Convergence in measure ($\mu (|f_n-f|>\epsilon )\to 0(\epsilon >0)$)

A sequence of real measurable functions converges in measure to $f$ when the displayed condition holds for every positive threshold. It controls the measure of the exceptional set at each index, rather than requiring eventual closeness on each sample path. On a probability space it is exactly convergence in probability. On a finite measure space, almost-everywhere convergence implies convergence in measure by continuity of measure on the decreasing tail events. The converse need not hold for the full sequence, but convergence in measure always admits an almost-everywhere convergent subsequence.

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.