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I apologize if I missed an existing question on this, perhaps with a different spelling of Khinchin's name. I feel like I'm missing something basic.

From Wikipedia, almost all real numbers have a continued fraction representation whose terms have a geometric mean of $K_0=2.685...$ From the definition of "almost all", I would understand that there is an at-most-countable set of counter-examples, ie real numbers with continued fraction representations whose terms have a different geometric mean.

But I also see here that continued fractions provide a homeomorphism between real numbers and and sequences of positive integers, seemingly confirming the intuition that the two sets should be isomorphic.

This seems to imply that there should be only a countable number of positive integer sequences with a geometric mean different from Khinchin's constant.

But this seems preposterous! If nothing else, we should be able to generate uncountably many sequences with a geometric mean of $2K_0$, by simply doubling the terms of any "normal" sequence with a mean of $K_0$.

Where did I go wrong?

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Here almost all means except on a set of measure zero; see here, for instance. All countable sets have measure zero, but not all measure zero sets are countable; the middle-thirds Cantor set is a well-known example of a set of measure zero whose cardinality is equal to that of $\Bbb R$.

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    $\begingroup$ I can see that but could you explain why his reasoning is wrong about doubling each $a_i$ and getting $2K_0$ as the geometric mean for just as many numbers? It seems reasonable. $\endgroup$ – Jam Sep 8 '16 at 0:23
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    $\begingroup$ @Jam: I’ve not worked it out in detail, but I suspect that it has to do with the rather complicated way that doubling each denominator of a continued fraction affects the real number represented by that continued fraction. $\endgroup$ – Brian M. Scott Sep 8 '16 at 0:32
  • $\begingroup$ Of course! I knew I made a simple error. $\endgroup$ – user3461142 Sep 8 '16 at 0:33
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    $\begingroup$ @Jam You will indeed get just as many numbers in the sense of cardinality, but that tells you nothing about the measure of the resulting set, which must be zero. $\endgroup$ – Jack M Sep 8 '16 at 10:19

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