[Measure Theory] Modes of convergence
For a sequence of functions \(\{ f_n \}\) on $X$, $f_n \to f$ as $n \to \infty$ can be taken in various ways.
Types of convergence
Almost everywhere convergence
We say a sequence of measurable functions $f_n$ converges almost everywhere to $f$ and write $$ \begin{aligned} f_n \to f \text{ a.e. } \end{aligned} $$ if there exists a set of measure 0 such that for $x$ not in this set we have $f_n (x) \to f(x)$.
Convergence in measure
If $\mu$ is a measure, we say a sequence of measurable functions $f_n$ converges in measure to $f$ if for any $\varepsilon > 0$ $$ \begin{aligned} \mu ( \{ x : \; | f_n (x) - f ( x ) | > \varepsilon \}) \to 0 \end{aligned} $$ as $n \to \infty$.
Cauchy in measure
If $\mu$ is a measure, we say a sequence of measurable functions $f_n$ is Cauchy in measure if for any $\varepsilon > 0$ $$ \begin{aligned} \mu ( \{ x : \; | f_n (x) - f_m ( x ) | > \varepsilon \}) \to 0 \end{aligned} $$ as $m, n \to \infty$. In other words, for any $\varepsilon > 0$ and $\delta > 0$, there exists $N$ such that if $m, n \geq N$, then $$ \begin{aligned} \mu ( \{ x : \; | f_n (x) - f_m ( x ) | > \varepsilon \}) < \delta. \end{aligned} $$
Convergence in $L^p$
Let $1 \leq p < \infty$. We say a sequence of measurable functions $f_n$ converges in $L^p$ to $f$ if $$ \begin{aligned} \int | f_n - f |^p \; d\mu \to 0 \end{aligned} $$ as $n \to \infty$.
Almost uniform convergence
We say a sequence of measurable functions $f_n$ almost uniform converges to $f$ if there exists a measurable set $A$ such that $\mu (A) < \varepsilon$ and $f_n \to f$ uniformly on $A^c$ for any $\varepsilon > 0$.
Typical counterexamples
The relationships between these various modes of convergence are not very intuitive and depend on the measure space, e.g. whether or not $\mu$ is a finite measure. But the following functions on $\mathbb{R}$ will be worth keeping in mind as useful counterexamples:
- $f_n = \frac{1}{n} \chi_{(0, n)}$
- $f_n = \chi_{(n, n+1)}$
- $f_n = n \chi_{[0, \frac{1}{n}]}$
- $f_1 = \chi_{[0, 1]}$, $f_2 = \chi_{[0, 1/2]}$, $f_3 = \chi_{[1/2, 1]}$, $f_4 = \chi_{[0, 1/4]}$, $f_5 = \chi_{[1/2, 3/4]}$, $f_6 = \chi_{[3/4, 1]}$, and in general $f_n = \chi_{[j/2^k, (j+1)/2^k]}$ where $n = 2^k + j$ with $j \in [0, 2^k)$
Relationships and Diagrams
Cauchy and M, A.E.
First, Cauchy in measure implies convergence in measure, and the existence of a subsequence of functions that almost everywhere converges:
-
Cauchy and M
There exists a measurable function $f$ such that $f_n \to f$ in measure. Also, if $f_n \to g$ in measure, then $g = f$ a.e. -
Cauchy and A.E.
There exists a subsequence $\{ f_{n_j} \}$ that converges to $f$ a.e.
