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RA-2

7-10 hours

Advanced

Measurable Functions & Approximations

Explore the theory of measurable functions, simple function approximations, Egorov's theorem on almost uniform convergence, and Lusin's theorem on almost continuity. These concepts form the foundation for Lebesgue integration.

Learning Objectives

By the end of this course, you will be able to:

Define measurable functions and understand their fundamental properties

Construct simple function approximations to measurable functions

Apply Egorov's theorem to understand the relationship between pointwise and uniform convergence

Use Lusin's theorem to approximate measurable functions by continuous functions

Understand almost-everywhere convergence and its implications

Prove that limits of measurable functions are measurable

Prerequisites

Before starting this course, you should be familiar with:

Measure foundations: outer measure, measurable sets, σ-algebras
Basic properties of functions: domain, range, composition
Pointwise and uniform convergence of sequences of functions
Continuous functions and their properties

Core Concepts

Fundamental definitions and properties

Definition 2.1: Measurable Function

Let $(X, \mathcal{M})$ be a measurable space. A function $f: X \to \mathbb{R}$ is called measurable if for every $a \in \mathbb{R}$ , the set

f^{-1}((a, \infty)) = \{x \in X : f(x) > a\}

is measurable (i.e., belongs to $\mathcal{M}$ ).

Note: This definition is equivalent to requiring that $f^{-1}(B)$ is measurable for all Borel sets $B$ , or for all open sets $B$ .

Definition 2.2: Simple Function

A function $\varphi: X \to \mathbb{R}$ is called a simple function if it takes only finitely many distinct values. That is, there exist distinct real numbers $c_1, c_2, \ldots, c_n$ and measurable sets $E_1, E_2, \ldots, E_n$ such that

\varphi(x) = \sum_{i=1}^n c_i \chi_{E_i}(x)

where $\chi_{E_i}$ is the characteristic function of $E_i$ .

Definition 2.3: Almost Everywhere

A property is said to hold almost everywhere (abbreviated a.e.) if it holds except possibly on a set of measure zero.

Two functions $f$ and $g$ are said to be equal almost everywhere if $\{x : f(x) \neq g(x)\}$ is a null set.

Theorem 2.1: Properties of Measurable Functions

Let $f$ and $g$ be measurable functions. Then:

$f + g$ is measurable
$fg$ is measurable
$\max(f, g)$ and $\min(f, g)$ are measurable
$|f|$ is measurable
If $g \neq 0$ a.e., then $f/g$ is measurable

Proof of Theorem 2.1:

Proof:

(1) For any $a$ , we have

\{x : (f+g)(x) > a\} = \bigcup_{r \in \mathbb{Q}} (\{x : f(x) > r\} \cap \{x : g(x) > a - r\})

Since $f$ and $g$ are measurable, each set in the union is measurable, and a countable union of measurable sets is measurable.

(2) Note that $fg = \frac{1}{4}[(f+g)^2 - (f-g)^2]$ . Since squares and differences of measurable functions are measurable, the result follows.

(3) $\max(f, g) = \frac{f+g+|f-g|}{2}$ and $\min(f, g) = \frac{f+g-|f-g|}{2}$ .

(4) $|f| = \max(f, -f)$ .

(5) $\{x : (f/g)(x) > a\} = \{x : f(x) > ag(x)\}$ , which can be written as a countable union similar to (1). ∎

∎

Theorem 2.2: Pointwise Limits of Measurable Functions

If $\{f_n\}$ is a sequence of measurable functions and $f_n \to f$ pointwise (or pointwise a.e.), then $f$ is measurable.

Proof of Theorem 2.2:

Proof:

For any $a$ , we have

\{x : f(x) > a\} = \bigcup_{k=1}^{\infty} \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} \left\{x : f_n(x) > a + \frac{1}{k}\right\}

This expresses the limit superior condition: $f(x) > a$ if and only if there exists $k$ such that for all $N$ , there exists $n \geq N$ with $f_n(x) > a + 1/k$ .

Since each $\{x : f_n(x) > a + 1/k\}$ is measurable, and measurable sets are closed under countable unions and intersections, the result follows. ∎

∎

Theorem 2.3: Simple Function Approximation

Let $f$ be a nonnegative measurable function. Then there exists an increasing sequence $\{\varphi_n\}$ of simple functions such that $\varphi_n \to f$ pointwise.

