Real Analysis, Folland Proposition 2.30 Modes of Convergence 
Proposition 2.30 - Suppose that $\{f_n\}$ is Cauchy in measure. Then there is a measurable function $f$ such that $f_n\rightarrow f$ in measure, and there is a subsequence $\{f_{n_j}\}$ that converges to $f$ a.e. Moreover, if also $f_n\rightarrow g$ in measure, then $g = f$ a.e.

Attempted proof - Suppose $\{f_n\}$ is Cauchy in measure. This means for every $\epsilon > 0$, $$\mu\left(\{x:|f_n(x) - f_m(x)|\geq \epsilon\}\right)\rightarrow 0 \ \ \text{as} \ \ m,n\rightarrow \infty$$ Thus we can choose a subsequence $\{g_j\} = \{f_{n_j}\}$ from $\{f_n\}$ such that if $E_j = \{x:|g_j(x) - g_{j+1}(x)| \geq 2^{-j}\}$, then $\mu(E_j)\leq 2^{-j}$. Let $F_k = \bigcup_{1}^{\infty}E_k$, then $\mu(F_k)\leq \sum_{j=k}^{\infty}\mu(E_j) \leq \sum_{j=k}^{\infty}2^{-j} = 2^{1-k}$, so we have $\mu(F_k)\leq 2^{1-k}$ by subadditivity of $\mu$. 
I am trying to take Folland's approach and expand upon it a bit to include more detail but after this part I am lost on where and why he goes where he goes. Any suggestions or alternative ways of proving this is greatly appreciated.
 A: Folland's book is very tersed, and this Theorem 2.30 is a good exemple of such tersed style.

Theorem 2.30 - Suppose that $\{f_n\}$ is Cauchy in measure. Then there is a measurable function $f$ such that $f_n\rightarrow f$ in measure, and there is a subsequence $\{f_{n_j}\}$ that converges to $f$ a.e. Moreover, if also $f_n\rightarrow g$ in measure, then $g = f$ a.e.

This theorem actually contains FIVE (relevant) "propositions":

P1. Suppose that $\{f_n\}$ is Cauchy in measure. Then there is a subsequence $\{f_{n_j}\}$ which is Cauchy a.e.
P2. Suppose that $\{f_n\}$ is Cauchy a.e.. Then there is a measurable function $f$ such that $f_n\rightarrow f$ a.e.
P3. Suppose that $\{f_n\}$ is Cauchy in measure and converges to $f$ a.e.. Then $\{f_n\}$ converges to $f$ measure.
P4. Suppose that $\{f_n\}$ is Cauchy in measure and there is a subsequence $\{f_{n_j}\}$ which converges to $f$ in measure. Then $\{f_n\}$ converges to $f$ measure.
P5. Suppose that $\{f_n\}$ converges in measure to $f$ and $g$. Then $g=f$ a.e.

