Convergence in Measure Implies Convergence a.e. of Subsequence

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Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $D \in \Sigma$.

Let $f : D \to \R$ be a $\Sigma$-measurable function.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable functions $f_n : D \to \R$ such that:

$\sequence {f_n}_{n \mathop \in \N}$ converges in measure to $f$.

Then there is a subsequence $\sequence {f_{n_k} }_{k \mathop \in \N}$ of $\sequence {f_n}_{n \mathop \in \N}$ that converges a.e. to $f$.


For each $n, k \ge 1$, define:

$\ds B_{n, k} = \set {x \in X : \size {\map {f_n} x - \map f x} > \frac 1 k}$

Since $\sequence {f_n}_{n \mathop \in \N}$ converges in measure to $f$, we have:

$\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \frac 1 k} } = 0$

Then from the definition of convergence, for each $k$ we can select $n_k$ such that:

$\ds \map \mu {\set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \frac 1 k} } < \frac 1 {2^k}$

That is:

$\ds \map \mu {B_{n_k, k} } < \frac 1 {2^k}$

We then have:

\(\ds \sum_{k \mathop = 1}^\infty \map \mu {B_{n_k, k} }\) \(=\) \(\ds \sum_{k \mathop = 1}^\infty \frac 1 {2^k}\)
\(\ds \) \(=\) \(\ds 1\) Sum of Infinite Geometric Progression
\(\ds \) \(<\) \(\ds \infty\)

By the Borel-Cantelli Lemma, we therefore have:

$\ds \map \mu {\limsup_{k \mathop \to \infty} B_{n_k, k} } = 0$

By Equivalence of Definitions of Limit Superior of Sequence of Sets:

$\map \mu {\set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k \text { for infinitely many } k} } = 0$


$N = \set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k \text { for infinitely many } k}$

Then, for $x \not \in N$, we have:

$\size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k$ holds for only finitely many $k$.

That is, there exists $K \in \N$ such that:

$\size {\map {f_{n_k} } x - \map f x} \le \dfrac 1 k$ for $k \ge K$.

Then from the Squeeze Theorem, we have:

$\size {\map {f_{n_k} } x - \map f x} \to 0$


$\map {f_{n_k} } x \to \map f x$

So if $x \notin N$:

$\map {f_{n_k} } x \to \map f x$

Since $\map \mu N = 0$:

$\sequence {f_{n_k}}_{n_k \mathop \in \N}$ converges a.e. to $f$.