see that the lengths has become $2$ times more after applying $T_2$. This will happen for all $I \in \mathcal I$ and therefore for $E$ as well. I need to know if you have further doubts here.
What is $T_3$? $T_3 x = x + b = x+3$. Geometrically, you imagine a set $E$ on the real line, and now just shift the whole set three units to the right. The set which you get now is $T_3(E)$. Shifting the set does not change the length : I gave the example of taking a notebook out of your bag. Here's another one : a plane which is $30$ meters long starts from London and flies to Paris. What is its length after it reached Paris? Still the same $30$ meters, right?
You have to think of $T_3$ like this : it just shifts the location of the set without really changing it's content. Therefore, the length does not change.
In the case of intervals, take $I = [4,8]$, then $T_3(I) = [7,11]$. Did the length change? Take $I = (34,+\infty)$, then $T_3(I) = (37,+\infty)$. Did the length change?
Now, for the grand finale on this example, we take $E = [1,2] \cup [4,6] \cup \{3\}$, so weird set. We are now going to compute what $T(E)$ is, what $\lambda^*(E)$ is, what $\lambda^*(T(E))$ is, and then confirm that it matches the formula.
What is $T(E)$? By definition, it is the set $\{T x : x \in E\}$. But, then I mentioned that $Tx = 2x+3$. So, $T(E) = \{2x+3 : x \in E\}$.
We break $T$ into $T_3T_2T_1$.
By definition, $T_1(E) = \{T_1x : x \in E\} = \{x : x \in E\}= E$.
Now, $T_2T_1(E) = T_2(E) = \{T_2 x : x \in E\} = \{2x : x \in E\} = [2,4] \cup [8,12]\cup \{6\}$, by just taking on each interval and multiplying by $3$.
Now, $TE = T_3T_2T_1E = T_3\{[2,4] \cup [8,12] \cup \{6\}\} = \{x + 3 : x \in [2,4] \cup [8,12] \cup {6}\} = [5,7] \cup [11,15] \cup \{9\}$.
By definition of $\lambda^*$, we note that $\lambda^* E = (2-1) + (6-4) + (3-3) = 3$. On the other hand, $\lambda^ TE = (7-5) \cup (15-11) \cup (9-9) = 6$, and $6 = 2 \times 3 = |a| \times \lambda^*E$.
Hence, we have verified the formula for this $T$ and this $E$.
Try the same with other $T$ and $E$.
EDIT 2 :
I go statement - by - statement on your queries.
If $I \in \mathcal I$ then $T_3(I) \in \mathcal I$.
The explanation for this statement is given. I have highlighted it in my explanation as (1). To be brief, I first looked at what happened if $I$ was just of the form $(x,y]$ or $(x,+\infty)$. Then, I showed that $T_3(I)$ as just $(x+b,y+b]$ or $(x+b,+\infty)$ respectively, and therefore is in $\mathcal I$. Now, if $I$ is a disjoint union of $(a_i,b_i]$, then I showed that $T_3(I)$ is a disjoint union of $(a_i+b,b_i-b]$ and hence remains in $\mathcal I$. I asked you to adapt this argument if instead of $(a_i,b_i]$ we had $(a_i,+\infty)$.
Moreover $\lambda_0$ is invariant under $T_3$.
Yes, this I have highlighted in (2). Just have to just observe the bullet points carefully. I have taken examples as well, the best being the long one above in EDIT 1.
How can I show that $\lambda^*(T_3(E)) \geq \lambda^*(E)$?
Just realized I confused $T_1$ with $T_3$ in my explanation above.
For this, just disassociate yourself from the whole theorem, and focus on just what this result says. Note that $T_3 x= x+b$ for some $b$ whose choice does not matter. Now, we showed that $\lambda^*(T_3 E) \geq \lambda(E)$. Write this in words : if you move any set $b$ units to the right, then its length is greater than or equal to what it was earlier. Nothing special about $b$ here : where did we use anything about $b$ in the proof? Nowhere.
Now, we just need to reverse : instead of moving $b$ units to the right, we move $\mathbf{-b}$ units to the right. Now, the same argument(that is, write down the argument that $\lambda^*(T_3 E) \geq \lambda^*E$, and just replace every $b$ by a $-b$) will show you that if $T'_3x = x-b$, then $\lambda^*(T'_3F) \geq \lambda^*(F)$ for any set $F$. In words : shifting any set $b$ units to the left , its length is greater than or equal to what it was earlier.
Here's the key point : if we move any set first $b$ units on the right, and then $b$ units to the left, we get back the same set! So, the length after shifting $b$ units to the right cannot be strictly greater than the length of the set, because then the length of the shifted set after shifting $b$ units to the left, must be strictly greater than the length of the shifted set, but then the latter is the length of the original set itself, so we have a contradiction, hence $T_3$ preserves the length exactly.
To take an example, take $E = [4,7]$ and $b = 2$. Then, applying the argument with $b= 2$ tells us that $\lambda^*([6,9]) \geq \lambda^*([4,7])$.
Now, applying the argument with $F=[6,9]$ and $b= -2$ tells us that $\lambda^*([4,7]) \geq \lambda^*([6,9])$.
