(Proof Verification) Show that the collection of all elementary cylinders is a semi-ring. Denote $\mathbb{R}^{\mathbb{T}}$ to be the set of all functions $x:\mathbb{T}\longrightarrow\mathbb{R}.$ Let $B_{1},\cdots, B_{n}\in\mathcal{B}(\mathbb{R})$, then we can define the elementary cylinder sets as $$E=\{x\in\mathbb{R}^{\mathbb{T}}:x_{t_{1}}\in B_{1},\cdots, x_{t_{n}}\in B_{n}\},$$ where $t_{1},\cdots, t_{n}\in\mathbb{T}$.
Denote the collection of all such elementary cylinders to be $\mathfrak{E}$. I want to show that 

$\mathfrak{E}$ forms a semi-algebra.

Many discussions have been made, for instance, 
here: The collection of cylinder sets is a semiring 
and here:Show that $\mathfrak{S}=\bigcup_{N=1}^{\infty}\mathfrak{Z}_N\cup\left\{\emptyset\right\}$ is a semi-ring.
However, it seems that given different context, the definition of cylinder sets and elementary cylinder are always different but also similar. And it seems that the closure under finite intersection is easy enough so it is rarely asked. 
I came up a proof with the closure under finite intersection, and mimicked the first link to show the complement is a finite disjoint union of elements in the collection.

Proof of closure under finite intersection:
Denote $\mathfrak{E}$ to be the collection of all elementary cylinder sets.  Firstly we need to note that for $E\in\mathfrak{E}$, we can always re-write that 
\begin{align*}
E&=\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}})\in B_{1}\times\cdots\times B_{n}\}\\
&=\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}},\cdots, x_{t_{m}})\in B_{1}\times\cdots\times B_{n}\times \mathbb{R}^{m-n}\},
\end{align*}
for any $t_{n+1},\cdots, t_{m}$. 
Thus, we can always assume two cylinder sets to have the same set of cutting points $t_{1},\cdots, t_{m}$, by taking the union of the indexing sets of the two cylinder sets and using the union as the new indexing set for both of them.
To show the closure under finite intersection, let $E_{1},E_{2}\in\mathfrak{E}$. Then, we can write $$E_{1}=\{x\in\mathbb{R}^{\mathbb{T}}:x_{t_{1}}\in A_{1},\cdots, x_{t_{m}}\in A_{m}\},$$ $$E_{2}=\{y\in\mathbb{R}^{\mathbb{T}}:y_{t_{1}}\in B_{1},\cdots, y_{t_{m}}\in B_{m}\},$$ for $A_{1},\cdots, A_{m}, B_{1},\cdots, B_{m}\in\mathcal{B}(\mathbb{R})$ so that $$E_{1}\cap E_{2}=\{z\in\mathbb{R}^{\mathbb{T}}:z_{t_{1}}\in A_{1}\cap B_{1},\cdots, z_{t_{m}}\in A_{m}\cap B_{m}\}.$$
Note that since $A_{1},\cdots, B_{m}$ are all Borel,  then the intersections $A_{1}\cap B_{1},\cdots, A_{m}\cap B_{m}\in\mathcal{B}(\mathbb{R})$, as well.
This implies $E_{1}\cap E_{2}$ is also an elementary cylinder, and thus $E_{1}\cap E_{2}\in\mathfrak{E}$. 

Proof of the last property:
To show the other property, it suffices to show that, for all $A_{1},\cdots, A_{m}, B_{1}\cdots, B_{m}\in\mathcal{B}(\mathbb{R})$, we can rewrite $(A_{1}\times\cdots\times A_{m})\setminus (B_{1}\times\cdots\times B_{m})$ into a finite disjoint union of sets, and each set in the union is a $m-$fold product of sets in $\mathcal{B}(\mathbb{R})$. 
This is sufficient due to the following reason: 

For any $E_{1}, E_{2}\in\mathfrak{E}$, we can write $$E_{1}=\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{m}})\in A_{1}\times\cdots\times A_{m}\},$$ $$E_{2}=\{y\in\mathbb{R}^{\mathbb{T}}:(y_{t_{1}},\cdots, y_{t_{m}})\in B_{1}\times\cdots\times B_{m}\}$$ for some $A_{1},\cdots, A_{m}, B_{1},\cdots, B_{m}\in\mathcal{B}(\mathbb{R})$, so that $$E_{1}\setminus E_{2}=\{z\in\mathbb{R}^{\mathbb{T}}:(z_{t_{1}},\cdots, z_{t_{m}})\in (A_{1}\times\cdots\times A_{m})\setminus (B_{1}\times\cdots\times B_{m})\}.$$ If we showed that $$(A_{1}\times\cdots\times A_{m})\setminus (B_{1}\times\cdots\times B_{m})=\bigcup_{n=1}^{\ell}C_{n},$$ where $C_{n}$ is a $m-$fold product of Borel set for all $n$, $C_{n}$ disjoint,  then we would be able to rewrite $$E_{1}\setminus E_{2}=\bigcup_{n=1}^{\ell}\{f_{n}\in\mathbb{R}^{\mathbb{T}}:(f_{n}(t_{1}),\cdots, f_{n}(t_{m}))\in C_{n}\},$$ where the union would be disjoint (since $C_{n}$ was disjoint), and each set in the union would be an elementary cylinder (since $C_{n}$ was a $k-$fold product of Borel sets).

