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( N$_{1}$$_{x}$ m − 2 $^{)}$N$_{2}$$_{x}$( N$_{1}$$_{y}$ m − 2 $^{)}$N$_{2}$$_{y}$ . Simple algebraic manipulations lead to the following expression for ( N 1 x m − $_{2}$) N$_{2}$$_{x}$ ( N − 1 m − $_{1}$) : 

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Since N$_{1}$$_{x}$/N converges almost surely to ∫ x$_{0}$ + R/ 4 x$_{0}$ f ( x ) dx , and similarly N$_{2}$$_{x}$/N con- verges almost surely to ∫ x$_{0}$ + R x$_{0}$ +3 R/ $_{4}$f ( x ) dx , then for m ≥ 3 finite it follows that ( N 1 x m − $_{2}$) N$_{2}$$_{x}$ ( N − 1 m − $_{1}$) converges almost surely to a positive constant. Similarly, ( N 1 y m − $_{2}$) N$_{2}$$_{y}$ ( N − 1 m − $_{1}$) converges almost surely to a positive constant. Therefore, | Γ | / { ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$} is bounded away from zero. 

Second, we shall consider the case that m → ∞ . The complement of Γ, Γ $^{C}$, is the set of contingency tables with no grid point in A$_{1}$ or in A$_{2}$ . An upper bound for | Γ $^{C}$| is: 

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Note that since | Γ | = { ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$}− | Γ $^{C}$| , it is enough to show that | Γ $^{C}$| / { ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$} converges to zero as N → ∞ . Simple algebraic manipulations lead to the following expression for ( N − 1 − N 1 x m − 1 ) ( N − 1 m − $_{1}$) : 

<img src='content_image/305103.jpg'>

Since N$_{1}$$_{x}$/N converges almost surely to a positive fraction ∫ x$_{0}$ + R/ 4 x$_{0}$ f ( x ) dx , it fol- lows that ( N − 1 − N 1 x m − 1 ) ( N − 1 m − $_{1}$) converges almost surely to zero. Similarly, since N$_{2}$$_{x}$/N , N$_{1}$$_{y}$/N and N$_{2}$$_{y}$/N converge almost surely to positive fractions, it follows that respectively, ( N − 1 − N 2 x m − 1 ) ( N − 1 m − $_{1}$) , ( N − 1 − N 1 y m − 1 ) ( N − 1 m − $_{1}$) , and ( N − 1 − N 2 y m − 1 ) ( N − 1 m − $_{1}$) converge almost surely to zero. Thus | Γ $^{C}$| / { ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$} converges almost surely to zero. 

Since lim$_{N}$$_{→∞}$ | Γ | / { ( N − 1 m − 1 )( N − 1 m − 1 $^{)}$} is bounded below by a positive constant, it follows 