Now, the two quantities must be equal.
The rest of your questions, and the proof for why $T_2$ and $T_1$ work out.
Let us adapt the argument that we gave for $T_3$, to $T_2$ and $T_1$. The point, however, is that won't be restricting ourselves to $\mathcal I$ anymore.
Let us adapt the argument we gave for $T_3$, to $T_2$ and $T_1$.
See, for $T_3$ we use the fact that $T_3(\cup_{j=1}^k I_j) = \cup_{j=1}^k T_3(I_j)$, where each $T_3(I_j) \in \mathcal I$ and hence the expression $\lambda_0(T_3(I_j))$ made sense.
Now, the point about $T_2$ , is that it also preserves $\mathcal I$.
Why? Let us note that $T_2((x,y]) = (|a|x,|a|y]$, and $T_2((z,+\infty)) = (|a|z,+\infty)$, which follow from the definition $T_2(x) = |a|x$.
Next, if we take a disjoint union of intervals $I = \cup (a_i,b_i]$, then $T_2 I = \cup (|a|a_i,|a|b_i]$ which is disjoint. Similarly if one of the $b_i$ was $+\infty$. Essentially, it follows that $T_2 I \subset \mathcal I$, so $T_2$ preserves $\mathcal I$.
Moreover, observe what happens to the lengths. $$\lambda_0(T_2(I) ) = \lambda_0(T_2(\cup_i (a_i,b_i]) ) = \lambda_0(\cup_i (|a|b_i,|a|a_i]) = \sum_i (|a|b_i - |a|a_i) = |a|\sum (b_i-a_i) = |a| \lambda_0(\cup_i (a_i,b_i]) =|a| \lambda_0(I)$$
Therefore, on $\mathcal I$ we have that $\lambda_0$ multiplies the length by $|a|$.
For $T_2$ now, we just need to “copy” the $T_3$ argument , making changes wherever necessary.
Start with arbitrary $\lambda^*E < \infty$ and an $\epsilon > 0$. Then, we know that there is a countable collection of intervals $(a_i,b_i]\subset I$ with $\sum (b_i-a_i) < \lambda^*(E) + \frac{\epsilon}{|a|}$(using the definition with a “modified” $\epsilon$, you will see why later) and $E \subset \cup I_k$. (Note that $b_i$ cannot be infinite if the above occurs for any $i$).
Now, simply note that :
$$
T_2(E) \subset T_2(\cup (a_i,b_i]) = \cup (|a|a_i,|b|a_i)
$$
And therefore ,
$$
\lambda^*(T_2(E) ) \leq \sum_{i} (|a|b_i-|a|a_i) = |a| \sum_i (b_i-a_i) \leq |a|(\lambda^*(E) + \frac{\epsilon}{a}) = |a|\lambda^*(E) + \epsilon
$$
From where it follows that $\lambda^*(T_2(E)) \leq |a| \lambda^*(E)$.
Now, what about the other direction? Let us first write what we derived in words : given a set, if you stretch it by any given $|a|$, then the new length is at most $|a|$ times the old length.
But then, you can stretch back by a factor $\frac 1{|a|}$ to get the original set itself. Now, if there was strict inequality in the length after stretching, it will be there in the stretch back as well, but we have come back to the original set, so we cannot have had strict inequality.
In short, use the reversal argument I used with $T_3$ here, with a changed $\alpha$. This will tell you that $\lambda^*(T_2(E) ) \geq |a| \lambda(E)$, which concludes the argument for $T_2$..
Now, you need to argue for the case when $\lambda^*(E) = \infty$, but I leave you to copy the argument just like I did.
For $T_1$ we have two possibilities : either $T_1x = x$ for all $x$, or $T_1x = -x$ for all $x$. The first possibility leads to $T_1 (E) = E$ for all $E$, so it is almost obvious that $\lambda^*$ is invariant under $T_1$.
In the other case, I had mentioned that $T_1$ does not preserve $I$ as mentioned earlier, so we will need a slight tweak. The point is, that thankfully for us, even though $T_1(I)$ need not be an element of $\mathcal I$ for a given $I$, we can still find an element $I’ \in \mathcal I$ so that $T_1(I) \subset I’$, but the sets are “almost” equal.
Let us write $-E$ for $T_1 E$, since $T_1x = -x$ for all $x$ now, so it is just like the taking the “negative” of a set.
And this , is how : Fix $\lambda^*(E) < \infty$ and an $\epsilon > 0$. There exist $(a_i,b_i]$ with $\sum (b_i - a_i) < \lambda^*(E) + \frac{\epsilon}{2} $ and $E \subset \cup (a_i,b_i]$.
Now, note the following carefully:
$$
E \subset \cup (a_i,b_i] \implies –E \subset \cup[-b_i,-a_i) \subset \cup ((-b_i - \frac{\epsilon}{2^i} , a_i])
$$
So because we cannot work with the intervals $[-b_i,a_i)$, we force them into slightly larger intervals we can work with, namely $(-b_i - \frac{\epsilon}{2^{i+1}} , a_i]$ which belong in $\mathcal I$.
