Pointwise convergence of Fourier series, Fefferman's article This is the first time I ask a question here, so sorry if I make any mistake in the way I ask it. I'm studing Fefferman's article Pointwise Convergence  of Fourier Series, and I have two questions:
1) In the first page, it is defined a function $n(\cdot)$. How can I prove that it is measurable? 
2) In the third page, I understand the proof of $||T_p||_2\leq CA_0(p)^{1/2}$. The author says that it is easy to prove the opposite inequality. How can I manage to do that?
3) The author considers intervals $I$ or $\omega$ and denotes by $I^*$ and $\omega^*$ the double of $I$ and $\omega$ respectively. What is exactly the double of an interval?
Thank you for your answers.
 A: For your first question, as Fefferman notes, it suffices to assume that $n(x)$ is bounded assuming the constants in the final result (inequality (1)) do not depend on this bound. This is equivalent to saying that it suffices to prove inequality (1) for trigonometric series of length $M$, as long as the constant in (1) doesn't depend on $M$. Now by the uncertainty principle (or, say, Bernstein's inequality) a trigonometric series of length $M$ is nearly constant on  intervals of length $1/M$. Using these observations, it isn't too hard to see that it suffices to consider integer-valued (simple) functions $n(x) : \mathbb{T} \rightarrow [1,2,\ldots,M]$ that are constant intervals of $1/M$ which are clearly measurable.
[For 2) can you post the definitions of $A_0(p)$ and $T_p$?]
A: I have a detailed answer for 1) (a professor of mine helped me with this). Take $M\in\mathbb{N}$ such that $n(x)\leq M$ for all $x\in [0,2\pi]$. Consider the sets 
$$E_1=\{x\in [0,2\pi]:\sup_{N\leq M}|S_Nf(x)|=|S_1f(x)|\},$$
$$E_2=\{x\in [0,2\pi]\backslash E_1:\sup_{N\leq M}|S_Nf(x)|=|S_2f(x)|\},$$
$$ \cdots $$
$$E_M=\{x\in [0,2\pi]\backslash E_{M-1}:\sup_{N\leq M}|S_Nf(x)|=|S_Mf(x)|\}.$$
These sets are measurable, as a consequence of the following result: ''if $A$ is measurable in $\mathbb{R}$ and $F,G:A\rightarrow\mathbb{R}$ are measurable, then $\{x\in A:F(x)=G(x)\}$ is measurable''.
Now notice that $$n=\sum_{j=1}^Mj\,\chi_{E_{j}},$$ which is measurable.
A: 1) This is not necessary for this particular question, but in my experience the Arsenin-Kunugui measurable selection theorem (see Kechris's book "Classical Descriptive Set Theory") is the "one size fits all" solution to measurable selection questions.
3) Double of an interval = interval with the same center and twice the length.
