searching help for understanding the proof to Kolmogorov's Theorem I am currently working on chapter 2 Section 3* p. 59 in Katznelsons "Introduction to harmonic analysis " trying to understand the nice proof of  Kolmogorov's theorem: 
"There exists a Fourier series diverging everywhere in $ L_1 $ "
My questions are probably basic, but I hope you could help me out in understanding those bits.
Thank you very much in advance 


okay, my thoughts on this section of the proof:
you assume, that there exists some positive integer $ k$ with measure $ \mu_k $  with property 3.11. So the supremum of the partial sum is bounded below by $k$.
And because of this property there exist an $ N_k \in \mathbb{Z} $ and a set $ E_k $ of normalized Lebesgue measure $ > 1- \frac{1}{k}$ so that 3.12 holds. 
So here you choose just a smaller set which you look at? Do I understand that correctly?
But why choosing especially $E_k$ with measure $1- \frac{1}{k}$?
Then you take the measure $ \mu_k$ and convolute it with the V.P. Kernel to get a trigonometric polynomial. But why does it make sense here to build a convolution to get that partial sum ??
I don't see that.
Then, by Lemma 3.3, E a set of divergence if and only if there exist trigonometric polynomials $ P_j$ in $ L_1 $ with  $ \sum || P_j ||_{L_1} < \infty $ so that $ \sup_j S^*(P_j,t)= \infty\in E$
I don't understand why choosing $P_j= j^{-2} \phi_{2^j} $ leads to that $ E = \cap_m \cup_{m \leq j} E_{2^j} $ is a set of divergence for $L_1 $ 
The next line is saying that because $ E$ is almost all $ T$ the theorem would follow from:

3.5  Theorem.Let $B $ be a homogeneous Banach space on $T$
    Assume  $B\supset C(T)$ ;  then either $T$ is a set of divergence for $B $ or the sets of divergence for $B$ are precisely the sets of measure zero.


Why is here $E$ almost all $ T$.
And why do you still need to show for the proof that the property 3.11 holds  for almost all $t \in T $?
That's quite confusing to me.
Okay, I have also a few questions on constructing those measures $ \mu_k$, but I feel like if I would add those as well it will be a little too much. So I would be super happy if I could get a concept of the ideas until this point, at first.
 A: In (3.11) you don't need the property "for almost all $t\in\mathbb{T}$", you only need measures such that "there exists an integer $N_\kappa$ and a set of ... ", but it happens that you can just prove the stronger statement "for almost all $t\in\mathbb{T}$" and the deduce the quantitative statement involving $N_\kappa$, $E_\kappa$ and so on. 
If you don't like $|E_\kappa|>1-\frac{1}{\kappa}$, then choose your favorite sequence $c_\kappa$ and let's see what happens. You want $E_\kappa$ to be almost all of $\mathbb{T}$, so why not to take an increasing sequence $c_\kappa$ such that $c_\kappa\to 1$ as $\kappa \to \infty$. Now you want to construct those polynomials $P_j$ that behave bad in a large set, $\textit{i.e.}\enspace\sup_j S^*(P_j,t) = \infty$ for every $t\in E$. You don't have polynomials, but measures $\mu_\kappa$ that behave bad. 
Recall that $(\mu*K)^\wedge(j) = \hat{\mu}(j)\hat{K}(j)$, so if $\hat{K}(j)$ is supported in $n\le j\le m$, the same holds for $(\mu*K)^\wedge(j)$. This is a standard way to "smooth" a function, to pass from a possibly very rough function $\mu$ to a nicer one $\mu*K$, and you will see it repeatedly in harmonic analysis. If you know the properties of $K$, then you get a very good control on the convolution, and you can claim things like $\lVert\varphi_\kappa\rVert_{L^1} \le 3$ in the situation at hand.
You have then polynomials $\varphi_\kappa$ and to apply Theorem 3.3 you need $\sum_\kappa\lVert\varphi_\kappa\rVert_{L^1} < \infty$, but this is not necessarily true, then you adjust the polynomials by taking $P_j = j^{-1000}\varphi_j$ to be completely sure that the series converges, or you simply use $P_j = j^{-2}\varphi_j$.
Now we have $S^*(P_j,t)>j^{-2}j = j^{-1}$ in a set $E_j$, but now the problem is that $\sup_j j^{-1}$ is finite, we cannot apply Lemma 3.3, so let's adjust $P_j$ again and say that $P_j = j^{-2}\varphi_{j^3}$. We have that $\sup_j S^*(P_j,t) = \infty$ if $t\in E_{j^3}$ for inifinitely many $j$, hence $\sup_j S^*(P_j,t) = \infty$ in $E := \cap_m\cup_{m\le j} E_{j^3}$ and $E$ is a set of divergence by Lemma 3.3.
To apply Theorem 3.5 I want $E$ to be large, why not of full measure? By the way, what is the measure of $E$? The sequence of sets $\cup_{m\le j} E_{j^3}$ is nested and decreasing, so $|E| = \lim_{m\to\infty}|\cup_{m\le j} E_{j^3}| \ge \lim_{m\to\infty}|E_{m^3}|\ge \lim_{m\to\infty}c_{m^3}= 1$. We can now apply Theorem 3.5.
The author has to choose values for the different quantities appearing in the proof, but sometimes the numerical value is not essential.
