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S Aug 11, 2013 at 1:48 history suggested Michael Albanese CC BY-SA 3.0
Replaced \\, by \, and \\# by \#.
Aug 11, 2013 at 1:43 review Suggested edits
S Aug 11, 2013 at 1:48
Jul 30, 2011 at 10:24 vote accept Ian Morris
Jul 19, 2011 at 10:12 comment added Ian Morris It occurred to me after sleeping that in the paper I'm reading, the extra assumption is made that the functions $f_n$ are essentially bounded above independently of $n$, and in this subcase we can just use the MCT applied to the sequence $\sup_{m \geq n} f_m$ to conclude that $\int f_N<0$. However the general case is of course extremely nice to know, and I think even answers an open question in that paper.
Jul 19, 2011 at 10:10 comment added Ian Morris Thanks! I'd noticed that the result is false for $f_n(x) \to +\infty$ using exactly Anthony's example. For some reason I let this make me believe that Atkinson's argument wouldn't be applicable and didn't bother checking. I've even got Atkinson's paper on my hard drive, so I feel a little stupid ;o)
Jul 19, 2011 at 7:44 comment added Anthony Quas This reminds me of a lemma of Tanny that I used recently: Suppose $T$ an ergodic measure-preserving transformation; $f$ a non-negative(not necessarily integrable) function. either (1) $f(T^nx)/n\to 0$ a.e. or (2) $\limsup f(T^nx)/n=\infty$ a.e. I eventually stumbled on a 6 line proof that looks a lot like this proof, apparently due to Feldman that appears in a paper of Lyons, Pemantle and Peres.
Jul 19, 2011 at 6:57 comment added Anthony Quas It's definitely false in the case $f_n(x)\to\infty$ as you see from the example $f_n(x)=sqrt n$.
Jul 19, 2011 at 5:05 history edited Vaughn Climenhaga CC BY-SA 3.0
added 6 characters in body
Jul 19, 2011 at 3:26 history answered Vaughn Climenhaga CC BY-SA 3.0