So, let us show the sufficient condition by an induction on $m$!
For $m=2$, note that we can write $$(A_{1}\times A_{2})\setminus(B_{1}\times B_{2})=\Big((A_{1}\setminus B_{1})\times A_{2}\Big)\cup\Big(A_{1}\times (A_{2}\setminus B_{2})\Big),$$ however this union is NOT disjoint since it has these two sets have an intersection $$C_{1}:=\Big((A_{1}\setminus B_{1})\times A_{2}\Big)\cap\Big(A_{1}\times (A_{2}\setminus B_{2})\Big)=(A_{1}\setminus B_{1})\times (A_{2}\setminus B_{2}).$$
But we can decompose as $$(A_{1}\times A_{2})\setminus(B_{1}\times B_{2})=C_{1}\cup\Big((A_{1}\setminus B_{1})\times (A_{2}\cap B_{2})\Big)\cup \Big((A_{1}\cap B_{1})\times (A_{2}\setminus B_{2})\Big).$$
Denote the second term and third them to be $C_{2}$ and $C_{3}$, respectively. Then, note that $C_{1}, C_{2}, C_{3}$ are all the product of two Borel sets, and they are disjoint. Hence, the desired condition holds for $m=2$. 
Suppose the desired condition holds for $m=k$ for some fixed $k>2$, then consider the case of $m=k+1$. Denote $A:=A_{1}\times\cdots\times A_{k}$ and $B:=B_{1}\times\cdots\times B_{k}$, then using the case of $m=2$, we have the following:
\begin{align*}
(A_{1}\times\cdots A_{k}\times A_{k+1})\setminus (B_{1}\times\cdot\times B_{k}\times B_{k+1})&=(A\times A_{k+1})\setminus (B\times B_{k+1})\\
&=D_{1}\cup D_{2}\cup D_{3},
\end{align*}
where $D_{1}:=(A\setminus B)\times (A_{k+1}\setminus B_{k+1})$, $D_{2}:=(A\setminus B)\times (A_{k+1}\cap B_{k+1})$, and $D_{3}:=(A\cap B)\times (A_{k+1}\setminus B_{k+1})$.
Note that $D_{1},D_{2},D_{3}$ are disjoint, so we only need to show each of $D_{1}, D_{2}, D_{3}$ is a $k+1-$fold product of Borel sets, or the finite disjoint union of them. 
By the induction hypothesis, we know that $$A\setminus B=\bigcup_{n=1}^{w}C_{n},$$ where $C_{n}$ disjoint and is a $k-$fold product of Borel sets.
Thus, we can rewrite $$D_{1}=\bigcup_{n=1}^{w}C_{n}\times (A_{k+1}\setminus B_{k+1})=\bigcup_{n=1}^{w}\Big(C_{n}\times (A_{k+1}\setminus B_{k+1})\Big),$$ and $$D_{2}=\bigcup_{n=1}^{w}C_{n}\times (A_{k+1}\cap B_{k+1})=\bigcup_{n=1}^{w}\Big(C_{n}\times (A_{k+1}\cap B_{k+1})\Big),$$ and thus $D_{1}$ and $D_{2}$ are finite disjoint union of sets, and each of the set in the union is a $k+1-$fold product of Borel sets. 
The desired property for $D_{3}$ is immediate since $$D_{3}=(A\cap B)\times (A_{k+1}\setminus B_{k+1})=(A_{1}\cap B_{1})\times \cdots\times(A_{k}\cap B_{k})\times (A_{k+1}\setminus B_{k+1}).$$
Thus, the desired property holds for $m=k+1$.
The result follows immediately. 
I am really not confident about my proof, so I really appreciate it if one could have a quick check of my proof.
Also, it would be really nice if someone has a shorter, alternative proof for the second part. My proof was really tedious :)
Thank you so so so much! I am really weak at measure theory..
 A: Your proofs seem fine to me, but let me suggest an alternative proof of the second claim that I believe is somewhat simpler and does not rely on induction.
Let $m\in\mathbb N$, and $A_i$ and $B_i$ be Borel subsets of $\mathbb R$ for every $i\in\{1,\ldots,m\}$. Define, for each $i\in\{1,\ldots,m\}$,
\begin{align*}
C_i&\equiv A_i\setminus B_i,\\
D_i&\equiv A_i\cap B_i.
\end{align*}
Think about what $(x_1,\ldots,x_m)\in(A_1\times\cdots\times A_m)\setminus(B_1\times\cdots\times B_m)$ really means for a moment. It means that $x_i\in A_i$ for all $i\in\{1,\ldots,m\}$, but $x_j\notin B_j$ for at least one $j\in\{1,\ldots,m\}$.
Viewed through this lens, I trust you to be able to prove the following:
\begin{align*}
(A_1\times\cdots\times A_m)\setminus(B_1\times\cdots\times B_m)=\bigcup (E_1\times\cdots\times E_m),\tag{$*$}
\end{align*}
where the union runs through sets of the form $E_1\times\cdots\times E_m$ such that $E_i\in\{C_i,D_i\}$ for each $i\in\{1,\ldots,m\}$, except that you do not include the set $D_1\times\cdots\times D_m$ in the union (that would mean $x_i\in B_i$ for all $i\in\{1,\ldots,m\}$ as well, which you do not want to allow).
The union on the right-hand side of ($*$) is clearly finite (more precisely, it consists of $2^m-1$ product sets) and also disjoint, because if you consider two different product sets $E_1'\times\cdots\times E_m'$ and $E_1''\times\cdots\times E_m''$, then there is at least one coordinate $j\in\{1,\ldots,m\}$ such that (without loss of generality) $E_j'=C_j$ and $E_j''=D_j$, making the intersection of $E_1'\times\cdots\times E_m'$ and $E_1''\times\cdots\times E_m''$ empty.
