# Instance where local convergence in measure implies global convergence in measure

Let $(X, \mathscr{A},\mu)$ be a measure space. Let $\gamma\subset L^0$ and suppose that for each $\epsilon>0$ and $\epsilon'>0$, there exists an $A\in\mathscr{A}$ with $\mu(A)<\infty$ such that for each $f \in \gamma$, we have that $\mu(A^c \cap \{|f| \geq \epsilon\})<\epsilon'$. Further, suppose that $f_n \rightarrow f$ in measure locally, and that $f, f_n \in \gamma$ for all $n$. Then I want to show that $f_n \rightarrow f$ in measure globally.

I know that since $f_n \rightarrow f$ in measure locally, then for any measurable $A$ with $\mu(A)<\infty$, there is an $N\in\mathbb{N}$ such that $\mu((A \cap \{|f_n-f| \geq \epsilon\})<\epsilon'$ for $n\geq N$. I need to show that convergence in measure is global, so I need to show that $N\in\mathbb{N}$ such that $\mu(\{|f_n-f| \geq \epsilon\})<\epsilon'$ for $n\geq N$. I know that $$\{|f_n-f| \geq \epsilon\}=\{X \cap|f_n-f| \geq \epsilon\}=\{(A \cup A^c) \cap|f_n-f| \geq \epsilon\}=$$ $$\{A\cap|f_n-f| \geq \epsilon\}\cup\{A^c \cap|f_n-f| \geq \epsilon\}$$

I can make the measure of the first set in the union small by the fact that we have local convergence in measure. For the second set in the union, I need to use the hypothesis that $\mu(A^c \cap \{|f| \geq \epsilon\})<\epsilon'$ but I am struggling with this part. I need there to be $|f_n-f|$ instead of $|f|$, which means that I will need to show $f_n -f \in \gamma$, but I am not sure how to do that or if I am going in the right direction. Any ideas on how I should proceed?

Hint: $$\{|f_n-f| \geq \varepsilon\} \subseteq \left\{|f| \geq \frac{\varepsilon}{2} \right\} \cup \left\{|f_n| \geq \frac{\varepsilon}{2}\right\}.$$