Proof of Theorem 2.3:

Proof:

For each $n$ , define

\varphi_n(x) = \sum_{k=0}^{n2^n - 1} \frac{k}{2^n} \chi_{E_{n,k}}(x) + n \chi_{F_n}(x)

where

E_{n,k} = \left\{x : \frac{k}{2^n} \leq f(x) < \frac{k+1}{2^n}\right\}, \quad F_n = \{x : f(x) \geq n\}

Each $\varphi_n$ is a simple function, the sequence is increasing, and $\varphi_n \to f$ pointwise. ∎

∎

Theorem 2.4: Egorov's Theorem

Let $(X, \mathcal{M}, \mu)$ be a measure space with $\mu(X) < \infty$ . Suppose $\{f_n\}$ is a sequence of measurable functions and $f_n \to f$ pointwise a.e. Then for every $\epsilon > 0$ , there exists a measurable set $E$ such that $\mu(E) < \epsilon$ and $f_n \to f$ uniformly on $X \setminus E$ .

Proof of Theorem 2.4:

Proof:

For $n, k \in \mathbb{N}$ , define

E_{n,k} = \bigcup_{m=n}^{\infty} \left\{x : |f_m(x) - f(x)| \geq \frac{1}{k}\right\}

Since $f_n \to f$ pointwise a.e., for each $k$ , we have $\mu(\bigcap_{n=1}^{\infty} E_{n,k}) = 0$ .

By continuity of measure, for each $k$ , there exists $n_k$ such that $\mu(E_{n_k, k}) < \epsilon/2^k$ .

Let $E = \bigcup_{k=1}^{\infty} E_{n_k, k}$ . Then $\mu(E) < \epsilon$ , and on $X \setminus E$ , we have uniform convergence. ∎

∎

Theorem 2.5: Lusin's Theorem

Let $f$ be a measurable function on $\mathbb{R}^n$ that is finite a.e. on a measurable set $E$ of finite measure. Then for every $\epsilon > 0$ , there exists a closed set $F$ with $F \subseteq E$ , $m(E \setminus F) < \epsilon$ , and $f$ is continuous on $F$ .

Proof of Theorem 2.5:

Proof:

First approximate $f$ by a simple function $\varphi$ such that $|f - \varphi| < \epsilon$ except on a set of small measure.

Then approximate each characteristic function in $\varphi$ by continuous functions using regularity of measure.

Finally, use Tietze's extension theorem to extend the continuous function to the closed set. The details involve careful approximation and measure theory techniques. ∎

∎

Example 2.1: Characteristic Function is Measurable

Problem: Show that the characteristic function $\chi_E$ of a measurable set $E$ is measurable.

Solution:

For any $a \in \mathbb{R}$ , we have

\chi_E^{-1}((a, \infty)) = \begin{cases} E & \text{if } a < 1 \\ \emptyset & \text{if } a \geq 1 \end{cases}

Since $E$ and $\emptyset$ are measurable, $\chi_E$ is measurable.

Example 2.2: Constructing Simple Function Approximation

Problem: Construct a simple function approximation for $f(x) = x^2$ on $[0, 4]$ .

Solution:

Using the construction in Theorem 2.3 with $n = 2$ , we partition $[0, 4]$ into intervals of length $1/4 = 0.25$ :

\varphi_2(x) = \sum_{k=0}^{15} \frac{k}{4} \chi_{[k/4, (k+1)/4)}(x) + 2 \chi_{[2, 4]}(x)

This gives a step function approximation that increases as $n$ increases, converging to $f(x) = x^2$ .

Corollary 2.1: Measurable Functions are Closed under Algebraic Operations

The collection of measurable functions forms a vector space and an algebra over $\mathbb{R}$ .

Theorem 2.6: Composition of Measurable Functions

Let $f: X \to \mathbb{R}$ be measurable and $g: \mathbb{R} \to \mathbb{R}$ be Borel measurable (i.e., $g^{-1}(B)$ is Borel for all Borel sets $B$ ). Then the composition $g \circ f$ is measurable.

In particular, if $g$ is continuous, then $g \circ f$ is measurable.

Proof of Theorem 2.6:

Proof:

For any Borel set $B$ , we have

(g \circ f)^{-1}(B) = f^{-1}(g^{-1}(B))

Since $g$ is Borel measurable, $g^{-1}(B)$ is a Borel set. Since $f$ is measurable, $f^{-1}(g^{-1}(B))$ is measurable.

Therefore, $(g \circ f)^{-1}(B)$ is measurable for all Borel sets $B$ , which implies $g \circ f$ is measurable.

