Function Convergence

In this article we discuss several important convergence properties of a sequence functions, $\{ f_n \}$, defined on the measure space $(X, a, \mu)$.

Definition (Pointwise Convergence) $\{ f_n \}$ converges to $f$ almost every (a.e.) if $\{ f_n \}$ converges to $f$ except on a set of measure zero, i.e, $\exists E \subset X$, s.t.,

$$\mu(X - E) = 0 \, \text{ and }\, \forall x \in E, \lim_n f_n(x) = f(x).$$

Definition (Uniform Convergence) $\{ f_n \}$ converges to $f$ almost uniformly if $\forall \epsilon > 0$. $\exists F \subset X$, s.t, $\mu(X - F) \le \epsilon$ and $\{ f_n \}$ converges uniformly to $f$ on $F$.

The following two theorems discuss the relation between pointwise and almost uniform convergence.

Theorem 1.[Almost Uniform $\rightarrow$ Pointwise] If $\{ f_n \}$ converges to $f$ almost uniformly, then $\{ f_n \}$ converge to $f$ almost everywhere.

Proof: By definition, $\forall n \in \mathbb{N}^+, \exists F_n \subset X, s.t., \mu(X - F_n) \le 1 / n$ and $\{ f_n \}$ converges uniformly to $f$ on $F_n$. But this immediately implies that $\{ f_n \}$ converges pointwise on $F_n$ and therefore it converges pointwise on $\cup_{n = 1}^\infty F_n$.

We claim that $\mu \Big( X - \big( \cup_{n = 1}^\infty F_n \big) \Big) = 0$, and hence $\{ f_n \}$ converges to $f$ almost everywhere. The claim holds as for each $n \in \mathbb{N}^+$,

Taking $n \rightarrow \infty$ finishes the proof.

$\square$

Theorem 2. [Pointwise $\rightarrow$ Almost Uniform][Egorov] Let $\{f_n\}$ be a sequence of measurable functions that converges pointwise to $f$ almost everywhere and suppose $\mu(X) \le \infty$, then $\{ f_n \}$ converges to $f$ almost uniformly.

Proof: Without lost of generality, we assume that $\{ f_n \}$ converge to $f$ everywhere on $X$.

For each $\epsilon > 0$, we need to find a set $F$ such that

1. $\{ f_n \} \rightarrow f$ uniformly on $F$ and,
2. $m(X - F) \le \epsilon$.

For each pair of $k, m \in \mathbb{N}$, consider the set $F_{k, m} = \cap_{n = m}^\infty \{ x : |f_n(x) - f(x)| \le 1 / k \}$. Now fix $k$ and note that $F_{k, m} \subset F_{k, m+1}$.

Next, $\forall x \in E$, $\lim_n f_n(x) = f(x)$. For fix $x$ and $k$, $\exists m > 0$, s.t., for all $n > m, |f_n(x) - f(x)| \le 1/k$. Hence $x \in F_{k, m}$.

We conclude that $F_{k, m}$ is a sequence of monotone increasing sets and $\lim_{m} F_{k,m} = X$. Thus we have $\lim_{m} F_{k,m}) = \mu(X) < \infty$. For arbitrary $\epsilon > 0$, we can find an integer $N_k$, s.t., $\mu(X - F_{k, N_k}) \le \epsilon / 2^k$. Now for convenience we abbreviate $F_{k, N_k}$ as $F_k$.

Observe that $\{ f_n \}$ converges uniformly on $F = \cap_{k = 1}^\infty F_k$, as for any $\lambda > 0, \exists k \in \mathbb{N}, s.t., 1 / k < \lambda$, and $x \in F$ implies $x \in F_k$. Therefore if $n > N_K$, $|f_n(x) - f(x)| \le 1/k \le \lambda$.

It remains to prove $\mu(X - F)$ is small enough. This is because of

$\square$

There are another two frequently used definitions for convergence.

