I shall attach my proof for the statement "Almost Uniform Convergence Imply Convergence in Measure" here. I know this is a lot longer than what it is necessary to be, but I want everything in the proof to be crystal clear to me. I also bolded the part that I am unsure about:
Let $(X, \mathcal{F}, \mu)$ be the measure space. Suppose $f_n \xrightarrow{a.u.} f$. That is, for all $\epsilon > 0$, there exists $E_\epsilon \in \mathcal{F}$ such that $\mu(E_\epsilon) < \epsilon$ and $$ \lim_{n \to \infty} \sup_{x \in X - E_\epsilon}|f_n(x) - f(x)| = 0. $$ We wish to show for all $c > 0$ that $$ \lim_{n \to \infty} \mu(\{ x \in X: |f_n(x) - f(x)| \geq c \}) = 0. $$ To this end, for all $c > 0$ and $n \in \mathbf{N}$, we define $A_c^n = \{ x \in X: |f_n(x) - f(x)| \geq c \}$. Then for all $c > 0$ and $\epsilon > 0$, we must have $A_c^n \cap (X - E_\epsilon) = \emptyset$ for $n$ large enough. To see this, let $\epsilon > 0$. Then we must have some $E_\epsilon \in \mathcal{F}$ such that $\mu(E_\epsilon) < \epsilon$ and $$ \lim_{n \to \infty} \sup_{x \in X - E_\epsilon} |f_n(x) - f(x)| = 0. $$ Therefore, for all $a > 0$, there exists $N(a, \epsilon) \in \mathbf{N}$ such that if $n \geq N(a, \epsilon)$, then $$ \sup_{x \in X - E_\epsilon} |f_n(x) - f(x)| < a. $$ Thus we must have if $n \geq N(a, \epsilon)$, then for all $x \in X - E_\epsilon$ that $$ |f_n(x) - f(x)| \leq \sup_{x \in X - E_\epsilon} |f_n(x) - f(x)| < a. $$ In particular, for all $c > 0$, there exists $N(c, \epsilon) \in \mathbf{N}$ such that if $n \geq N(c, \epsilon)$, then for all $x \in X - E_\epsilon$ that $$ |f_n(x) - f(x)| < c. $$ Thus for all $\epsilon > 0$ and $c > 0$, there exists $N(c, \epsilon) \in \mathbf{N}$ such that if $n \geq N(c, \epsilon)$, then $x \not \in A_c^n$. In particular, this implies for all $\epsilon > 0$ and $c > 0$, there exists $N(c, \epsilon) \in \mathbf{N}$ such that if $n \geq N(c, \epsilon)$, then $A_c^n \subseteq E_\epsilon$. Therefore, we have for all $\epsilon > 0$ and $c > 0$, there exists $N(\epsilon, c) \in \mathbf{N}$ such that if $n \geq N(\epsilon, c)$, then $$ \mu(A_c^n) \leq \mu(E_\epsilon) < \epsilon. $$ Thus, we have for all $c > 0$ and $\epsilon > 0$ that $$ \lim_{n \to \infty} \mu(A_c^n) \leq \epsilon. $$ Thus we have for all $c > 0$ that $$ 0 \leq \lim_{n \to \infty} \mu(A_c^n) = \lim_{n \to \infty} \mu(\{ x \in X: |f_n(x) - f(x)| \geq c \}) \leq 0. $$ In particular, we must have for all $c > 0$ that $$ \lim_{n \to \infty} \mu(\{ x \in X: |f_n(x) - f(x)| \geq c \}) = 0. $$ This proves that $f_n \xrightarrow{\mu} f$ by definition.
My Question:
Is the part that I marked above correct? I am unsure because our $N$ is dependent on $\epsilon > 0$. However, since $\epsilon$ is fixed and given, I think the inequality should be fine by, for example, this result: Comparison for Sequences. How about other parts of the proof? Is the proof correct?