If $g$ is continuous, then $g$ is Borel measurable (since continuous functions map Borel sets to Borel sets), so the result follows. ∎

∎

Theorem 2.7: Detailed Construction of Simple Function Approximation

Let $f$ be a nonnegative measurable function. For each $n \in \mathbb{N}$ , define the simple function $\varphi_n$ by:

\varphi_n(x) = \begin{cases} \frac{k}{2^n} & \text{if } \frac{k}{2^n} \leq f(x) < \frac{k+1}{2^n}, \, k = 0, 1, \ldots, n2^n - 1 \\ n & \text{if } f(x) \geq n \end{cases}

Then $\{\varphi_n\}$ is an increasing sequence of simple functions converging pointwise to $f$ , and the convergence is uniform on any set where $f$ is bounded.

Proof of Theorem 2.7:

Proof:

For each $n$ , the function $\varphi_n$ takes only finitely many values (at most $n2^n + 1$ values), so it is simple.

To show the sequence is increasing, fix $x$ and $n$ . If $f(x) \geq n$ , then $\varphi_n(x) = n$ and $\varphi_{n+1}(x) \geq n$ , so $\varphi_{n+1}(x) \geq \varphi_n(x)$ .

If $f(x) < n$ , then $\varphi_n(x) = \frac{k}{2^n}$ for some $k$ . Since the partition at step $n+1$ is finer, we have $\varphi_{n+1}(x) \geq \varphi_n(x)$ .

For pointwise convergence, fix $x$ . If $f(x) = \infty$ , then $\varphi_n(x) = n \to \infty$ . If $f(x) < \infty$ , then for large $n$ , we have $f(x) < n$ , and

0 \leq f(x) - \varphi_n(x) < \frac{1}{2^n}

so $\varphi_n(x) \to f(x)$ .

If $f$ is bounded by $M$ , then for $n > M$ , the approximation error is at most $\frac{1}{2^n}$ uniformly, giving uniform convergence. ∎

∎

Theorem 2.8: Supremum and Infimum of Measurable Functions

Let $\{f_n\}$ be a sequence of measurable functions. Then:

$\sup_n f_n$ is measurable
$\inf_n f_n$ is measurable
$\limsup_{n \to \infty} f_n$ is measurable
$\liminf_{n \to \infty} f_n$ is measurable

Proof of Theorem 2.8:

Proof:

(1) For any $a$ , we have

\{x : \sup_n f_n(x) > a\} = \bigcup_{n=1}^{\infty} \{x : f_n(x) > a\}

Since each $\{x : f_n(x) > a\}$ is measurable, the countable union is measurable.

(2) Similarly, $\inf_n f_n = -\sup_n (-f_n)$ , so it is measurable.

(3) Note that $\limsup_{n \to \infty} f_n = \inf_{n \geq 1} \sup_{k \geq n} f_k$ . By (1) and (2), this is measurable.

(4) Similarly, $\liminf_{n \to \infty} f_n = \sup_{n \geq 1} \inf_{k \geq n} f_k$ is measurable. ∎

∎

Example 2.3: Application of Egorov's Theorem: Constructing a Counterexample

Problem: Show that the condition $\mu(X) < \infty$ in Egorov's theorem is necessary by constructing a counterexample on an infinite measure space.

Solution:

Let $X = \mathbb{R}$ with Lebesgue measure, and define

f_n(x) = \chi_{[n, n+1]}(x) = \begin{cases} 1 & \text{if } x \in [n, n+1] \\ 0 & \text{otherwise} \end{cases}

Then $f_n \to 0$ pointwise everywhere, since for any fixed $x$ , eventually $f_n(x) = 0$ .

However, for any set $E$ with $m(E) < \infty$ , the convergence is not uniform on $\mathbb{R} \setminus E$ . In fact, for any $\epsilon > 0$ , no matter how large the exceptional set $E$ is (as long as it has finite measure), there will be points in $\mathbb{R} \setminus E$ where $f_n = 1$ for arbitrarily large $n$ .

This shows that Egorov's theorem fails when the measure space is infinite, demonstrating the necessity of the finite measure condition.

Example 2.4: Application of Lusin's Theorem: A Specific Function

Problem: Consider the function $f(x) = \chi_{\mathbb{Q}}(x)$ on $[0,1]$ (the characteristic function of the rationals). Apply Lusin's theorem to approximate this function by a continuous function.

Solution:

The function $f$ is measurable (since $\mathbb{Q}$ is countable and hence measurable) but is discontinuous everywhere.

By Lusin's theorem, for any $\epsilon > 0$ , there exists a closed set $F$ with $m([0,1] \setminus F) < \epsilon$ such that $f$ is continuous on $F$ .