$\color{red}{\mathbf{Proof.}}$
Construct a subsequence \(\{ g_j \} = \{ f_{n_j} \}\) of \(\{ f_n \}\) such that
\[\begin{aligned} \mu (E_j = \{ x: \; | g_j (x) - g_{j+1} (x) | \geq 2^{-j} \} ) \leq 2^{-j}. \end{aligned}\]Let $F_k = \bigcup_{j=k}^\infty E_j$, then $\mu (F_k) \leq \sum_{j=k}^\infty 2^{-j} = 2^{1-k}$. And if $x \notin F_k$, for $i \geq j \geq k$ we have
\[\begin{aligned} \left|g_j(x)-g_i(x)\right| \leq \sum_{l=j}^{i-1}\left|g_{l+1}(x)-g_l(x)\right| \leq \sum_{l=j}^{i-1} 2^{-l} \leq 2^{1-j} \quad (\star) \end{aligned}\]Thus ${ g_j }$ is pointwise Cauchy on $F_k^c$. Let $F = \bigcap_{k = 1}^\infty F_k = \underset{j}{\text{ lim sup }} E_j$. Then $\mu (F) = 0$. Now, define
\[\begin{aligned} f(x) = \begin{cases} \lim_{j \to \infty} g_j (x) & \quad \text{ for } x \notin F \\ 0 & \quad \text{ otherwise } \\ \end{cases} \end{aligned}\]then $f$ is obviously measurable and $g_j \to f$ a.e. Also, for $x \notin F_k$, from $(\star)$ we have
\[\begin{aligned} | g_j (x) - f(x) | \leq 2^{1-j} \end{aligned}\]for $j \geq k$. Thus
\[\begin{aligned} F_k^c \subset \{x: \; | g_j (x) - f(x) | \leq 2^{1-j} \} \end{aligned}\]and since $\mu (F_k) \to 0$ as $k \to \infty$, it follows that $g_j \to f$ in measure. Then, we conclude that $f_n \to f$ in measure because:
\[\begin{aligned} \left\{x:\left|f_n(x)-f(x)\right| \geq \epsilon\right\} \subset\left\{x:\left|f_n(x)-g_j(x)\right| \geq \frac{1}{2} \epsilon\right\} \cup\left\{x:\left|g_j(x)-f(x)\right| \geq \frac{1}{2} \epsilon\right\} \end{aligned}\]and the sets on the right both have small measure when $n$ and $j$ are large.
Likewise, if $f_n \to g$ in measure,
\[\begin{aligned} \left\{x:\left|f(x)-g(x)\right| \geq \epsilon\right\} \subset\left\{x:\left|f(x)-f_n(x)\right| \geq \frac{1}{2} \epsilon\right\} \cup\left\{x:\left|f_n(x)-g(x)\right| \geq \frac{1}{2} \epsilon\right\} \end{aligned}\]for all $n$, hence
\[\begin{aligned} \mu ( \{ x: \; | f(x) - g(x) | \geq \varepsilon \}) = 0 \end{aligned}\]for any $\varepsilon > 0$. Letting $\varepsilon$ tend to zero through some sequence of values, we obtain that $f = g$ a.e.
\[\tag*{$\blacksquare$}\]A.E. and M
One should notice that definitions of almost everywhere convergence and convergence in measure have subtle difference. Formally, almost everywhere convergence is saying that
\[\begin{aligned} \mu ( \{ x : \; \lim_{n \to \infty} f_n (x) \neq f(x) \}) = 0 \end{aligned}\]For simplicity, let’s just denote \(\{ f_n \to f \} = \{ x \; : \; \lim_{n \to \infty} f_n (x) = f(x) \}\) and construct
\[\begin{aligned} A_{mn} = \{ x : \; | f_n (x) - f(x) | < \frac{1}{m} \}. \end{aligned}\]Then we can easily show that
\[\begin{aligned} \{ f_n \to f \} = \bigcap_{m=1}^\infty \underset{n}{\text{ lim inf }} A_{mn}. \end{aligned}\]$\mathbf{Proof.}$
It is obvious from
\[\begin{aligned} f_n (x) \to f(x) & \leftrightarrow \forall m \in \mathbb{N}, \exists n_0 \in \mathbb{N} \; | f_n (x) - f(x) | < \frac{1}{m} \text{ for all } n \geq n_0 \\ & \leftrightarrow \bigcap_{m=1}^\infty \left( \bigcap_{n = n_0}^\infty \underset{n}{\text{ lim inf }} A_{mn} \right) \subseteq \bigcap_{m=1}^\infty \left( \bigcup_{n_0}^\infty \bigcap_{n = n_0}^\infty \underset{n}{\text{ lim inf }} A_{mn} \right) = \bigcap_{m=1}^\infty \underset{n}{\text{ lim inf }} A_{mn} \end{aligned}\] \[\tag*{$\blacksquare$}\]Just for our intuition, let assume that $\mu (X) < \infty$. Then, we obtain the equivalent definition of a.e. convergence in the similar form with convergence in measure.
