Convergence in measure and almost everywhere

Question

In a finite measure space, let $\{f_{n}\}$ be a sequence of measurable functions. Show that $f_{n} \rightarrow f$ in measure if and only if every subsequence $\{f_{n_{k}}\}$ contains a subsequence $\{f_{n_{k_{j}}}\}$, that converges almost everywhere to f.

I can prove from converge in measure to converge almost everywhere, but I don't know how to write it down for the other direction.

Answer 1

Assume $\{f_n\}$ converges pointwise a.e. to $f$. Fix $\varepsilon > 0$ and define: $$ E_N = \{x \in X \mid \exists n > N : |f_n(x) - f(x)| > \varepsilon\} $$

We have $E_1 \supset E_2 \supset \cdots$ and $\mu(E_1) \le \mu(X) < \infty$. Furthermore, since $\{f_n\}$ converges pointwise a.e. to $f$, we have $\mu\left(\bigcap_{N \in \mathbb{N}} E_N\right) = 0$.

Hence, $\lim_{N \to \infty} \mu(E_N) = 0$ and we can find an $N$ for which $\mu(E_N) < \varepsilon$.

It's clear that: $$ \forall n > N : \{x \in X \mid |f_n(x) - f(x)| > \varepsilon\} \subset E_N $$

Hence, $\{f_n\}$ converges to $f$ in measure.

Answer 2

If not, then for some $\varepsilon>0$ and c>0, there exist a subsequence of $f_n(x)$, for example $f_{n_k}(x)$, which satisfies $\forall n_k$,
$m\{x \in X : |f_{n_k}(x) - f(x)| > \varepsilon\}>c $ (*)

Since $f_{n_k}(x)$ has a subsequence $g_n(x)$, which converges almost everywhere to $f(x)$, and that the problem is discussed in a finite measure space, we can say $g_n(x)$ converges to $f(x)$ in measure. However, that contradicts the existence of $f_{n_k}(x)$. $g_n(x)$ is a subsequence of $f_{n_k}(x)$, but it doesn't follow (*).

Answer 3

I'll use the hint from Kolmogorov's book to suggest two possible proofs.

$\textbf{Hint}$. Let $\left \lbrace \delta_n \right \rbrace$ be a sequence of positive numbers such that \begin{align*} \lim_{n\rightarrow \infty}\delta_n=0, \end{align*} and let $\left \lbrace \epsilon_n \right \rbrace$ be a sequence of positive numbers such that \begin{align*} \sum_{n=1}^{\infty} \epsilon_n < \infty . \end{align*} Let $\left \lbrace n_k \right \rbrace$ be a sequence of positive integers such that $n_k>n_{k-1}$ and \begin{align*} \mu \left \lbrace x: \left| f_{n_k}(x) - f(x) \right| \geq \delta_k\right\rbrace < \epsilon_k \qquad (k=1,2,...). \end{align*}

Moreover, let

\begin{align*} R_i = \bigcup_{k=i}^{\infty} \left \lbrace x: \left| f_{n_k}(x) - f(x) \right| \geq \delta_k \right\rbrace, \qquad Q=\bigcap_{i=1}^{\infty}R_i. \end{align*}

Then $\mu(R_i)\rightarrow\mu(Q)$ as $i\rightarrow \infty$, since $R_1\supset R_2 \supset \cdot\cdot\cdot$. On the other hand, \begin{align*} \mu(R_i) < \sum_{n=1}^{\infty} \epsilon_n, \end{align*} and hence $\mu(R_i)\rightarrow 0$, so that $\mu(Q)=0$.

Now show that $\left\lbrace f_{n_k} \right\rbrace$ converges to $f$ on $E-Q$.

$\textbf{Alternative 1}$

Let $B_k=\left\lbrace x: \left| f_{n_k}(x) - f(x) \right| \geq \delta_k \right\rbrace$.

Note that \begin{align*} \forall x\in (E-Q) \implies x \notin Q \implies x \notin \bigcap_{i=1}^{\infty}R_i \implies x \notin R_i \, \text{ for some } i \implies x \notin B_k \, \text{ for } k \geq i . \end{align*}

In fact this implies that there is no $k\geq i$ for which $\left| f_{n_k}(x) - f(x) \right|\geq \delta_k$ holds in $E-Q$, since otherwise it would be contained in every $R_i$ and hence in $Q$.

So we have that in $E-Q$ there is no $k\geq i $ such that $\left| f_{n_k}(x) - f(x) \right| \geq \delta_k$ implying that the opposite is true, i.e., for all $k\geq i$, $\left| f_{n_k}(x) - f(x) \right| < \delta_k$, where $\delta_k \rightarrow 0$ as $k\rightarrow \infty$ (by its definition). So we have that $f_n(x)\rightarrow f(x), \, \forall x \in (E-Q)$, and finally that $f_n(x)\rightarrow f(x) \, a.e.$ since the set $Q$ has measure zero.

$\textbf{Alternative 2}$

Let $B_k=\left\lbrace x: \left| f_{n_k}(x) - f(x) \right| \geq \delta_k \right\rbrace$.

We have \begin{align*} (E-Q) &= E-\bigcap_{i=1}^{\infty}R_i=\bigcup_{i=1}^{\infty}(E-R_i)=\bigcup_{i=1}^{\infty}R_i^{\complement} \\ &=\bigcup_{i=1}^{\infty}\left( \bigcup_{k=i}^{\infty}B_k\right)^\complement \bigcup_{i=1}^{\infty}\bigcap_{k=i}^{\infty}B_k^{\complement} \end{align*}

\begin{equation} (E-Q)=\bigcup_{i=1}^{\infty}\bigcap_{k=i}^{\infty} \left\lbrace x: \left| f_{n_k}(x) - f(x) \right| < \delta_k \right\rbrace. \end{equation}

From the last equation, we can read that in this set there exists an integer $i$ such that for all $k\geq i$, for which $\left| f_{n_k}(x) - f(x) \right| $ is less than $\delta_k$ which get arbitrarily small, implying the convergence of $f_{n_k}$ to $f$ on $(E-Q)$.

Answer 4

"Convergence almost everywhere implies convergence in measure (for a finite measure)" follows almost trivially from the following lemma (e.g., lemma 2.5.4 of the Ash and Dade text):

Suppose $\mu$ is a finite measure. Then $f_n \to f$ pointwise $\mu$-a.e. if and only if $$ \forall \epsilon > 0, \lim_{n \to \infty} \mu( \bigcup_{n \geq k} B_{k \epsilon}) = 0, $$ where $B_{k\epsilon}$ is the set where the convergence is violated by the amount $\epsilon$ after index $k$, $$ B_{k \epsilon} := \{x: |f_k(x) - f(x)| > \epsilon \}. $$

(we actually only need the $\implies$ direction).

Now to prove convergence in measure, fix any $\epsilon > 0$, then $$ \lim_{n \to \infty} \mu(\{x: |f_n(x) - f(x)| > \epsilon \}) = \lim_{n \to \infty} \mu(B_{n\epsilon}) \leq \lim_{n \to \infty} \mu(\bigcup_{n \geq k} B_{k \epsilon}) = 0 $$ since $B_{n \epsilon} \subset \bigcup_{n \geq k} B_{k \epsilon}$. That's it!