[126.1.1] The proof given below follows Ref. [30].
[126.1.2] Suppose and
with
and
both nondegenerate.
[126.1.3] Then it must be shown that there exist
and
such
that
![]() |
(182) |
[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
![]() |
(183) |
[126.2.4] Then for
![]() |
(184) |
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
![]() |
(185) |
[page 127, §0]
if is chosen sufficiently large.
[127.0.1] By monotonicity of
it follows that
![]() |
(186) |
[127.0.2] If is chosen so that
is continuous at
the points
and
, then
![]() |
(187) |
[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
![]() |
(188) |
[127.1.3] On the other hand because
one has for sufficiently large
that
![]() |
(189) |
where is chosen such that the distribution function
is continuous at the points
and
.
[127.1.4] Hence by monotonicity
![]() |
(190) |
and for
![]() |
(191) |
[127.1.5] Because is a point of continuity for
and
is arbitrary it follows that
![]() |
(192) |
and hence proving the assertion.