$\color{green}{\mathbf{Proof.}}$
Suppose $f_n \to f$ a.e. Then from the previous discussion, done by sandwich theorem:
\[\begin{aligned} \mu (\{ f_n \to f \}) \leq \mu (\underset{n}{\text{ lim inf }} A_{nm}) \end{aligned}\]Now suppose $\mu (\underset{n}{\text{ lim inf }} A_{nm}) = \mu (X)$. Then
\[\begin{aligned} \mu (\{ f_n \to f \}) = \mu \left( \bigcap_{m=1}^\infty \underset{n}{\text{ lim inf }} A_{mn} \right) \end{aligned}\]As $\underset{n}{\text{ lim inf }} A_{mn}$ is decreasing with respect to $m$ and $\mu$ is finite, by measure’s continuity from above
\[\begin{aligned} \mu \left( \bigcap_{m=1}^\infty \underset{n}{\text{ lim inf }} A_{mn} \right) = \lim_{m \to \infty} \mu \left( \underset{n}{\text{ lim inf }} A_{mn} \right) = \mu (X) \end{aligned}\] \[\tag*{$\blacksquare$}\]$f_n \to f$ a.e. if and only if for all $\varepsilon > 0$, $$ \begin{aligned} \mu \left( \{ x : \; | f_k (x) - f (x) | \geq \varepsilon \text{ for at least one } k \geq n \}\right) \to 0 \end{aligned} $$ as $n \to \infty$. In other words, $$ \begin{aligned} \mu \left( \bigcup_{k = n}^\infty \{ x : \; | f_k (x) - f (x) | \geq \varepsilon \}\right) \to 0 \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
And this might be much comprehensible with probability measure and real-world analogy. But be aware that our discussion is only valid on finite measure. For example, let $X = \mathbb{R}$ and let $f_n = \chi_{(n, n+1)}$. Then $f_n \to 0$ a.e. but $f_n$ does not converge in measure.
In summary, we have the following relationships between two mathematical concepts of convergence:
-
In finite measure, a.e. convergence implies convergence in measure
Suppose $\mu$ is a finite measure. If $f_n \to f$ a.e., then $f_n$ converges to $f$ in measure. -
Partial converse
If $\mu$ is a measure, not necessarily finite, and $f_n \to f$ in measure, there is a subsequence $n_j$ such that $f_{n_j} \to f$ a.e.
$\color{red}{\mathbf{Proof.}}$
(1) Let any $\varepsilon > 0$ be given. If
\[\begin{aligned} A_n = \{ x \; : \; | f_n (x) - f(x) | > \varepsilon \} \end{aligned}\]then $\chi_{A_n} \to 0$ a.e. Since $\mu$ is finite $\chi_{A_n}$ is ensured to be integrable and by the dominated convergence theorem
\[\begin{aligned} \mu (A_n) = \int \chi_{A_n} (x) \mu (dx) \to 0. \end{aligned}\](2) Suppose $f_n \to f$ in measure. Let $n_1 = 1$, and choose $n_j > n_{j-1}$ by induction so that
\[\begin{aligned} \mu \left( \left\{ x : \; | f_{n_j} (x) - f (x) | > \frac{1}{j} \right\} \right) \leq \frac{1}{2^j} \end{aligned}\]Then let
\[\begin{aligned} A_j = \left\{ x : \; | f_{n_j} (x) - f (x) | > \frac{1}{j} \right\}. \end{aligned}\]If we set $A = \underset{j}{\text{ lim sup }} A_j = \bigcap_{k=1}^\infty \bigcup_{j=k}^\infty A_j$, then
\[\begin{aligned} \mu (A) = \lim_{k \to \infty} \mu \left(\bigcup_{j=k}^\infty A_j \right) \leq \lim_{k \to \infty} \sum_{j = k}^\infty \mu(A_j) \leq \lim_{k \to \infty} 2^{-k + 1} = 0. \end{aligned}\]Therefore $A$ has measure $0$. If $x \notin A$, then $x \notin \bigcup_{j=k}^\infty A_j$ for some $k$ and so $| f_{n_j} (x) - f (x) | \leq 1/j$ for for $j \geq k$. This implies $f_{n_j} \to f$ on $A^c$.
\[\tag*{$\blacksquare$}\]But in some cases, convergence in measure becomes the sufficient is a sufficient condition for a.e. convergence.
$\color{red}{\mathbf{Proof.}}$
Define an equivalence class $\sim$ on $X$:
\[\begin{aligned} x \sim y \text{ if and only if } x \in A \leftrightarrow y \in A \quad \forall \; A \in \mathcal{A} \end{aligned}\]For each $x \in X$, let $A_x$ be a collection of equivalence class.
