Timeline for Is Conway's base-13 function measurable?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2010 at 17:59 | comment | added | Robin Chapman | The probability that the base-13 expansion of a uniformly distributed random number in (0,1) has contains only digits 0-9 starting at the N-th digit is $\prod_{n=N}^\infty(10/13)=0$. This is also a Borel set (intersection of a countable sequence of Borel sets defined by the $n$-th digit). A countable union of measure zero Borel sets is a measure zero Borel set. | |
| Jul 20, 2010 at 15:41 | comment | added | Noah Stein | the base $10^{k+1}$ representation. | |
| Jul 20, 2010 at 15:40 | comment | added | Noah Stein | Well we don't even really need the full version of normality, just that every digit occurs infinitely often. To see this it is enough to note if we select a number in [0,1) at random according to Lebesgue measure, then its digits in base n are i.i.d. uniform over $\{0,\ldots, n-1\}$. Therefore the probability that any given digit doesn't occur is zero. There are countably many digits and bases, so a.s. in any base all digits will occur. This means they will all occur infinitely often, because if say $5$ only occurred $k$ times then $5\cdots 5$ ($k+1$ times) would never occur in | |
| Jul 20, 2010 at 15:00 | comment | added | Willie Wong | Ah. I got as far as the lack of normality of the tredecimal expansion. But I forgot that real numbers are almost surely normal. (Now off to tracking down Borel's original proof of this fact, since I've never actually seen it.) | |
| Jul 20, 2010 at 14:58 | vote | accept | Willie Wong | ||
| Jul 20, 2010 at 14:43 | history | answered | Noah Stein | CC BY-SA 2.5 |