sup Λ ,k ∣ ( ε$_{1}$ ) ′ k − iε$_{k}$$_{2}$ ∣ . For 0 < ζ$_{4}$ find N$_{2}$ such that 

Bound Pr(sup Λ | ˆ κ$_{kn}$ − κ$_{k}$ | > ζ$_{5}$ ) ≤ 

Pr(sup Λ ∣ ((ˆ ε$_{1}$$_{n}$ ) ′ k − i ˆ ε$_{2}$$_{kn}$ ) − (( ε$_{1}$ ) ′ k − iε$_{k}$$_{2}$ ) ∣ > ζ$_{4}$ ) < ζ$_{2}$ for n > N$_{2}$. 

Pr(sup Λ ∣ ˆ ε − 1 1 n ∣ ∣ ((ˆ ε$_{1}$$_{n}$ ) ′ k − i ˆ ε$_{2}$$_{n}$ − ( ε$_{1}$ ) ′ k + iε$_{2}$ ∣ +sup Λ ∣ ˆ ε − 1 1 n − ε − 1 1 ∣ ∣ ( ε$_{1}$ ) ′ k − iε$_{k}$$_{2}$ ∣ > ζ$_{5}$ ) ≤ 

Pr(sup Λ ∣ ˆ ε − 1 1 n ∣ > ζ − 1 1 ) + Pr ( sup Λ ∣ ((ˆ ε$_{1}$$_{n}$ ) ′ k − i ˆ ε$_{2}$$_{kn}$ − ( ε$_{1}$ ) ′ k + iε$_{k}$$_{2}$ ∣ > ζ$_{5}$ζ$_{1}$ ) + Pr(sup Λ ∣ ˆ ε − 1 1 n − ε − 1 1 ∣ > ζ$_{5}$/ ¯ B ) . 

If ζ$_{5}$ = min { ζ$_{1}$ ¯ B, ζ$_{4}$/ζ$_{1}$ } the probability as n > max { N$_{1}$, N$_{2}$ } is less than 4 ζ$_{2}$. 

Then Pr(sup Λ ∣ ∫ s 0 ∑ d k $_{=1}$ˆ κ$_{kn}$ ( t ) dt$_{k}$ − ∫ s 0 ∑ d k $_{=1}$κ$_{k}$ ( t ) dt$_{k}$ ∣ > ζ$_{6}$ ) ≤ 

Pr(sup Λ ∫ s 0 ∑ d k $_{=1}$| ˆ κ$_{kn}$ ( t ) − κ$_{k}$ ( t ) | dt$_{k}$ > ζ$_{6}$ ) ≤ Pr ( sup Λ | ˆ κ$_{kn}$ ( t ) − κ$_{k}$ ( t ) | > ζ$_{6}$/µ (Λ) ) , 

where µ (Λ) is the measure of the compact set Λ . For ζ$_{6}$ = µ (Λ) ζ$_{5}$ then the probability is less than 4 ζ$_{2}$. 

Consider now on Λ the function ˆ φ − 1 n ( s ) = exp( − ∫ s 0 ∑ d k $_{=1}$ˆ κ$_{kn}$ ( t ) dt$_{k}$ ) . De- fine ˜ B = sup Λ V (1 + s $^{2}$) $^{m}$; then sup Λ ∣ φ − $^{1}$( s ) ∣ < ˜ B. 