Since $\mathbb{Q} \cap [0,1]$ has measure zero, we can take $F = [0,1] \setminus (\mathbb{Q} \cap [0,1])$ (the irrationals in $[0,1]$ ). However, this set is not closed. Instead, we can remove a small open neighborhood around each rational point.

More precisely, enumerate $\mathbb{Q} \cap [0,1] = \{q_1, q_2, \ldots\}$ . For each $q_i$ , remove an open interval $(q_i - \delta_i, q_i + \delta_i)$ where $\sum \delta_i < \epsilon$ . The complement is closed, and on this closed set, $f$ is identically 0, hence continuous.

This demonstrates that even a function that is discontinuous everywhere can be made continuous on a set of nearly full measure.

Example 2.5: Step-by-Step Construction of Simple Function Approximation

Problem: Construct the first three approximations $\varphi_1, \varphi_2, \varphi_3$ for the function $f(x) = x^2$ on $[0, 4]$ using the method in Theorem 2.7.

Solution:

Step 1: $n = 1$

We partition $[0, 1)$ into intervals of length $\frac{1}{2}$ : $[0, \frac{1}{2}), [\frac{1}{2}, 1)$ . For $x \in [0, 1)$ , we have $f(x) = x^2 < 1$ , so:

\varphi_1(x) = \begin{cases} 0 & \text{if } 0 \leq x^2 < \frac{1}{2}, \text{ i.e., } 0 \leq x < \frac{1}{\sqrt{2}} \\ \frac{1}{2} & \text{if } \frac{1}{2} \leq x^2 < 1, \text{ i.e., } \frac{1}{\sqrt{2}} \leq x < 1 \\ 1 & \text{if } x^2 \geq 1, \text{ i.e., } x \geq 1 \end{cases}

Step 2: $n = 2$

We partition $[0, 2)$ into intervals of length $\frac{1}{4}$ : $[0, \frac{1}{4}), [\frac{1}{4}, \frac{1}{2}), \ldots, [\frac{7}{4}, 2)$ . This gives a finer approximation:

\varphi_2(x) = \begin{cases} \frac{k}{4} & \text{if } \frac{k}{4} \leq x^2 < \frac{k+1}{4}, \, k = 0, 1, \ldots, 7 \\ 2 & \text{if } x^2 \geq 2, \text{ i.e., } x \geq \sqrt{2} \end{cases}

Step 3: $n = 3$

We partition $[0, 3)$ into intervals of length $\frac{1}{8}$ , giving an even finer approximation with 24 steps in $[0, 3)$ and the constant value 3 for $x \geq \sqrt{3}$ .

As $n$ increases, $\varphi_n$ becomes a better approximation to $f(x) = x^2$ , with the error bounded by $\frac{1}{2^n}$ on $[0, n)$ .

Example 2.6: A Measurable Function That Is Not Continuous

Problem: Construct a measurable function on $[0,1]$ that is not continuous at any point.

Solution:

Consider the function $f(x) = \chi_{\mathbb{Q} \cap [0,1]}(x)$ , the characteristic function of the rationals in $[0,1]$ .

This function is measurable because $\mathbb{Q} \cap [0,1]$ is countable and hence measurable.

However, $f$ is discontinuous at every point. To see this, fix any $x_0 \in [0,1]$ . If $x_0 \in \mathbb{Q}$ , then $f(x_0) = 1$ , but in any neighborhood of $x_0$ , there are irrationals where $f$ takes the value 0. If $x_0 \notin \mathbb{Q}$ , then $f(x_0) = 0$ , but in any neighborhood of $x_0$ , there are rationals where $f$ takes the value 1.

This example shows that measurability does not imply continuity, even though Lusin's theorem tells us that measurable functions are "almost continuous" in a measure-theoretic sense.

Example 2.7: Pointwise Convergence That Is Not Uniform

Problem: Give an example of a sequence of measurable functions that converges pointwise almost everywhere but not uniformly, even on sets of finite measure.

Solution:

On $[0,1]$ with Lebesgue measure, define

f_n(x) = \begin{cases} n & \text{if } x \in (0, \frac{1}{n}] \\ 0 & \text{otherwise} \end{cases}

Then $f_n \to 0$ pointwise almost everywhere. Indeed, for any $x > 0$ , we have $f_n(x) = 0$ for all $n > 1/x$ . The only point where convergence fails is $x = 0$ , which is a null set.

However, the convergence is not uniform on $[0,1]$ . For any $n$ , we have $f_n(\frac{1}{2n}) = n$ , so $\sup_{x \in [0,1]} |f_n(x) - 0| = n$ , which does not tend to 0.