(1) $A_x$ is measurable and no proper subset of $A_x$ except empty set is in $\mathcal{A}$
To show $A_x \in \mathcal{A}$, consider each $y \in A_x^c$. It implies that there exists $B_y \in \mathcal{A}$ such that $y \in B_y$ but $x \notin B_y$. Then, it can be easily shown that
\[\begin{aligned} A_x = \left(\bigcup_{y \in X - A_x} B_y\right)^c \end{aligned}\]Also, suppose $A \in \mathcal{A}$, $A \subset A_x$ and $A \neq \emptyset$. Choose $y \in A$ and we can select $z \in A_x - A$. As $y, z \in A_x$ and $A \in \mathcal{A}$, $y \in A_x$ and $z \in A_x$, which is contradiction.
(2) Every $A \in \mathcal{A}$ can be written as countable disjoint union of $A_x$
It is obivous that $A = \bigcup_{x \in A} A_x$.
(3)$f_n$ is constant on $A_x$
For any $c \in \mathbb{R}$, as $f_n$ is measurable, \(f_n^{-1} (\{ c \}) \in \mathcal{A}\). Therefore
\[\begin{aligned} f_n^{-1} (\{ c \}) = \bigcup_{f(x) = c} A_x \end{aligned}\]So, for any given $A_y$, find $f_n (y)$. Then $A_y \subset f_n^{-1} ({ f_n (y) })$.
Hence, without loss of generality, we may assume that $\mathcal{A} = \mathcal{P} (X)$, so that a singleton set is measurable.
Collect $y_k \in X$ such that \(\mu (\{y_k\}) > 0\) and construct
\[\begin{aligned} Y = \{ y_k | k \in \mathbb{N}, \mu(\{ y_k \}) > 0 \}. \end{aligned}\]Note that $Y$ is also countable. Then $\mu (X) = \mu (Y)$, therefore it suffices to show the a.e convergence on $Y$, i.e. $f_n (y_k) \to f (y_k)$ for each $k \in \mathbb{N}$.
As $f_n \in f$ in measure, there exists a sufficiently large $N \in \mathbb{N}$ such that
\[\begin{aligned} \mu ( \{ x : | f_n (x) - f (x) | > \varepsilon \} ) < \mu (\{ y_k \}) \end{aligned}\]for all $n \geq N$ as \(\mu (\{ y_k \}) > 0\). This implies
\[\begin{aligned} y_k \notin \{ x : | f_n (x) - f (x) | > \varepsilon \} \end{aligned}\]for all $n \geq N$. In other words,
\[\begin{aligned} | f_n (y_k) - f (y_k) | \leq \varepsilon \quad \text{ for all } n \geq N. \end{aligned}\] \[\tag*{$\blacksquare$}\]$L^p$ and M
In this section, we compare convergence in $L^p$ to convergence in measure. Before we start, we need an easy lemma known as Chebyshev’s inequality.
If $1 \leq p < \infty$, then $$ \begin{aligned} \mu \left( \{ x : \; | f(x) | \geq a \} \right) \leq \frac{\int |f|^p \; d\mu}{a^p}. \end{aligned} $$
$\color{green}{\mathbf{Proof.}}$
Let
\[\begin{aligned} A = \{ x: \; | f(x) | \geq a \}. \end{aligned}\]Since $\chi_A \leq | f |^p \chi_A / a^p$, we have
\[\begin{aligned} \mu(A) \leq \frac{1}{a^p} \int_A |f|^p \; d\mu \leq \frac{1}{a^p} \int |f|^p \; d\mu \end{aligned}\] \[\tag*{$\blacksquare$}\]From Chebyshev’s inequality, it is now obvious that convergence in $L^p$ implies convergence in measure:
$\color{red}{\mathbf{Proof.}}$
Let any $\varepsilon > 0$ be given. Then
\[\begin{aligned} \mu \left( \{ x: \; | f_n (x) - f (x) | \geq \varepsilon \} \right) \leq \frac{1}{\varepsilon^p} \int |f_n - f|^p \; d \mu \end{aligned}\]and done by sandwich theorem.
\[\tag*{$\blacksquare$}\]Note that the converse has a counterexample. Let $f_n = n^2 \chi_{(0, 1/n)}$ and $f = 0$. Then $f_n$ converges to $f$ in measure (and a.e.) but not $L^p$.
$L^p$ and A.E.
Indeed, there are no general relationships between $L^p$ and a.e. convergence.
A.E., but not $L^p$
Let $f_n = n^2 \chi_{(0, 1/n)}$ on $[0, 1]$ and let $\mu$ be Lebesgue measure. Then $f_n$ converges to $0$ a.e. and in measure, but doesn’t converge in $L^p$ for any $p \geq 1$.
