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So for given $x \in [0,1]$, if we let $f_n(x)$ be the fraction of 1's occurring in the first $n$ binary digits of the binary expansion for $x$ (where we always assume an infinite trailing string of 0 instead of 1 for certain rationals $x$), then the measure of the set $X_c \subset [0,1]$ for which $\lim_{n \to \infty} f_n(x) = c$ for $x \in X_c$ is 1 if $c = 1/2$, and then obviously $0$ otherwise. However this is a non-trivial fact that the measure of $X_{1/2}$ is 1. A consequence, however, is that the measure of the set $Y$ for which the limit does not exist is also $0$. Is there any easy direct way to prove this consequence, without showing that the measure of $X_{1/2}$ is 1? It seems easier to show that the measure is $0$ for $X_c$ when $c \neq 1/2$.

user2566092
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  • What's the easy proof you have in mind that $X_c$ has measure $0$ for $c\ne1/2$? The reason I ask is there is a proof that $X_{1/2}$ has measure $1$ that's much easier than the proofs a lot of people know; it could be that it's even easier than the proofs for $c\ne1/2$ that you're thinking of... Give us an idea how the proof for $c\ne1/2$ goes to give us an idea how easy it has to be to count as easy. – David C. Ullrich Aug 28 '15 at 21:16
  • @DavidC.Ullrich I said it SEEMS easier, and I would have to think to perhaps get a rigorous proof, but basically my idea is that if $|c_1 - 1/2| < |c_2 - 1/2|$, then the measure of $X_{c_1}$ should be greater than or equal to measure of $X_{c_2}$. Thus, if any $X_c$ has positive measure for $c \neq 1/2$, then we would get a contradiction because there are uncountably many $c_1$ between $c$ and $1/2$. That's of course assuming we know the $X_c$ are measurable and also, assuming my intuition can be made rigorous for comparing $X_{c_1}$ to $X_{c_2}$. – user2566092 Aug 28 '15 at 21:33
  • The "fact" you want to use does not seem obvious. Strong law of large numbers, on the other hand... – Did Aug 28 '15 at 21:36
  • @Did I agree, it depends on how easy it is to translate a certain intuition into proof. However what is definitely true is that all but countably many of the $X_c$ would have to have measure zero. But what would make those choices of $c$ so special? Why would there be arbitrarily close choices of $c_1,c_2$ to fixed $c$ where $c_1 < c < c_2 < 1/2$ such that $X_c$ has positive measure but $X_{c_1}$ and $X_{c_2}$ have measure zero? – user2566092 Aug 28 '15 at 21:41
  • @DavidC.Ullrich I try to avoid being pedantic, but even if the easiest route to show the complement of $X_{1/2}$ in its entriety has measure 0 is to show that $X_{1/2}$ has measure 1, the question remains whether there is an easier way to show my set $Y$ in question has measure $0$. Maybe it's actually "harder" to directly show $X_c$ has measure $0$ for $c \neq 1/2$. But my fundamental question was whether there's an "easier" way to show $Y$ has measure $0$ (if it's measurable, which I'm not requesting to prove), compared to whatever the easiest proof is for showing $X_{1/2}$ has measure 1. – user2566092 Aug 28 '15 at 22:15
  • @DavidC.Ullrich Also I may be in the camp that doesn't know your "easy" proof you allude to for showing $X_{1/2}$ has measure 1. If you can enlighten me, and I think it's easy enough that it can't be beaten as far as directly proving $Y$ has measure 0, I'll probably just delete this question, or leave it and upvote your answer if you post it as an answer. – user2566092 Aug 28 '15 at 22:18
  • @user2566092 Thought you'd never ask. See http://math.stackexchange.com/questions/1413157/law-of-large-numbers-utility-difficulty-of-various-versions . And thanks for the excuse... – David C. Ullrich Aug 28 '15 at 22:53
  • @DavidC.Ullrich Thanks, I upvoted your linked question. I'll leave this post here since you referenced it. – user2566092 Aug 28 '15 at 23:01

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