Folland's proof is actually a concatenation of tersed proofs of those five propositions.
Let us prove each of those propositions individually and then, let us see how they combine to prove Theorem 2.30.
Proof of P1:
Suppose $\{f_n\}$ is Cauchy in measure. This means for every $\epsilon > 0$, $$\mu\left(\{x:|f_n(x) - f_m(x)|\geq \epsilon\}\right)\rightarrow 0 \ \ \text{as} \ \ m,n\rightarrow \infty$$
Thus we can choose a subsequence $\{f_{n_j}\}$ from $\{f_n\}$ such that
$$\mu\left ( \left \{x:|f_{n_j}(x) - f_{n_{j+1}}(x)| \geq \frac{1}{2^{j}}\right \} \right ) \leq  \frac{1}{2^{j}}$$
Let, for $k \in \mathbb{N}$ and $k\geq 1$,
$$F_k = \bigcup_{j=k}^{\infty}\left \{x:|f_{n_j}(x) - f_{n_{j+1}}(x)| \geq  \frac{1}{2^{j}}\right \}$$
then
$$\mu(F_k)\leq \sum_{j=k}^{\infty}\mu\left (\left \{x:|f_{n_j}(x) - f_{n_{j+1}}(x)| \geq  \frac{1}{2^{j}}\right \}\right) \leq \sum_{j=k}^{\infty}\frac{1}{2^{j}} = \frac{1}{2^{k-1}}$$ so we have $\mu(F_k)\leq \frac{1}{2^{k-1}}$.
Let
$$ F=\bigcap_{k=1}^\infty F_k $$
Since, for all $k\geq 1 $, $F \subset F_k$, we have that, for all $k\geq 1 $, $\mu(F) \leq \frac{1}{2^{k-1}}$. So we have that $ \mu(F) =0$.
Now if $x \notin F $, then there is $k$ such that $x \notin F_k$. Given $\epsilon>0$, let $M\in \mathbb{N}$ such that $M\geq k$ and $\frac{1}{2^{M-1}} <\epsilon$. For all $i, j \in \mathbb{N}$, if $i\geq j \geq M$, then
$$|f_{n_j} -f_{n_i}| \leq \sum_{l=j}^{i-1} |f_{n_l} -f_{n_{l+1}}|\leq  \sum_{l=j}^{i-1}\frac{1}{2^l}\leq \frac{1}{2^{j-1}}\leq \frac{1}{2^{M-1}} \leq \epsilon$$
So the subsequence $\{f_{n_j}\}$ is Cauchy a.e.
Proof of P2:
Suppose that $\{f_n\}$ is Cauchy a.e.. The there is a measurable set $F$, such that $\mu(F)=0$ and $\{f_n\}$ is pointwise Cauchy on $F^c$. That means for every $x \notin F$, $f_n(x)$ is a Cauchy sequence in $\mathbb{C}$, and so is convergent.
For every $x \notin F$, let us define $f(x)$ to be $\lim_{n\to \infty}f_n(x)$. If $x \in F$, let us define $f(x)=0$.
Then we have that $f_n \to f$ a.e  and  $f$ is measurable.
(Remark: Note that $F$ is a measurable set, so $F^c$ is a measurable set as well. We have that $f|_{F^c}$ is the pointwise limit of measurable functions $f_n|_{F^c}$, so $f|_{F^c}$ is a measurable function. Also, it is easy to see that $f|_{F}$ (which is identically zero) is a measurable function. So $f$ is measurable.)
Proof of P3:
Suppose that $\{f_n\}$ is Cauchy in measure and converges to $f$ a.e.. Since  $\{f_n\}$ is Cauchy in measure, we have that, for every $\epsilon > 0$ and $\delta>0$, there is $N\in \mathbb{N}$, such that, for all $n,m \geq N$ $$\mu\left(\{x:|f_n(x) - f_m(x)|\geq \epsilon/2\}\right)<\delta$$
Note that, given any $n\geq N$, if $|f_n(y) - f(y)|\geq \epsilon$, then there is $M\geq N$ such that for all $m\geq M$, $|f_n(y) - f_m(y)|\geq \epsilon/2$.
In other words: given any $n\geq N$, if $y\in\{x: |f_n(x) - f(x)|\geq \epsilon\}$, then $y\in \bigcup_{M=N}^\infty\bigcap_{m=M}^\infty \{x:|f_n(x) - f_m(x)|\geq \epsilon/2\}$.
Yet in other words: given any $n\geq N$,
$$ \{x: |f_n(x) - f(x)|\geq \epsilon\} \subset \bigcup_{M=N}^\infty\bigcap_{m=M}^\infty \{x:|f_n(x) - f_m(x)|\geq \epsilon/2\}$$
