Completeness of convergence in measure on a finite measure space ($d(f,g)=\int \min \{1,|f-g|\}d\mu ,\mu (E)<\infty$)

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, $d(f_n,f)\le \epsilon \mu (E)+\mu (|f_n-f|>\epsilon )$ for $0<\epsilon <1$, while $\mu (|f_n-f|>\epsilon )\le d(f_n,f)/\min \{1,\epsilon \}$. 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.