We work in the framework of a measurable sapce $X$ with a complete and finite measure $\mu$. We say that a sequence of functions $\left( f_{n}\right)$, defined on a measurable space, converge in measure to a measurable function $f$ if \begin{equation*} \forall \varepsilon >0:\lim_{n\rightarrow +\infty }\mu \left( \left\{ x\in X\mid \left\vert f_{n}\left( x\right) -f\left( x\right) \right\vert \geq \varepsilon \right\} \right) =0, \end{equation*} where $\mu$ is a measure on $X$. We denote this convergence with $f_{n}% \overset{\mu }{\rightarrow }f$. Prove that \begin{equation*} f_{n}\overset{\mu }{\rightarrow }f \quad \Leftrightarrow \quad f_{n}\overset{d}{\rightarrow }f, \end{equation*} where $d$ is a pseudo-metric on measurable space $X$ defined by \begin{equation*} d\left( f,g\right) :=\min \left\{ 1,\inf_{\alpha >0}\mu \left( \left\{ x\in X\mid \left\vert f\left( x\right) -g\left( x\right) \right\vert \geq \alpha \right\} \right) \right\} . \end{equation*}
My attempt was to write and rewrite the definition of the convergences and then, using the characterization with $\varepsilon $ for the convergence of a sequence, to get at least one implication. But I get stuck. It seems that there is no link between this two types of convergence. Thank you for any help.
First suppose that $f_n \overset{\mu}\to f$. We want to show that $d(f_n,f) \to 0$. Clearly, we have $$d(f_n,f) \leq \inf_{\alpha>0} \mu(\{x \in X \mid |f_n(x) - f(x)| \ge \alpha\}) \leq \mu(\{x \in X \mid |f_n(x) - f(x)| \ge 1\}) \to 0$$ by taking $\varepsilon = 1$ in the definition of $f_n \overset{\mu} \to f$, which gives the desired result.
The other implication isn't true because you can attain the $\inf$ by making $\alpha$ large in the definition of $d$, but the definition of convergence in measure is stricter for $\varepsilon$ close to $0$. For example, let $f_n(x) = 1$ almost surely and $f(x) = 0$ almost surely. Then $$\mu(\{x \in X \mid |f_n(x) - f(x)| \geq 2\}) = 0$$ since $|f_n(x) - f(x)| = 1$ almost surely. This means $d(f_n,f) = 0$ for every $n$. However, it is clear the $f_n$ does not converge in measure to $f$ since $$\mu(\{x \in X \mid |f_n(x) - f(x)| \geq \frac12\}) = 1.$$