= Completeness of convergence in measure on a finite measure space
{title2=$d(f,g)=\int\min\{1,|f-g|\}\,d\mu,\qquad\mu(E)<\infty$}
Modulo equality almost everywhere, the displayed metric induces <convergence in measure> and is complete. A sequence Cauchy in measure has a strictly increasing subsequence $f_{n_k}$ with $\mu(|f_{n_{k+1}}-f_{n_k}|>2^{-k})<2^{-k}$. Subadditivity makes the limsup of these exceptional sets null, so the subsequence has summable pointwise successive differences almost everywhere. It therefore has a finite measurable limit. Almost-everywhere convergence implies convergence in measure for finite measure, and the triangle inequality plus the original Cauchy property extends convergence to the full sequence.
For the metric equivalence, $d(f_n,f)\leq\varepsilon\mu(E)+\mu(|f_n-f|>\varepsilon)$ for $0<\varepsilon<1$, while $\mu(|f_n-f|>\varepsilon)\leq d(f_n,f)/\min\{1,\varepsilon\}$. The same estimates compare the two Cauchy properties. Define the subsequence limit as zero where no finite limit exists; that exceptional set is measurable by the countable Cauchy criterion, so completeness of the underlying measure is unnecessary.
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