The cube $[0, 1]^n$ is admissible, and $v([0, 1]^n ) = 1$. Prove it Using Darboux sums.
Hint: $L_N (\mathbb{1}[0,1]^n ) = 1$ and $U_N (\mathbb{1}[0,1]^n ) = 2^{-nN} (2^N + 2)^n$
So I get that using the hint $U_N (\mathbb{1}[0,1]^n )\rightarrow1$ when $N \rightarrow \infty$ and then $\mathbb{1}[0,1]^n$ is integrable and the cube is admissable and that solves it, however I would love to see a proof of the hint, I think it would help me grasp the concept of darboux sums in n-dimensions better.
Any help?