content_image/319228.jpg content_image/319229.jpg content_image/319227.jpg content_image/319226.jpg
Therefore, 

<img src='content_image/319228.jpg'>

<img src='content_image/319229.jpg'>

where the last equality follows from equations (B.3) and (B.5). 

By the definition of the Riemann integral, expression (B.10) can be made arbitrarily close to − H$_{UV}$ . Specifically, there exists a 0 < d ( ϵ ) < 1 such that if all cells satisfy r$_{h}$ ( C ) − r$_{l}$ ( C ) N < d and s$_{h}$ ( C ) − s$_{l}$ ( C ) N < d , then 

<img src='content_image/319227.jpg'>

Therefore, it follows that for any partition I ∈ Π for which all cells satisfy r$_{h}$ ( C ) − r$_{l}$ ( C ) N < d and s$_{h}$ ( C ) − s$_{l}$ ( C ) N < d , then we have 

<img src='content_image/319226.jpg'>

It remains to show that the contribution of the fraction of partitions that do not satisfy r$_{h}$ ( C ) − r$_{l}$ ( C ) N < d and s$_{h}$ ( C ) − s$_{l}$ ( C ) N < d goes to zero as N → ∞ . Since the probability of selecting an x -value (or y -value) for a partition that will have a cell larger than d is 

