Given a function $m': \mathcal{E}_o \to [0,1)$ that is translation invariant and additive show $m'([0,1)^d)=n^dm'([0,\frac{1}{n})^d)$.
I believe I am simply misunderstanding some of the definitions, because my proposed proof is close but obviously imperfect:
\begin{align} m'([0,1)^d)&=m'(\bigsqcup^{n}_{j=1}\frac{1}{n}[j-1,j)^d) \\ &= \sum^{n}_{j=1}m'(\frac{1}{n}[j-1,j)^d)\\ &= \sum^{n}_{j=1}m'(\frac{1}{n}[j-1,j)^d +\left(\frac{1-j}{n}\right)^d)\\ &= \sum^{n}_{j=1}m'([0,\frac{1}{n})^d)\\ &= nm'([0,\frac{1}{n})^d) \end{align}
It is not true that $[0, 1)^d = \bigcup_{j = 1}^{n}\frac{1}{n}[j - 1, j)^d$. You are missing the cross terms. For example when $d = 2$, $$(A \cup B)^2 = ((A \times A) \cup (A \times B)) \cup ((B \times A) \cup (B \times B)).$$