There are the definitions which we need for the theorem:
There are the theorem $2.20$ and proof of its part $(a)$ and $(b)$:
There are the proofs of part $(c)$, $(d)$ and $(e)$:
The proofs of $(c)$, $(d)$, and $(e)$ will use the following observation: If $\lambda$ is a positive Borel measure on $R^k$ and $\lambda(E)$ $=$ $m(E)$ for all boxes $E$, then the same equality holds for all open sets $E$, by property $2.19(d)$, and therefore for all Borel sets $E$, since $\lambda$ and $m$ are regular (Theorem $2.18$).
To prove $(c)$, fix $x$ $\in$ $R^k$ and define $\lambda(E)$ $=$ $m(E+x)$.
It is clear that $\lambda$ is then a measure; by $(a)$, $\lambda(E)$ $=$ $m(E)$ for all boxes, hence $m(E+x)$ $=$ $m(E)$ for all Borel sets $E$.
The same equality holds for every $E$ $\in$ $\mathfrak M$, because of $(b)$.
I don't understand why is it clear in $(c)$ that $\lambda$ is a measure. I also don't understand how does $(a)$ imply, that $\lambda(E)$ $=$ $m(E)$ for all boxes?
Any help would be appreciated.
We can demonstrate this using the second part of (e), that is, when $T$ is invertible, $\Delta(T) = |\det T|$ (this is not obvious, but it can be proven by proving it for $T$ that generate $GL(n, \mathbb{R})$, e.g. $T$ being each of the three elementary row operations).
Suppose $Y \subset \mathbb{R}^n$ is a linear subspace of dimension $k < n$. By basic linear algebra, there is an orthogonal matrix $A$ such that $A(Y) = \mathbb{R}^{k} \times \{0\}^{n - k}$. In particular, $A : \mathbb{R}^n \to \mathbb{R}^n$ is an isometric isomorphism, so $A$ and $A^{-1}$ are both continuous, hence measurable. Thus $Y = A^{-1}AY = A^{-1}(\mathbb{R}^{k} \times \{0\}^{n - k})$ is Borel measurable. Since $\det A = \pm 1$, we get $$m(Y) = m(A(Y)) = m(\mathbb{R}^{k} \times \{0\}^{n - k}) = m(\mathbb{R}^k) \cdot m(\{0\}^{n - k}) = \infty \cdot 0 = 0.$$