Note that $\{\bigcap_{m=M}^\infty \{x:|f_n(x) - f_m(x)|\geq \epsilon/2\}\}_M$ is a monotone non-decreasing sequence of measurable sets and for any any $M\geq N$
$$\mu \left (\bigcap_{m=M}^\infty \{x:|f_n(x) - f_m(x)|\geq \epsilon/2\} \right)\leq \mu(\{x:|f_n(x) - f_M(x)|\geq \epsilon/2\})\leq \delta$$
So using the continuity of the measure from below, we have
$$ \mu(\{x: |f_n(x) - f(x)|\geq \epsilon\}) \leq \mu\left( \bigcup_{M=N}^\infty\bigcap_{m=M}^\infty \{x:|f_n(x) - f_m(x)|\geq \epsilon/2\} \right) =\\=\lim_{M \to \infty} \mu\left(\bigcap_{m=M}^\infty \{x:|f_n(x) - f_m(x)|\geq \epsilon/2\} \right)\leq \delta$$
So, we proved that,
for every $\epsilon > 0$ and $\delta>0$, there is $N\in \mathbb{N}$, such that, for all $n \geq N$ $$\mu\left(\{x:|f_n(x) - f(x)|\geq \epsilon\}\right)<\delta$$
So, $\{f_n\}$ converges to $f$ measure.
Proof of P4:
It is an imediate consequence of
$$\{x:|f_n(x) - f(x)|\geq \epsilon\} \subset \{x:|f_n(x) - f_{n_j}(x)|\geq \epsilon/2\} \cup  \{x:|f_{n_j}(x) - f(x)|\geq \epsilon/2\}$$
The set $\{x:|f_n(x) - f_{n_j}(x)|\geq \epsilon/2\}$ can have arbitraly small measure, because $\{f_n\}_n$ is Cauchy in measure.
The set $ \{x:|f_{n_j}(x) - f(x)|\geq \epsilon/2\}$ can have arbitraly small measure, because the subsequence $\{f_{n_j}\}$ converges to $f$ in measure.
Proof of P5:
Suppose that $\{f_n\}$ converges in measure to $f$ and $g$. Then, for any $\epsilon>0$,
$$ \{x : |f(x)-g(x)|\geq \epsilon \} \subset \{x:|f(x) - f_n(x)|\geq \epsilon/2\} \cup  \{x:|f_n(x) - g(x)|\geq \epsilon/2\}$$
Given any $\delta >0$ for $n$ sufuciently large we have
$$\mu(\{x:|f(x) - f_n(x)|\geq \epsilon/2\})\leq \delta/2 $$
and
$$\mu(\{x:|f_n(x) - g(x)|\geq \epsilon/2\})\leq \delta/2 $$
So, for any $\epsilon>0$, for all $\delta>0$
$$\mu( \{x : |f(x)-g(x)|\geq \epsilon \} )\leq \delta$$
So, for any $\epsilon>0$, $$\mu( \{x : |f(x)-g(x)|\geq \epsilon \} )= 0$$
But $$ \{x : |f(x)-g(x)|\geq 0 \}= \bigcup_{k=1}^\infty \{x : |f(x)-g(x)|\geq 1/k\}$$
So, $\mu(\{x : |f(x)-g(x)|\geq 0 \})=0$. So $f=g$ a.e.
Proof of Theorem 2.30: (following the logic in proof presented Folland's book)
By P1, there is a subsequence $\{f_{n_j}\}$ which is Cauchy a.e.
By P2, since $\{f_{n_j}\}$ which is Cauchy a.e., there is a measurable function $f$ such that $f_{n_j}\rightarrow f$ a.e.
Since  $\{f_{n_j}\}$  is a subsequence of  $\{f_n\}$, we have that  $\{f_{n_j}\}$  is Cauchy in measure and since  $f_{n_j}\rightarrow f$ a.e., by P3, we have that  $\{f_{n_j}\}$ converges to $f$ measure.
By P4, since $\{f_n\}$ is Cauchy in measure and the subsequence $\{f_{n_j}\}$
converges to $f$ in measure, we have that $\{f_n\}$ converges to $f$ measure.
By P5, we have that the limit for the convergence in measure is unique a.e.
Remark: In thersed proof, Follands concatenate the proofs of those five points. In particular, to prove P3, he uses estimates made for P1, which helps to make the proof shorter.
A: The interest of having the $2^{-k}$ upper bound is that you can sum it, and then use Borel-Cantelli...
