Let $g: [0, \infty) \rightarrow \mathbb{R}$ be continuous, increasing, and bounded with $g(0)= 0$ and $g(x) > 0$ for $x > 0$. Let $(X, \mathcal{A}, \mu)$ be a finite measure space and $f_{n}: X \rightarrow \mathbb{C}$ measurable functions. Show that $f_{n} \rightarrow 0$ in measure (what does this mean?) $\iff$ $\int g\circ |f_{n}| d\mu \rightarrow 0$.
Attempt: ($\implies$) As $f_{n} \rightarrow 0$ in measure, for a.e. $x \in X$ and for any $\epsilon > 0$ there is $N > 0$ with $|f_{n}(x)| < \epsilon$ when $n > N$. I think I can say that $\mu(\{|f_{n}| > \epsilon\}) = 0$. Now for any $a \in [0, \infty)$ and $\epsilon > 0$ there is $\delta > 0$ so that $|g(x)| < \epsilon$ when $|a| < \delta$. I want to say something like $\mu(\{ |f_{n}| > \delta \}) \rightarrow 0$ as $\delta \rightarrow 0$. In this case, I want to do something like
\begin{align} \int g\circ |f_{n}| d\mu &= \int_{|f_{n}| > \delta} g\circ |f_{n}| d\mu + \int_{|f_{n}| < \delta} g\circ|f_{n}| d\mu \\ &\leq M\mu(\{|f_{n}| > \delta\}) + \epsilon\mu(\{|f_{n}| < \delta\}) \\ &\longrightarrow 0 + 0 = 0 \end{align}
as $\delta, \epsilon \rightarrow 0$ (here, $M$ is the bound on $g$). I'd like some guidance on this argument to clean it up, or if I'm even going in the right direction.
($\Longleftarrow$) My idea is to show the contrapositive by using a very similar argument as in ($\implies$). So I'd assume $f_{n} \rightarrow f \neq 0$ (but possibly $\infty$?) and hence $\int g\circ |f_{n}| d\mu \rightarrow \int g\circ |f| d\mu \neq 0$. But I am more stuck on this part than the last.
$\implies$ Let $M$ denote a bound on $g$. Let $\epsilon >0$ be fixed for the moment.
Note that $$\begin{align}0\leq \int g(|f_n|) d\mu &= \int g(|f_n|)1_{|f_n|\geq \epsilon} d\mu + \int g(|f_n|)1_{|f_n|< \epsilon} d\mu\\ &\leq M \mu(|f_n|\geq \epsilon)+g(\epsilon)(\mu(X)-\mu(|f_n|> \epsilon))\end{align}$$ Taking $\limsup_n$ on both sides yields $$0\leq \limsup_n \int g(|f_n|) d\mu \leq g(\epsilon)\mu(X)$$ Now, letting $\epsilon \to 0$ and using the continuity of $g$ yields $0\leq \limsup_n \int g(|f_n|) d\mu \leq 0$, hence $\limsup_n \int g(|f_n|) d\mu=0$. Since $\int g(|f_n|) d\mu \geq 0$, this implies that $\lim_n \int g(|f_n|) d\mu = 0$.
$\Longleftarrow$ Let $\epsilon >0$. Since $\displaystyle \int g(|f_n|) d\mu = \int g(|f_n|)1_{|f_n|\geq \epsilon} d\mu + \int g(|f_n|)1_{|f_n|< \epsilon} d\mu$, $$\displaystyle \int g(|f_n|) d\mu \geq \int g(|f_n|)1_{|f_n|\geq \epsilon} d\mu\geq g(\epsilon) \mu(|f_n|\geq \epsilon)$$ Hence $\mu(|f_n|\geq \epsilon)\to 0$.