Let $f_n = \chi_{[j/2^k, (j+1)/2^k]}$ where $n = 2^k + j$ with $j \in [0, 2^k)$. Then $f_n$ converges a.e., but not in $L^p$.
$L^p$, but not A.E.
Let \(S = \{ e^{i\theta} : 0 \leq \theta < 2\pi \}\) be the unit circle in the complex plane and define
\[\begin{aligned} \mu (A) = m ( \{ \theta \in [0, 2\pi) : e^{i\theta} \in A \} ) \end{aligned}\]to be arclength measure on $S$, where $m$ is Lebesgue measure on $[0, 2\pi)$. Let $X = S$ and $f_n (x) = \chi_{F_n} (x)$ where
\[\begin{aligned} F_n = \left\{ e^{i \theta} : \sum_{j=1}^n \frac{1}{j} \leq \theta \leq \sum_{j=1}^{n + 1} \frac{1}{j} \right\}. \end{aligned}\]Let $f(e^{i\theta}) = 0$ for all $\theta$. Then $\mu (F_n) \leq 1 / (n+1) \to 0$, hence $f_n \to f$ in measure. Also, since $f_n$ is either $1$ or $0$,
\[\begin{aligned} \int | f_n - f |^p \; d\mu = \int \chi_{F_n} \; d\mu = \mu (F_n) \to 0. \end{aligned}\]But since $\sum_{j=1}^\infty 1 / j = \infty$, each point of $S$ is in infinitely many $F_n$, and each point of $S$ is in $S - F_n$ for infinitely many $n$, so $f_n$ doesn’t converge to $f$ any point.
A.U. and A.E., M
It is not hard to see that almost uniform convergence implies a.e. convergence and convergence in measure.
$\color{red}{\mathbf{Proof.}}$
A.E. Convergence
For each $n \in \mathbb{N}$, there exists a measurable $A_n$ such that $\mu (A_n) < \frac{1}{n}$ and $f_n \to f$ uniformly on $A_n^c$.
Set $A = \bigcap_{n = 1}^\infty A_n$. Then, $\mu (A) = 0$ and for any $x \in A^c = \bigcup_{n=1}^\infty A_n^c$, there exists $n \in \mathbb{N}$ such that $x \in A_n^c$ so that $f_n (x) \to f(x)$.
Convergence in measure
Let any $\varepsilon > 0$ be given. Then there exists a measurable $A$ with $\mu (A) < \varepsilon$ such that $f_n \to f$ uniformly on $A^c$.
For any $x \in A^c$, there exists $N \in \mathbb{N}$ independent of $x$ such that $| f_n (x) - f (x) | < \varepsilon$ for all $n \geq N$. In other words, for all $n \geq N$,
\[\begin{aligned} A \supset \{| f_n - f | < \varepsilon \} \end{aligned}\]Thus
\[\begin{aligned} \mu ( \{| f_n - f | < \varepsilon \} ) < \varepsilon \text{ for all } n \geq N. \end{aligned}\] \[\tag*{$\blacksquare$}\]Conversely, if we add some conditions on a.e. convergence, we can ensure almost uniform convergence. The following theorem is known as Egorov’s theorem:
Suppose $\mu$ is a finite measure and $f_n \to f$ a.e. Then $f_n$ almost uniform converges to $f$.
$\color{red}{\mathbf{Proof.}}$
Let
\[\begin{aligned} A_{nk} = \bigcup_{m = n}^\infty \{ x : \; | f_m(x) - f(x)| > 1/k \}. \end{aligned}\]For fixed $k$, $A_{nk}$ decreases as $n$ increases. And the intersection $\bigcap_n A_{nk}$ has measure $0$ because $\mu$ is finite measure and for almost every $x$, $| f_m (x) - f(x) | \leq 1/k$ if $m$ is sufficiently large. Therefore $\mu (A_{nk}) \to 0$ as $n \to \infty$. And we can find an integer $n_k$ such that $\mu(A_{n_k k}) < \varepsilon 2^{-k}$. Let
\[\begin{aligned} A = \bigcup_{k=1}^\infty A_{n_k k}. \end{aligned}\]Hence $\mu (A) < \varepsilon$. If $x \notin A$, then $A \notin A_{n_k k}$, and so $| f_n(x) - f(x) | \leq 1/k$ if $n \geq n_{k}$. Thus $f_n \to f$ uniformly on $A^c$.
\[\tag*{$\blacksquare$}\]Reference
[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley & Sons, 1999.
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