[126.1.1] The proof given below follows Ref. . [126.1.2] Suppose and with and both nondegenerate. [126.1.3] Then it must be shown that there exist and such that
[126.1.4] Pick a sequence of integers such that and exist with and . [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 and .
[126.2.1] First it will be shown that . [126.2.2] Suppose . [126.2.3] Let
[126.2.4] Then for
and hence for every it follows that because with . [126.2.5] For , on the other hand, and hence for . [126.2.6] Thus the assumption contradicts to being nondegenerate.
[126.3.1] It follows that also must be finite. [126.3.2] In fact if then while for follows .
[126.4.1] Suppose now that . [126.4.2] Then for every and
[page 127, §0] if is chosen sufficiently large. [127.0.1] By monotonicity of it follows that
[127.0.2] If is chosen so that is continuous at the points and , then
[127.0.3] Because was arbitrary it follows that and . [127.0.4] Hence is degenerate, contrary to the conditions above.
[127.1.1] Finally, let be such that is continuous at the point , and that is continuous at . [127.1.2] Then
[127.1.3] On the other hand because one has for sufficiently large that
where is chosen such that the distribution function is continuous at the points and . [127.1.4] Hence by monotonicity
[127.1.5] Because is a point of continuity for and is arbitrary it follows that
and hence proving the assertion.