Here is a fact : $\sum_{i=1}^\infty \frac{\epsilon}{2^{i+1}} = \frac{\epsilon}{2}$.
From here, a similar thing may be done :
$$
\lambda^*(-E) \leq \sum_i (-a_i + (b_i + \frac{\epsilon}{2^{i+1}})) = \sum_{i} (b_i-a_i) + \sum_i \frac{\epsilon}{2^{i+1}} < \lambda^*(E) + \frac{\epsilon}{2} + \frac{\epsilon}{2} = \lambda^*(E) + \epsilon
$$
Now, since $E = - (-E)$, you can switch the roles of $E$ and $-E$ above (which is like the reverse argument, but much easier to see) so the equality follows.
Now, I leave you to see what happens if $\lambda^*(E)$ is infinite.
Best Answer
Question 0.
Monotone: Suppose $F_1 \subseteq F_2$ for $F_1,F_2 \in \mathcal{I}_0$. We wish to show $\lambda_0(F_1) \le \lambda_0(F_2)$. This is trivial if $F_2 = (a,+\infty)$ for some $a \in \mathbb{R}$, so suppose otherwise: $F_2 = (a,b]$ for some $-\infty \le a \le b <+\infty$. Since $F_1 \subseteq F_2$, $F_1 = (c,d]$ for some $c \ge a$ and $d \le b$. But then $\lambda_0(F_1) = d-c \le b-a = \lambda_0(F_2)$.
Additive: Suppose $F_1,F_2 \in \mathcal{I}_0$ with $F_1\cup F_2 \in \mathcal{I}_0$ and $F_1 \cap F_2 = \emptyset$. We wish to show $\lambda_0(F_1\cup F_2) = \lambda_0(F_1)+\lambda_0(F_2)$. Since we have shown $\lambda_0$ is monotone, the equality is trivial if $F_1 = (a,+\infty)$ for some $a \in \mathbb{R}$ or $F_2 = (b,+\infty)$ for some $b \in \mathbb{R}$. So suppose $F_1 = (a,b]$ and $F_2 = (c,d]$ for some $-\infty \le a,b,c,d < + \infty$, $a \le b$, $c \le d$. Without loss of generality, say $a \le c$. Since $F_1 \cup F_2 \in \mathcal{I}_0$, it must be that $c \le b$. And since $F_1\cap F_2 = \emptyset$, it must be that $c=b$. Then $\lambda_0((a,b]\cup(c,d]) = \lambda_0((a,d]) = d-a = (d-c)+(b-a) = \lambda_0(F_2)+\lambda_0(F_1)$, as desired.
.
Question 1. This is a good question. The answer to it is very much dependent on the $\mathcal{I}_0$ we are working with presently. The answer is that we may (re)order the $F_l$'s so that $\cup_{l=1}^n F_l \in \mathcal{I}_0$ for each $n$. More precisely, there are $(\tilde{F_l})_{l=1}^\infty$ such that $\{\tilde{F_l}\} = \{F_l\}$ and $\cup_{l=1}^n \tilde{F_l} \in \mathcal{I}_0$ for each $n \ge 1$ and Below is a proof.
Each $F_l$ is of the form $(a_l,b_l]$ or $(a_l,+\infty)$ and the union $\cup_{l=1}^\infty F_l$ is of the form $(a,b]$ or $(a,+\infty)$. Let's first suppose that $\cup_{l=1}^\infty F_l = (a,b]$. This means each $F_l = (a_l,b_l]$ for some $-\infty \le a_l \le b_l < +\infty$. Since $b \in \cup_{l=1}^\infty F_{l}$, $b \in F_{l_1}$ for some $l_1$. Let $\tilde{F_1} = F_{l_1}$. Note $\tilde{F_1} \in \mathcal{I}_0$. Now $(\cup_{l=1}^\infty F_l)\setminus \tilde{F_1} = (a,a_{l_1}]$, so there is some $l_2$ with $a_{l_1} \in F_{l_2}$. We must have $b_{l_2} = a_{l_1}$, so if we let $\tilde{F_2} = F_{l_2}$, we then have $(\cup_{l=1}^\infty F_l)\setminus(\tilde{F_1}\cup \tilde{F_2}) = (a,a_{l_2}]$. Note that $\cup_{l=1}^2 \tilde{F_l} = (a_{l_2},b_{l_1}] \in \mathcal{I}_0$. We can keep constructing $\tilde{F_l}$ in this manner. It is clear that we will have $\{\tilde{F_l}\} = \{F_l\}$ and $\cup_{l=1}^n \tilde{F_l} \in \mathcal{I}_0$ for each $n \ge 1$. I leave the case of $\cup_{l=1}^\infty F_l = (a,+\infty)$ to you.
.
Final question.
The case of $F = (a,b]$ is correct (and well-written). The case of $F = (a,+\infty)$ is not correct. The $H$ you defined does not have $cl(H_n) \subseteq F$ since we don't even have $H_n \subseteq F$ since $a-\frac{n}{2} \not \in F$ (for example). You should instead look at, for example, $H_n = (a+\frac{1}{n},n]$.