About the proof that lebesgue measure is a premeasure. let $\lambda^n$ be $n$-dimensional Lebesgue measure. I am having trouble understanding the proof that this "mapping" is a premeasure on the set of half open rectangles $\lambda^n ([a,b))= \prod_{i=1}^n(b_i-a_i)$. The proof uses induction over the dimension $n$. I understood how it was shown that $\lambda$ is a premeasure. So now assume $\lambda^n$ is a premeasure the goal is to show $\lambda^{n+1}$ is too, which can be simply reduced to showing that $\lambda^{n+1}$ is $\sigma$-additive. 
Let $I_j=I_1^j \times I_j^n \in \mathcal J^1 \times \mathcal J^n=\mathcal J^{n+1}$ be mutually disjoint and $\bigcup_{j \in \Bbb N}I_j=I\in \mathcal J^{n+1}$(disjoint union). Ok so now to the first thing I can't understand : Since $I \in \mathcal J^{n+1}$ we know that $$\bigcup_{j\in\Bbb{N}}I_j^d \in \mathcal J^d, \ d=1,n$$
What does he mean by this, and what does $d=1,n$ mean?
Define $\hat{I}_1^d=I_1^d$ and $\hat{I}_{j+1}^d=I_{j+1}^d \setminus (I_1\cup \ldots\cup I_j^d)$. The $\hat{I}_j^d$ are disjoint and
$$\bigcup_{j=1}^NI_j^d=\bigcup_{j=1}^N\hat{I}_j^d$$ 
for all $N \in \Bbb{N}$ and $d=1,n$ (I still don't get what this notation is, the second union is disjoint). 
Now since $\mathcal J^d$ is a semi-ring each $\hat{I}_j^d$ is a finite union of disjoint union of disjoint rectangles. Hence there are disjoint sets $\widetilde{I}_k^1\in \mathcal J^1$ and $\widetilde{I}_l^n\in \mathcal J^n$ such that(all union below are disjoint) 
$$I=\bigcup_{j \in \Bbb{N}}(I_j^1 \times I_j^n)=\text{why?}\bigcup_{k \in \Bbb{N}}\bigcup_{l \in \Bbb{N}}(\widetilde{I}_k^1 \times \widetilde{I}_l^1)=\bigcup_{k \in \Bbb{N}}\widetilde{I}_k^1 \times \bigcup_{l \in \Bbb{N}}\widetilde {I}_l^n \in \mathcal J^1 \times \mathcal J^d $$
Is $ \mathcal J^d$ a typo?
Also why is $\lambda^{n+1}(\bigcup_{k \in \Bbb{N}}\widetilde{I}_k^1 \times \bigcup_{l \in \Bbb{N}}\widetilde {I}_l^n )=\lambda^1(\bigcup_{k \in \Bbb{N}}\widetilde{I}_k^1) \cdot \lambda^n (\bigcup_{l \in \Bbb{N}}\widetilde {I}_l^n) $? (this is just stated without any proof and doesn't really seem direct)
And finally why is $$\sum_{j\in \Bbb{N}}\mathop{\sum\sum}_{(k,l):(\widetilde{I}_k^1 \times \widetilde{I}_l^n)\subset (I_j^1 \times I_j^n)}\lambda^{n+1}(\widetilde{I}_k^1 \times \widetilde{I}_l^n)=\sum_{k \in \Bbb{N}}\sum_{l \in \Bbb{N}} \lambda^{n+1}(\widetilde{I}_k^1 \times \widetilde{I}_l^n)$$
Such hardness much handwavery.
 A: In the following we use
$\bigsqcup_{j \in \mathbb{N}}I_j$
to denote the disjoint union of $I_j,j\in\mathbb{N}$. We go through the stated parts of the proof and consider OPs questions.

Let 
  $I_j=\color{blue}{I_j^1} \times I_j^n \in \mathcal J^1 \times \mathcal J^n=\mathcal J^{n+1}$
  be mutually disjoint and $\bigsqcup_{j \in \mathbb N}I_j=I\in \mathcal J^{n+1}$.

We split the $(n+1)$-dimensional interval $I_j$ into a one-dimensional and an $n$-dimensional part in order to prepare for the inductive step. Note that we have $I_j=\color{blue}{I_j^1} \times I_j^n$ and not $I_1^j$.

Since $I \in \mathcal J^{n+1}$ we know that $$\bigcup_{j\in\mathbb{N}}I_j^d \in \mathcal J^d, \ d=1,n$$

The usage of $d=1,n$ is just a shorthand for
\begin{align*}
\bigcup_{j\in\mathbb{N}}I_j^1 \in \mathcal J^1\qquad\text{and}\qquad\bigcup_{j\in\mathbb{N}}I_j^n \in \mathcal J^n\tag{1}
\end{align*}
which follows from the representation $I_j=I_j^1 \times I_j^n \in \mathcal J^1 \times \mathcal J^n=\mathcal J^{n+1},j\in \mathbb{N}$.
The representation in (1) is not a disjoint union, but we need mutually disjoint intervals to ease further calculations. So, we introduce:

Define $\hat{I}_1^d=I_1^d$ and $\hat{I}_{j+1}^d=I_{j+1}^d \setminus (I_1\cup \ldots\cup I_j^d)$. The $\hat{I}_j^d$ are disjoint and
  \begin{align*}
\bigcup_{j=1}^NI_j^d=\bigsqcup_{j=1}^N\hat{I}_j^d\tag{2}
\end{align*}
  for all $N \in \mathbb{N}$ and $d=1,n$. 

