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The Cauchy assumption lets us choose $n_k$ recursively, increasing strictly, so that $\mu (|f_n-f_m|>2^{-k})<2^{-k}$ for all $m,n\ge n_k$. In particular the measurable eventshave summable measures. The measure of ${\bigcup }_{k\ge K}{E}_k$ is at most ${\sum }_{k\ge K}{2}^{-k}\to 0$, so $\mathrm{limsup}{E}_k$ is a null set. Outside it, the successive differences are eventually bounded by a summable geometric series. Hence $f_{n_k}$ is pointwise Cauchy and has a finite real limit.

Define $f$ to be that limit wherever the sequence converges, and zero elsewhere. The convergence set is measurable by the countable Cauchy criterion, and the limit is measurable as a limit of measurable functions on that set. Thus there is no need to assume the measure is complete. Part (a)(ii), or the corresponding finite-measure implication, gives $f_{n_k}\to f$ in measure.

For the full sequence useChoose a sufficiently late subsequence index so the second term is small; the Cauchy property then bounds the first uniformly for all large $n$. Therefore $f_n\to f$ in measure. This proves completeness of convergence in measure on a finite measure space.