I'm reading Folland's book, and I had a hard time following one of the proofs. I'm trying to understand the proof for:
Suppose that $\{f_n\} \subset L^1$, and that $\{f_n\}$ is cauchy in measure. Then there is a measurable function $f$ such that $f_n \to f$ in measure, and there is a subsequence $\{f_{n_j}\}$ that converges to $f$ a.e.. Moreover, if also $f_n \to g$ in measure, then $g = f$ a.e..
Here is my attempted proof, in that I modified the written proof in the bits that I couldn't understand. Is this proof correct? (I'm especially worried about the $f^{-1}(B) = \left(\left(\bigcap_{l=k}^\infty \bigcup_{j=l}^\infty g_j^{-1}(B)\right) \cap F^c \right) \cup F \text{ if } 0 \in B$ part, and any error corrections are greatly appreciated.) $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
As $f_n$ is cauchy in measure, we can choose a subsequence $\{g_j\} = \{f_{n_j}\}$ of $\{f_n\}$ such that if $E_j = \{ x:\norm{g_j(x) - g_{j+1}(x)} \ge 2^{-j}\}$, then $\mu(E_j) \le 2^{-j}$. If $F_k = \cup_{j=k}^\infty E_j$, then $\mu(F_k) \le \sum_k^\infty 2^{-j} = 2^{1-k}$. If $x$ is not in $F_k$, for $i \ge j \ge k$ (note that $\norm{g_j(x) - g_{j+1}(x)} \le 2^{-j}$ in this case) $$\norm{g_j(x) - g_i(x)} \le \sum_{l = j}^{i-1} \norm{g_{l+1}(x) - g_l(x)} \le \sum_{l = j}^{i-1} 2^{-1} \le 2^{1-j}. $$ Thus $\{g_j\}$ is pointwise cauchy on $F_k^c$. Let $F = \bigcap_1^\infty F_k = \limsup E_j$. Then as $\mu(F_k) \le 2^{1-k}$, $\mu(F) = 0$. If we set $f(x) = \lim g_j(x)$ for $x \notin F$ and $f(x) = 0$ for $x \in F$, then $f$ is measurable (explained below) and $g_j \to f$ a.e. (as $F$ is a null set).
$x \in F^c$, $x \in \bigcup_{k=1}^\infty F_k^c$, note that $j \ge k$ by definition. So on such $x$, $g_j(x)$ is cauchy, so it converges.
For any borel set $B$, $$f^{-1}(B) = \left(\left(\bigcap_{l=k}^\infty \bigcup_{j=l}^\infty g_j^{-1}(B)\right) \cap F^c \right) \cup F \text{ if } 0 \in B$$ $$f^{-1}(B) = \left(\left(\bigcap_{l=k}^\infty \bigcup_{j=l}^\infty g_j^{-1}(B)\right) \cap F^c \right) \cup \emptyset \text{ if } 0 \notin B$$ For justification, $ a \in \left(\left(\bigcap_{l=k}^\infty \bigcup_{j=l}^\infty g_j^{-1}(B)\right) \cap F^c \right)$ $\iff$ $a \in g_j^{-1}(B)$ for infinitely many $j$. We need only show $g_j^{-1}(B)$ is a measurable set to conclude $f$ measurable. But this is immediate; as $g_j \in L^1$, there exists measurable and integrable $h_j$ and $h_j = g_j$ a.e., say on set $N$. Then $$ g_j^{-1}(B) = (h_j^{-1}(B) \cap N) \cup N_B$$ where $N_B \subset N$, so is a null set in a complete measure space. The measurability of $g_j$ thus follows from measurability of $h_j$.
From $\norm{g_j(x) - g_i(x)} \le 2^{1-j}$ that we proved above we obtain $\norm{g_j(x) - f(x)} \le 2^{1-j}$ for $x \notin F_k$ and $j \ge k$ (As $\norm{\cdot}$ is continuous, we can pass through limits, i.e. $\lim_{i \to \infty}$). $\{ \norm{g_j - f} \ge \epsilon\}$ is no larger than $F_k$ for large enough $k$, but as $\mu(F_k) \to 0$ as $k \to \infty$, we have that $g_j \to f$ in measure. Noting that $$ \{ x: \norm{f_n(x) -f(x)} \ge \epsilon\} \subset \{ x: \norm{f_n(x) - g_j(x)} \ge \frac{\epsilon}{2}\}\bigcup \{ x: \norm{g_j(x) - f(x)} \ge \frac{\epsilon}{2}\}$$ we conclude $f_n \to f$ in measure (As $f_n$ is cauchy in measure, $\{ x: \norm{f_n(x) - g_j(x)} \ge \frac{\epsilon}{2}\}$ has very small measure for large $n,j$ and as $g_j \to f$ in measure, $\{ x: \norm{g_j(x) - f(x)} \ge \frac{\epsilon}{2}\}$ has very small measure for large $j$.).
Finally, suppose $f_n \to g$ in measure. Similar to above, $$ \{ x: \norm{f(x) -g(x)} \ge \epsilon\} \subset \{ x: \norm{f_n(x) - f(x)} \ge \frac{\epsilon}{2}\}\bigcup \{ x: \norm{g(x) - f_n(x)} \ge \frac{\epsilon}{2}\}$$ for all n. Now we have $f_n \to f$ in measure as well, and using a similar process to above, we conclude that $\mu(\{ x: \norm{f(x) -g(x)} \ge \epsilon\}) = 0$ for all $\epsilon$ (by letting $n$ grow large). Letting $\epsilon$ tend to $0$ through some sequence of values, we conclude that $f=g$ a.e..