most recent 30 from http://mathoverflow.net 2013-05-26T03:32:23Z http://mathoverflow.net/feeds/user/21215 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts/87829#87829 Andrew Stewart 2012-02-07T19:21:16Z 2012-02-08T00:25:18Z

<p>Here's something that's pretty neat: find a measurable subset $A$ of $[0,1]$ such that for any subinterval $I$ of $[0,1]$, the Lebesgue measure $\mu(A\cap I)$ has $0 &lt; \mu(A\cap I) &lt; \mu(I)$. There's an explicit construction of such a set in Rudin, who describes such sets as "well-distributed". Balint Virag (and maybe others) found a very slick probabilistic construction.</p> <p>Let $X_1, X_2, \ldots$ be i.i.d. coin flips, i.e. $X_1$ is $1$ with probability $1/2$ and $-1$ with probability $1/2$. Consider the (random) series</p> <p>$$S:=\sum_{n=1}^\infty X_n/n.$$</p> <p>By the Kolmogorov three-series theorem, it converges almost surely. However, it's a simple exercise to see that for any $a$, the event ${S > a}$ has non-trivial measure: for $a>0$, there's a positive chance of the first $e^a$ terms of the series being positive, so the $e^a$-th partial sum is positive, and the tail is independent and positive or negative with equal probability, due to symmetry. For $a\leq 0$, it's trivial, again because of symmetry.</p> <p>A common way of realizing i.i.d. coin flips on the unit interval is as Rademacher functions: for $x\in[0,1]$, let ${b_n}$ be its binary expansion, and $X_n(x) = (-1)^{b_n}$. Realized this way, the random sum $S$ becomes an almost everywhere finite measurable function from $[0,1]$ to $\R$. It only takes a bit more work to see that the set ${S>a}$ is exactly a well-distributed set.</p> <p>Alex Bloemendal has written this up in a short note, but I'm not sure if he's published it anywhere.</p>

http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts/87829#87829 Andrew Stewart 2012-02-08T00:20:47Z 2012-02-08T00:20:47Z

D'oh. Thanks...