Problems in understanding Lacey's proof of Carleson's theorem. I am new to the Stackexchange community so do let me know if I can improve my question in any way. 
Right, I have just started reading  Michael Lacey's proof of Carleson's theorem (http://people.math.gatech.edu/~lacey/research/esi.pdf), and have already hit a problem.
In Proposition 1.4 (Page 4), I don't understand a few things:
1) Why has the author proceeded in this method for establishing that the desired set is closed? 
My instincts would have said, take a sequence in the set of functions for which the results hold, say they converge in the norm to some function and finally prove that the said function is member of the desired set. 
I think I can, however, see how the two might be equivalent: if one proves the desired set is closed the criterion that Lacey is trying to establish follows. Vice-versa if one proves what Lacey sets out to prove in the first line of the proof then the set must be closed.
2) This is just to confirm my own instincts, but I believe the result follows from an application of Chebyshev's Inequality.
I would be grateful for any help/pointers and as above please don't hesitate to let me know if I haven't complied with any other rules of the forum.
Thank you.
 A: (1) I take it you're asking why he used this for showing that the set of functions satisfying 1.2 is closed? This is a standard technique; 1.4 is a "maximal function inequality", and maximal function inequalities are simply how people prove almost-everywhere convergence results.
How would you do it? You say "take a sequence of functions satisfying 1.2 which converge in norm and show the limit satisfies 1.2". How do you show that, exactly? You've left out a lot of details... (Hint: The fact that the set satisfying 1.2 is closed is precisely equivalent to Carleson's theorem. So if you actually do see a proof that's a simple as you're implying you're going to be famous.)
Not to, um, but if you're not familiar with the notion of maximal functions versus almost everywhere convergence it may well be that you're in over your head here. (Note I happen to be very familiar with this notion, but I'm in way over my head.)
(2) No, it would be a consequence of Chebyshev if we knew that $Cf$ was in $L^2$. We don't know that.
