Convergence in $\mathcal{M}$ implies convergence in measure on sets of finite measure Let $(\Omega,\mathcal{A},\mu)$ be a $\sigma$-finite measure space such that $\Omega= \bigcup_{n=1}^\infty A_n$ where $\mu(A_n)<\infty$ for all $n \in \mathbb{N}$. Denote by $\mathcal{M}$ the set of all scalar valued $\mu$-measurable functions on $\Omega$ that are finite $\mu$-a.e. I know that $d(f,g)= \sum_{n=1}^{\infty} \frac{1}{2^n} \cdot \frac{1}{\mu(A_n)} \cdot \int_{A_n}\frac{|f-g|}{1+|f-g|}d\mu$ defines a metric in $\mathcal{M}$ and I want to prove that if $d(f_m,f) \rightarrow 0$, then $f_m \rightarrow f$ in measure on sets of finite measure (I have already proved the converse).
My attempt
We assume that $d(f_m,f) \rightarrow 0$ as $m \rightarrow + \infty$.
Let $\varepsilon>0$ and $D \subset \Omega$ have finite measure.
We can write $\{x \in D:|f_m(x)-f(x)| \geq \varepsilon\}= \bigcup_{n=1}^{\infty}(A_n \cap \{x \in D:|f_m(x)-f(x)|\geq \varepsilon\})$ so that
\begin{equation}
\begin{split}
\frac{\varepsilon}{1+ \varepsilon} \mu(\{x \in D:|f_m(x)-f(x)| \geq \varepsilon\}) &\leq \sum_{n=1}^{\infty} \frac{\varepsilon}{1+ \varepsilon}\mu(A_n \cap \{x \in D:|f_m(x)-f(x)| \geq \varepsilon\})\\
&= \sum_{n=1}^{\infty} \int_{A_n \cap \{x \in D:|f_m(x)-f(x)| \geq \varepsilon\}} \frac{\varepsilon}{1+ \varepsilon} d\mu\\
&\leq \sum_{n=1}^{\infty} \int_{A_n } \frac{|f_m-f|}{1+ |f_m-f|} d\mu
\end{split}
\end{equation}
where I don't know how to continue due to the lack of the terms $\frac{1}{2^n} \cdot \frac{1}{\mu(A_n)}$. If those were inside the sum, then the assumption $d(f_m,f) \rightarrow 0$ would  give the desired result.
 A: Let $\epsilon,\delta >0$
Then $d(f_m,f) \to 0 \Rightarrow \int_{A_n}\frac{|f_m-f|}{1+|f_m-f|}d\mu \to 0,\forall n \in \Bbb{N}... (*)$
Assume that $A_n$ are chosen to be disjoint.

Indeed,we can make such an assumption, because if we take $B=A_1$ and $B_n=A_n\setminus \bigcup_{2\leq k<n}A_k$ and the metric $r(f,g)$ similar to $d$ with $B_n$ in place of $A_n$,then we can prove, using $(*)$ and the monotonicity of the integral, that $d(f_m,f) \to 0 \Rightarrow r(f_m,f) \to 0$

Also we can assume that $\mu(A_n)>0 ,\forall n \in \Bbb{N}$
Then exists $N \geq 1$ such that $\sum_{n \geq N}\mu(A_n \cap D)<\epsilon$ since $D$ has finite measure.
In general we have  $\{|f_m-f| \geq \delta\} \subseteq \{\frac{|f_m-f|}{1+|f_m-f|}\geq \frac{\delta}{1+\delta}\}$
Thus $$\mu(x \in D:|f_m-f| \geq \delta)=\sum_{n=1}^{\infty}\mu(x \in A_n \cap D:|f_m-f| \geq \delta\})$$ $$ \leq \frac{\delta+1}{\delta}\sum_{n=1}^{N-1} \int_{A_n \cap D} \frac{|f_m-f|}{1+|f_m-f|}d\mu +\sum_{n \geq N}\mu(A_n \cap D)$$ $$< \frac{\delta+1}{\delta}\sum_{n=1}^{N-1} \int_{A_n \cap D} \frac{|f_m-f|}{1+|f_m-f|}d\mu+\epsilon$$
Thus $\limsup_m \mu(x \in D:|f_m-f| \geq \delta)\leq \epsilon$
Since $\epsilon$ is arbitrary we conclude that $\lim_m \mu(x \in D:|f_m-f| \geq \delta)=0$
