Finite version of Szemerédi implies Furstenberg's theorem? I want to prove that theorem 1 implies theorem 2.
Theorem 1: Let $k \geq 3$ be an integer and let $0<\delta\leq 1$. Then there is a positive integer $N(k,\delta)$ such that for any $N \geq N(k,\delta)$ an arbitrary set $A \subseteq [1,N]$ with $|A|\geq \delta N $ contains an arithmetic progression of lenghth $k$.
Theorem 2: Let $(X,\mathcal{B},\mu)$ a measure space. Let $T$ be a self-map of $X$ preserving the measure $\mu$ and let $k \geq 3$. Then for any measurable set $E$ with $\mu(E)>0$ there is an integer $n >0$ such that $$\mu(E \cap T^{-n}E \cap \dots \cap T^{-(k-1)n}E)>0 $$
I have a proof that is not very clear to me in some steps. Here it is:  
Let $N$ be a sufficiently large positive integer (we shall specify this number below). For any $x \in X$ we consider the set $\Lambda(x)=\{l \in [1,N]: T^{l}x \in E\}$. So we have $$ \int_X |\Lambda(x)|d\mu=N\mu(E) $$
Let $M=\{x \in X: |\Lambda(x)| \geq N\mu(E)/2\}$.It follows that $\mu(M) \geq \mu(E)/2$. Let $N=N(k,\mu(E)/2)$. So, for theorem 1, for any $x \in M$ the set $\Lambda(x)$ contains an arithmetic progression $\{a(x)+b(x)j\}_{j=0}^{k-1}$ of length $k$. Thus, to any point $x \in M$ we have assigned a pair of numbers $(a(x),b(x))$. Now, there is a step that I don't understand:
"Since for any $x \in X$ we have $(a(x),b(x)) \in [1,N]^{2}$, there is a set $M' \subseteq M$ such that $\mu(M')\geq \mu(E)/(2N^{2})$, and to any point of this set we have assigned the same pair $(a,b)$. In this case $\mu(\bigcap_{j=0}^{k-1} T^{-(a+bj)}E)\geq \mu(M')>0$."
Where the subset $M'$ comes from?
 A: (This does not address your question directly but it is too long for a comment.)  
Here is a way to prove that Theorem 1 implies Theorem 2.   First, prove that Theorem 2 is equivalent to Theorem 2':
Theorem 2'.  Let $k\in \mathbb N$. If $(X,\mu,T)$ is a probability measure preserving system and $E\subset X$ has positive measure, then there is an $n\in \mathbb N$ such that $E\cap T^{-n}E\cap \cdots \cap T^{-(k-1)n}E=\varnothing$.
Proof.  Deducing Theorem 2' from Theorem 2 is straightforward, so we only prove that Theorem 2' implies Theorem 2.  We prove the contrapositive -- we prove that if Theorem 2 is false then Theorem 2' is false as well.  So we assume that there is a measure preserving system and a set $E\subset X$ having positive measure such that for all $n\in \mathbb N$, the set $I_n:=E\cap T^{-n}E\cap \cdots \cap T^{-(k-1)n}E$ has measure $0$.  Then $I:=\bigcup_{n\in \mathbb N} I_n$  has measure $0$, and $E':=E\setminus I$ has $\mu(E')=\mu(E)>0$.  We will prove that $I_n':=E'\cap T^{-n}E'\cap \cdots \cap T^{-(k-1)n}E'=\varnothing$ for each $n\in \mathbb N$.  We do so by proving that $I_n'\subset E'\cap I_n,$ which is empty by the definition of $E'$. So assume that $x\in I_n'$.   Then $x\in E'$ and $x\in E$ (since $E\subset E')$, and likewise $T^nx,\dots, T^{(k-1)n}x\in E$, so $x\in I_n$.  Thus $x\in E'\cap I_n$, showing that $I_n'=\varnothing$ for every $n\in \mathbb N$.   This shows that $E'$ is a set of positive measure and $E'\cap T^{-n}E'\cap \cdots \cap T^{-(k-1)n}E'=\varnothing$ for all $n\in \mathbb N$.    $\square$
Second, prove the following lemma. 
Lemma. If $(X,\mu,T)$ is a measure preserving system and $\mu(E)>0$, then there is an orbit $\{T^nx:n\in \mathbb N\}$ such that $E_x:=\{n\in \mathbb N:T^nx\in E\}$ has asymptotic density greater than or equal to $\mu(E)$, meaning $d(E_x):=\lim_{N\to \infty} \frac{|E\cap \{1,\dots,N\}|}{N} \geq \mu(E)$.  
The lemma can be proved by applying the pointwise ergodic theorem to the the characteristic function $1_E$: we have that $F(x):=\lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N f(T^nx)$ exists for $\mu$-almost every $x$, and the dominated convergence theorem implies $\int F\, d\mu=\int 1_E\, d\mu =\mu(E).$  Thus there is a point $x$ where the limit $F(x)$ is greater than or equal to $\mu(E)$.
Now Theorem 2' can be deduced from Theorem 1: assuming $(X,\mu,T)$ is a measure preserving system and $E\subset X$ has positive measure, the Lemma will imply that for some $x$, $|E_x\cap \{1,\dots, N\}|>\mu(E)N/2$ for infinitely many $N$.  Then Theorem 1 implies $E_x$ contains an arithmetic progression $\{a, a+n,a+2n,\dots,a+(k-1)n\}$.  By the definition of $E_x$, this means $\{T^{a}x,T^{a+n}x,\dots,T^{a+(k-1)n}x\}\subset E$, so that $T^{a}x\in E\cap \dots \cap T^{-(k-1)n}E$, meaning the intersection is nonempty. 
