Therefore, 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 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 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 58 | ( − H I UV + H I U + H I V + H$_{UV}$ ) | < ϵ/ 3 . | ∑ C ∈C ( I ) r$_{h}$ ( C ) − r$_{l}$ ( C ) N s$_{h}$ ( C ) − s$_{l}$ ( C ) N c ( u$_{C}$,v$_{C}$ ) log c ( u$_{C}$,v$_{C}$ ) + H$_{UV}$ | < ϵ/ 3 . − H I UV = ∑ C ∈C ( I ) c ( u$_{C}$,v$_{C}$ ) r$_{h}$ ( C ) − r$_{l}$ ( C ) N s$_{h}$ ( C ) − s$_{l}$ ( C ) N log$^{(}$ c ( u$_{C}$,v$_{C}$ ) r$_{h}$ ( C ) − r$_{l}$ ( C ) N s$_{h}$ ( C ) − s$_{l}$ ( C ) N ) = ∑ C ∈C ( I ) r$_{h}$ ( C ) − r$_{l}$ ( C ) N s$_{h}$ ( C ) − s$_{l}$ ( C ) N c ( u$_{C}$,v$_{C}$ ) log c ( u$_{C}$,v$_{C}$ ) (B.10) + ∑ C ∈C ( I ) Pr ( r$_{l}$ ( C ) ≤ U ≤ r$_{h}$ ( C ) ,s$_{l}$ ( C ) ≤ V ≤ s$_{h}$ ( C )) log r$_{h}$ ( C ) − r$_{l}$ ( C ) N + ∑ C ∈C ( I ) Pr ( r$_{l}$ ( C ) ≤ U ≤ r$_{h}$ ( C ) ,s$_{l}$ ( C ) ≤ V ≤ s$_{h}$ ( C )) log s$_{h}$ ( C ) − s$_{l}$ ( C ) N = ∑ C ∈C ( I ) r$_{h}$ ( C ) − r$_{l}$ ( C ) N s$_{h}$ ( C ) − s$_{l}$ ( C ) N c ( u$_{C}$,v$_{C}$ ) log c ( u$_{C}$,v$_{C}$ ) − H I U − H I $_{V}$, (B.11)