Definition (Converges in measure) $\{ f_n \}$ converges to $f$ in measure if $\forall \epsilon > 0$, $\forall \delta > 0$, $\exists N \in \mathbb{N}$, s.t, for $n > N$, $\mu( \{ x : |f_n (x) - f(x)| \ge \epsilon \} ) < \delta, i.e,$,

$$\lim_{n} \mu( \{ x : |f_n(x) - f(x)| \ge \epsilon \}) = 0.$$

Definition (Cauchy in measure) $\{ f_n \}$ is Cauchy in measure if $\forall \epsilon > 0$, $\forall \delta > 0$, $\exists N \in \mathbb{N}$, s.t, for $n, m > N$, $\mu( \{ x : |f_n (x) - f_m(x)| \ge \epsilon \} ) < \delta, i.e,$,

$$\lim_{n, m} \mu( \{ x : |f_n(x) - f_m(x)| \ge \epsilon \}) = 0.$$

It is easy to see that if $\{ f_n \}$ converge in measure, then it is Cauchy in measure. Later we will show that the converse is also true.

Remark: Converging in measure does not imply converge almost every where. To see this, consider a countable sequence of functions defined on $[0, 1]$.

1. $f_{1, 1} = \mathbf{1}_{[0, 1/2]}$, $f_{1, 2} = \mathbf{1}_{[1/2, 1]}$
2. $f_{2, 1} = \mathbf{1}_{[0, 1/4]}$, $f_{2, 2} = \mathbf{1}_{[1/4, 1/2]}$, $f_{2, 3} = \mathbf{1}_{[1/2, 3/4]}$, $f_{2, 4} = \mathbf{1}_{[3/4, 1]}$
3. ....
4. $f_{n, 1} = \mathbf{1}_{ [0, 1/2^n] }$, ..., $f_{n, i} = \mathbf{1}_{ [(i - 1) / 2^{n}, i /2^n] }$, ..., $f_{n, n} = \mathbf{1}_{ [1 - 1/2^{n}, 1] }$
5. ...

We can list the functions as $\{ f_{1, 1}, f_{1, 2}, f_{2, 1}, ...\}$ and relabel them concisely as $\{g_1, g_2, g_3, ... \}$. Apparently, $\{ g_n \}$ converges to $g(x) \equiv 0$ in measure. But for any $x \in [0, 1]$, there are infinite number of functions from the sequence $\{ g_n \}$ such that $g_n(x) = 1$ and so $\{ g_n \} \nrightarrow g, a.e$.

However, we have the following theorem.

Theorem 3 If $\{ f_n \}$ is Cauchy is Cauchy in measure, then it has a sub-sequence that converges to a measurable function $f$ almost uniformly.

Key Observation: if we look at the previous example, $f_{n, i}$ only deviates from $g(x) \equiv 0$ by a small interval $[( i - 1) / 2^n, i / 2^n]$ with length $1 / 2^n$. However, there are $2^n$ functions with such deviation and the intervals of deviation do not overlap. This motivates filtering out a sub-sequence.

Proof: Since $\{ f_n \}$ is Cauchy, $\forall k > 0$, we can find an integer $n_k > 0, s.t., \forall n, m \ge n_k, \mu( \{ x: |f_n(x) - f_m(x)| \ge 1 / 2^k \} ) \le 1/ 2^k$. We choose such a increasing sequence $(n_k)$ satisfying

and the corresponding sub-sequence $\{ f_{n_k} \}$ of $\{ f_n \}$.

Let

Then $\{ f_{n_k} \}$ is uniformly Cauchy in $E_n$ and therefore converge uniformly. It follows that there is a measurable function defined on

such that $\{ f_{n_k} \}$ converges uniformly on each $E_n$. We can extend the domain of $f$ to $X$ by defining $f(x) = 0$ for $x \in X - F$. Note that $f$ is measurable on $X$.

Finally,

It follows that $\mu(F) = \mu(X)$ and $( f_{n_k} )$ converge almost uniformly on $F$.

$\square$

Corollary If $\{ f_n \}$ is Cauchy is Cauchy in measure, then it converges to a measurable function $f$ in measure.

Reference

1. Avner Friedman, Foundation of Modern Analysis.