Theorem 24.6.6.label Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a complete metric space, then:
- (1)
For any $\seq{f_n}\subset L^{0}(X; Y)$ that is Cauchy in measure, there exists $f \in L^{0}(X; Y)$ and a subsequence $\seq{n_k}$ such that $f_{n_k}\to f$ almost everywhere and in measure.
- (2)
$L^{0}(X; Y)$ equipped with the uniform structure of convergence in measure is complete.
Proof, [[Theorem 2.30, Fol99]. ] (1): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
In this case, for any $K \in \natp$ and $j \ge k \ge K$,
By monotonicity and subadditivity,
so
By the First Borel-Cantelli Lemma,
Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges almost everywhere to a Borel measurable function $f \in L^{0}(X; Y)$.
Finally, for each $K \in \natp$,
so $f_{n_k}\to f$ in measure as well.
(2): Since the uniform structure of convergence in measure on $L^{0}(X; Y)$ is defined by the Ky Fan metric, completeness follows from (1) and Proposition 8.1.4.$\square$
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