= Solution
{c}
The Cauchy assumption lets us choose $n_k$ recursively, increasing strictly, so that $\mu(|f_n-f_m|>2^{-k})<2^{-k}$ for all $m,n\geq n_k$. In particular the measurable events
$$
E_k=\{|f_{n_{k+1}}-f_{n_k}|>2^{-k}\}
$$
have summable measures. The measure of $\bigcup_{k\geq K}E_k$ is at most $\sum_{k\geq K}2^{-k}\to0$, so $\limsup E_k$ 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 use
$$
\mu(|f_n-f|>\varepsilon)\leq
\mu(|f_n-f_{n_k}|>\varepsilon/2)+\mu(|f_{n_k}-f|>\varepsilon/2).
$$
Choose a sufficiently late subsequence index so the second term is small; the Cauchy property then bounds the first uniformly for all large $n$. Therefore \b[$f_n\to f$ in measure]. This proves <completeness of convergence in measure on a finite measure space>.
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