# 5 Appendix: Proof of Proposition 2.2

[126.1.1] The proof given below follows Ref. [30].
[126.1.2] Suppose limn→∞μns=μs and
limn→∞μnans+bn=νs with μs and νs
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

u=supx:lim supn→∞anx+bn<∞. | | (183) |

[126.2.4] Then for v<x<u

lim supn→∞anv+bn≤lim supn→∞v-xan+lim supn→∞anx+bn, | | (184) |

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

[126.3.1] It follows that also b must be finite.
[126.3.2] In fact if limn→∞anx+bn=∞ then νx=1
while for limn→∞anx+bn=-∞ follows
νx=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

μnb-ε≤μnanx+bn≤μnb+ε. | | (186) |

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

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

[127.1.1] Finally, let x be such that μx is continuous at
the point ax+b, and that νx is continuous at x.
[127.1.2] Then

limn→∞μnanx+bn=νx. | | (188) |

[127.1.3] On the other hand because limn→∞anx+bn=ax+b
one has for sufficiently large n that

ax+b-ε≤anx+bn≤ax+b+ε, | | (189) |

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

μnax+b-ε≤μnanx+bn≤μnax+b+ε | | (190) |

and for n→∞

μax+b-ε≤lim infn→∞μnanx+bn≤lim supn→∞μnanx+bn≤μax+b+ε. | | (191) |

[127.1.5] Because ax+b is a point of continuity for μx
and ε is arbitrary it follows that

limn→∞μnanx+bn=μax+b | | (192) |

and hence νx=μax+b proving the assertion.