... which again is a short-hand for
\begin{align*}
\bigcup_{j=1}^NI_j^1=\bigsqcup_{j=1}^N\hat{I}_j^1\qquad\text{and}\qquad \bigcup_{j=1}^NI_j^n=\bigsqcup_{j=1}^N\hat{I}_j^n
\end{align*}
From now on we recall that $d=1,n$ is a short-hand notation. This way the author can write a single statement regarding $d$ and we consider them to be two statements, one with $d=1$ and one with $d=n$. For instance the statement $\mathcal J^d$ is a semi-ring is to read as $\mathcal{J}^1$ and $\mathcal{J}^n$ are semi-rings.

Now since $\mathcal J^d$ is a semi-ring each $\hat{I}_j^d$ is a finite union of disjoint union of disjoint rectangles. Hence there are disjoint sets $\widetilde{I}_k^1\in \mathcal J^1$ and $\widetilde{I}_l^n\in \mathcal J^n$ such that
$$I=\bigsqcup_{j \in \mathbb{N}}(I_j^1 \times I_j^n)=\text{why?}\bigsqcup_{k \in \mathbb{N}}\bigsqcup_{l \in \mathbb{N}}(\widetilde{I}_k^1 \times \color{blue}{\widetilde{I}_l^n})=\bigsqcup_{k \in \mathbb{N}}\widetilde{I}_k^1 \times \bigsqcup_{l \in \mathbb{N}}\widetilde {I}_l^n \in \mathcal J^1 \times \color{blue}{\mathcal J^d} $$

Note that we have $\bigsqcup_{k \in \mathbb{N}}\bigsqcup_{l \in \mathbb{N}}(\widetilde{I}_k^1 \times \color{blue}{\widetilde{I}_l^n})$ and not $\bigsqcup_{k \in \mathbb{N}}\bigsqcup_{l \in \mathbb{N}}(\widetilde{I}_k^1 \times \color{blue}{\widetilde{I}_l^d})$.
Is $\color{blue}{\mathcal J^d}$ a typo? Yes, it is a typo in the book and should be read as $\color{blue}{\mathcal J^n}$.
Ad why?: See the start of chapter 6 and recall that property (S3) of the definition of a semi-ring $\mathcal{S}$ tells us that the difference $S\setminus T$ of each two elements $S,T\in\mathcal{S}$ can be represented as finite disjoint union of elements of $\mathcal{S}$. 

The complete statement of the last paragraph reads as:
Since $\mathcal J^d$ is a semi-ring, the property (S3) shows that each $\hat{I}_j^d$ is a finite union of disjoint union of disjoint rectangles. Hence there exist disjoint sets $\widetilde{I}_k^1\in \mathcal J^1$ and $\widetilde{I}_l^n\in \mathcal J^n$ such that
$$I=\bigsqcup_{j \in \mathbb{N}}(I_j^1 \times I_j^n)=\bigsqcup_{k \in \mathbb{N}}\bigsqcup_{l \in \mathbb{N}}(\widetilde{I}_k^1 \times \widetilde{I}_l^n)=\bigsqcup_{k \in \mathbb{N}}\widetilde{I}_k^1 \times \bigsqcup_{l \in \mathbb{N}}\widetilde {I}_l^n \in \mathcal J^1 \times \mathcal J^n $$

The equality chain holds due to the paragraph above addressing the property (S3) and the relationship (2).

Sigma-additivity of $\lambda^{n+1}$:
The following equality chain is valid essentially due to the sigma-additivity of $\lambda^{n+1}$:
  \begin{align*}
\lambda^{n+1}\left(\bigsqcup_{j \in \mathbb{N}}I_j\right)
&=\lambda^{n+1}\left(\bigsqcup_{k \in \mathbb{N}}I_j^1\times\bigsqcup_{k \in \mathbb{N}}I_l^n\right)\\
&=\lambda^{1}\left(\bigsqcup_{k \in \mathbb{N}}I_j^1\right)
\cdot\lambda^{n}\left(\bigsqcup_{k \in \mathbb{N}}I_j^n\right)\tag{3}\\
&=\sum_{k\in\mathbb{N}}\sum_{l\in\mathbb{N}}\lambda^1(\widetilde{I}_k^1)\cdot\lambda^n(\widetilde{I}_l^n)\\
&=\sum_{k\in\mathbb{N}}\sum_{l\in\mathbb{N}}\lambda^{n+1}(\widetilde{I}_k^1\times\widetilde{I}_l^n)
\end{align*}

Note that as stated in the book (3) is valid due to the very definition of $\lambda^{n+1}$ as a product on $\mathcal J^{n+1}$: If $I=I^1\times I^{n+1}\in(\mathcal J^{1}\times\mathcal J^{n})=\mathcal J^{n+1}$ then
\begin{align*}
\lambda^{n+1}(I)=\lambda^{n+1}(I^1\times I^n)=\lambda^{1}(I^1)\cdot\lambda^{n}(I^n)
\end{align*}
The last equality should now be in reach.
