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5 Appendix: Proof of Proposition 2.2

[126.1.1] The proof given below follows Ref. [30]. [126.1.2] Suppose \lim _{{n\to\infty}}\mu _{n}(s)=\mu(s) and \lim _{{n\to\infty}}\mu _{n}(a_{n}s+b_{n})=\nu(s) with \mu(s) and \nu(s) both nondegenerate. [126.1.3] Then it must be shown that there exist a>0 and b such that

\mu(s)=\nu(as+b). (182)

[126.1.4] Pick a sequence of integers n_{1}<n_{2}<\ldots<n_{k}<\ldots such that \lim _{{k\to\infty}}a_{{n_{k}}}=a and \lim _{{k\to\infty}}b_{{n_{k}}}=b exist with 0\leq a\leq\infty and -\infty\leq b\leq\infty. [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 \lim _{{k\to\infty}}a_{k}=a and \lim _{{k\to\infty}}b_{k}=b.

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

u=\sup\{ x:\limsup _{{n\to\infty}}(a_{n}x+b_{n})<\infty\}. (183)

[126.2.4] Then for v<x<u

\limsup _{{n\to\infty}}(a_{n}v+b_{n})\leq\limsup _{{n\to\infty}}(v-x)a_{n}+\limsup _{{n\to\infty}}(a_{n}x+b_{n}), (184)

and hence for every v<u it follows that \nu(v)=0 because (a_{n}v+b_{n})\to-\infty with n\to\infty. [126.2.5] For v>u, on the other hand, \limsup(a_{n}v+b_{n})=\infty and hence \nu(v)=1 for v>u. [126.2.6] Thus the assumption a=\infty contradicts to \nu(s) being nondegenerate.

[126.3.1] It follows that also b must be finite. [126.3.2] In fact if \lim _{{n\to\infty}}(a_{n}x+b_{n})=\infty then \nu(x)=1 while for \lim _{{n\to\infty}}(a_{n}x+b_{n})=-\infty follows \nu(x)=0.

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

b-\varepsilon\leq a_{n}x+b_{n}\leq b+\varepsilon (185)

[page 127, §0]    if n is chosen sufficiently large. [127.0.1] By monotonicity of \mu _{n} it follows that

\mu _{n}(b-\varepsilon)\leq\mu _{n}(a_{n}x+b_{n})\leq\mu _{n}(b+\varepsilon). (186)

[127.0.2] If \varepsilon is chosen so that \mu(x) is continuous at the points b-\varepsilon and b+\varepsilon, then

\mu(b-\varepsilon)\leq\nu(x)\leq\mu(b+\varepsilon). (187)

[127.0.3] Because x was arbitrary it follows that \mu(b-\varepsilon)=0 and \mu(b+\varepsilon)=1. [127.0.4] Hence \mu(x) is degenerate, contrary to the conditions above.

[127.1.1] Finally, let x be such that \mu(x) is continuous at the point ax+b, and that \nu(x) is continuous at x. [127.1.2] Then

\lim _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})=\nu(x). (188)

[127.1.3] On the other hand because \lim _{{n\to\infty}}(a_{n}x+b_{n})=ax+b one has for sufficiently large n that

ax+b-\varepsilon\leq a_{n}x+b_{n}\leq ax+b+\varepsilon, (189)

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

\mu _{n}(ax+b-\varepsilon)\leq\mu _{n}(a_{n}x+b_{n})\leq\mu _{n}(ax+b+\varepsilon) (190)

and for n\to\infty

\mu(ax+b-\varepsilon)\leq\liminf _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})\leq\limsup _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})\leq\mu(ax+b+\varepsilon). (191)

[127.1.5] Because ax+b is a point of continuity for \mu(x) and \varepsilon is arbitrary it follows that

\lim _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})=\mu(ax+b) (192)

and hence \nu(x)=\mu(ax+b) proving the assertion.