I saw this question in measure theory course:
Let $m′:L(R^d)→[0,∞]$ be a measure on the Lebesgue measurable sets, satisfying:
• Unit box measure: $m′([0,1]^d)=1$.
• Translation invariance: $m′(x+A)=m′(A)$ for all $A∈L(R^d)$ and $x∈R^d$.
Show that $m′$ agrees with the Lebesgue measure on boxes, i.e., $m′(B)=m(B)$ for any box $B⊆R^d$.
I was able to show that for any box with integer length edges, I just used the 2 properties of $m'$ and the additivity property of a measure. But how can I show it for general boxes?
Best Answer
Suppose for the moment that $m'([0,1)^d) = m'([0,1]^d) = 1$. Then we could divide the unit half-open hypercube into a disjoint collection of such cubes that look like $[k_1/n, (k_1+1)/n)\times\dotsm\times [k_d/n, (k_d+1)/n)$ where $k_1,\dotsc, k_d \in \{0,\dotsc, n-1\}$. If $Q_k$ denotes the half-open hypercube with $k=(k_1,\dotsc, k_d)$, then $$\sum_k m'(Q_k) = m'([0,1)^d) = 1,$$ yet by translation invariance all the $m'(Q_k)$ are equal, and since there are $n^d$ of them we get $m'(Q) = 1/n^d$ for any half-open hypercube with side lengths $1/n$.
Piecing together such half-open cubes one can show $m'([0,s_1)\times\dotsm[0,s_d)) = s_1\dotsm s_d$ for rational $s_1,\dotsc, s_d$. Then any half-open box $B$ can be written as the limit from below of an increasing sequence of rational side-length half-open boxes; continuity of measure from below allows us to conclude $m'(B)=m(B)$.
The only thing left to check is that the deleted faces are $m'$-measure zero. To this end, suppose to the contrary that there is a face of the closed unit hypercube with nonzero $m'$-measure $c$ with $c>0$. Say without loss of generality it is $\{1\}\times[0,1]^{d-1}$. This cannot happen though by translation invariance. Indeed, pick $n$ so large that $cn>1$, and also pick $n$ distinct points $x_1,\dotsc, x_n\in (0,1]$. We have $$E:=\bigcup_{k=1}^n \{x_k\}\times [0,1]^{d-1}\subset [0,1]^d$$ yet $m'(E) = cn$ while $m'([0,1]^d) = 1$, a contradiction. Thus any face of the unit hypercube (and, by extension, any face of any box) has $m'$-measure zero, and our work above suffices for the original closed box problem.