5 Appendix: Proof of Proposition 2.2

[126.1.1] The proof given below follows Ref. [30]. [126.1.2] Suppose limn→∞⁡μn⁢s=μ⁢s and limn→∞⁡μn⁢an⁢s+bn=ν⁢s with μ⁢s and ν⁢s both nondegenerate. [126.1.3] Then it must be shown that there exist a>0 and b such that

[126.1.4] Pick a sequence of integers n1<n2<…<nk<… such that limk→∞⁡ank=a and limk→∞⁡bnk=b exist with 0≤a≤∞ and -∞≤b≤∞. [126.1.5] Consider this sequence of indices from now on as fixed. [126.1.6] Then, to simplify the notation, suppose without loss of generality that limk→∞⁡ak=a and limk→∞⁡bk=b.

[126.2.1] First it will be shown that 0<a<∞. [126.2.2] Suppose a=∞. [126.2.3] Let

[126.2.4] Then for v<x<u

and hence for every v<u it follows that ν⁢v=0 because an⁢v+bn→-∞ with n→∞. [126.2.5] For v>u, on the other hand, lim sup⁡an⁢v+bn=∞ and hence ν⁢v=1 for v>u. [126.2.6] Thus the assumption a=∞ contradicts to ν⁢s being nondegenerate.

[126.3.1] It follows that also b must be finite. [126.3.2] In fact if limn→∞⁡an⁢x+bn=∞ then ν⁢x=1 while for limn→∞⁡an⁢x+bn=-∞ follows ν⁢x=0.

[126.4.1] Suppose now that a=0. [126.4.2] Then for every x and ε>0

[page 127, §0]    if n is chosen sufficiently large. [127.0.1] By monotonicity of μn it follows that

[127.0.2] If ε is chosen so that μ⁢x is continuous at the points b-ε and b+ε, then

[127.0.3] Because x was arbitrary it follows that μ⁢b-ε=0 and μ⁢b+ε=1. [127.0.4] Hence μ⁢x is degenerate, contrary to the conditions above.

[127.1.1] Finally, let x be such that μ⁢x is continuous at the point a⁢x+b, and that ν⁢x is continuous at x. [127.1.2] Then

[127.1.3] On the other hand because limn→∞⁡an⁢x+bn=a⁢x+b one has for sufficiently large n that

where ε>0 is chosen such that the distribution function μ is continuous at the points a⁢x+b-ε and a⁢x+b+ε. [127.1.4] Hence by monotonicity

and for n→∞

[127.1.5] Because a⁢x+b is a point of continuity for μ⁢x and ε is arbitrary it follows that

and hence ν⁢x=μ⁢a⁢x+b proving the assertion.