Uniqueness of convergence in measure

Let $(X,\mathcal{M},\mu)$ be a measure space. If $\mu(X)<\infty$, then the metric $$\rho(f,g)=\int \frac{|f-g|}{1+|f-g|}$$ forms a metric on the space of measurable functions $f,g:X\rightarrow \mathbb{C}$ when we equivalence out a.e. equal function. Then we can show that $f_n \rightarrow f$ in $\rho$ if and only if $f_n \rightarrow f$ in measure.

Let $f_n \rightarrow f$ in measure over the space of measurable functions with a.e. equal functions equivalenced. Can I show that is $f$ unique even when $\mu(X)=\infty$? (By uniqueness, I mean $f_n\rightarrow f,g$ where $f = g$ a.e. is false)

Answer

Let $(X, \mathcal M, \mu)$ be a measure space, $f_n \colon X \to \mathbb C$ for $n \in \mathbb N$, $f,g \colon X \to \mathbb C$ such that $f_n \to f$ and $f_n \to g$ in measure. We will show that $f = g$ $\mu$-a. e.

$\def\abs#1{\left|#1\right|}$Let $k\in \mathbb N$ and $\epsilon > 0$, choose $n \in \mathbb N$ such that
$$\mu\left(\abs{f_n - f} > \frac 1{2k}\right) + \mu\left(\abs{f_n - g} > \frac 1{2k}\right) < \epsilon$$
As
$$\{\abs{f-g}> k^{-1}\} \subseteq \{\abs{f_n - f} > (2k)^{-1}\} \cup \{\abs{f_n -g} > (2k)^{-1}\}$$
by the triangle inequality, we get
$$\begin{align*} \mu(\abs{f-g} > k^{-1}) &\le \mu\left(\abs{f_n - f} > \frac 1{2k}\right) + \mu\left(\abs{f_n - g} > \frac 1{2k}\right)\\ &\le \epsilon \end{align*}$$
As $\epsilon$ was arbitrary, $\mu(\abs{f-g} > k^{-1}) = 0$. Therefore
$$\mu(\abs{f-g} > 0) = \mu \left(\bigcup_k \{\abs{f-g} > k^{-1}\}\right) \le \sum_k \mu(\abs{f-g} > k^{-1}) = 0.$$
That is, $f = g$ $\mu$-a. e.