This example illustrates why Egorov's theorem is remarkable: it tells us that even though uniform convergence may fail on the whole set, we can always find a large subset (all but a set of small measure) where uniform convergence holds.

Corollary 2.2: Limits of Measurable Functions

If $\{f_n\}$ is a sequence of measurable functions and $f_n \to f$ pointwise (or pointwise a.e.), then $f$ is measurable.

This follows immediately from Theorem 2.8, since $f = \limsup f_n = \liminf f_n$ when the limit exists.

Corollary 2.3: Equivalence of Measurable and Borel Functions

A function $f$ is measurable if and only if $f^{-1}(B)$ is measurable for all Borel sets $B$ .

This is because the Borel σ-algebra is generated by intervals of the form $(a, \infty)$ , and measurability can be checked on a generating family.

Remark 2.2: Historical Context of Egorov and Lusin Theorems

Egorov's theorem (1911) and Lusin's theorem (1912) were fundamental results in the early development of measure theory and integration theory.

Egorov's theorem was proved by Dmitri Egorov, a Russian mathematician who was a student of Lebesgue. It bridges the gap between pointwise convergence (which is natural but weak) and uniform convergence (which is strong but often too restrictive). The theorem shows that on finite measure spaces, these two notions are "almost" equivalent.

Lusin's theorem was proved by Nikolai Luzin (also spelled Lusin), another Russian mathematician. It demonstrates a remarkable fact: measurable functions, which can be very irregular, are "almost continuous" in a precise measure-theoretic sense. This result connects measure theory with topology and is crucial for many applications.

Together, these theorems form part of "Littlewood's three principles," which summarize the approximation properties in measure theory: measurable sets are nearly open, measurable functions are nearly continuous, and pointwise convergence is nearly uniform.

Remark 2.3: Connection to Topology

There is a deep connection between measurability and topology:

Continuous functions are measurable: Since continuous functions map open sets to open sets, and open sets are Borel, continuous functions are Borel measurable and hence measurable.
Lusin's theorem: Shows that measurable functions can be approximated by continuous functions, establishing a bridge between measure theory and topology.
Borel σ-algebra: The smallest σ-algebra containing all open sets. All Borel sets are measurable, but there exist non-Borel measurable sets (requiring the Axiom of Choice to construct).
Regularity: The regularity properties of measure (approximation by open/closed sets) connect measure theory with the topology of the underlying space.

These connections are essential for understanding how measure theory generalizes and extends classical analysis.

Remark 2.4: Applications in Probability Theory

Measurable functions play a central role in probability theory:

Random variables: In probability theory, random variables are defined as measurable functions from a probability space to $\mathbb{R}$ . The measurability condition ensures that events like $\{X > a\}$ have well-defined probabilities.
Almost sure convergence: The concept of "almost everywhere" in measure theory corresponds to "almost surely" in probability theory. Egorov's theorem has probabilistic analogs.
Simple function approximation: The approximation of measurable functions by simple functions is crucial for defining the expectation (integral) of random variables.
Convergence in measure: Various modes of convergence (pointwise, uniform, in measure) have probabilistic interpretations (almost sure convergence, convergence in probability, etc.).

The theory of measurable functions provides the rigorous foundation for modern probability theory and stochastic processes.

Example 2.8: Constructing Measurable Functions from Simple Functions

Problem: Show that any simple function can be written as a linear combination of characteristic functions of disjoint measurable sets.

Solution:

Let $\varphi = \sum_{i=1}^n c_i \chi_{E_i}$ be a simple function. If the sets $E_i$ are not disjoint, we can refine them.

For example, if $n = 2$ and $E_1$ and $E_2$ overlap, define:

F_1 = E_1 \setminus E_2, \quad F_2 = E_1 \cap E_2, \quad F_3 = E_2 \setminus E_1

Then $\varphi = c_1 \chi_{F_1} + (c_1 + c_2) \chi_{F_2} + c_2 \chi_{F_3}$ , where the $F_i$ are disjoint.

This canonical form is useful for defining the integral of simple functions.

Example 2.9: Egorov's Theorem: Detailed Application

Problem: Let $f_n(x) = x^n$ on $[0,1]$ . Apply Egorov's theorem to show that the convergence $f_n \to 0$ (pointwise on $[0,1)$ ) can be made uniform on a large subset.

Solution:

We have $f_n(x) \to 0$ for $x \in [0, 1)$ and $f_n(1) = 1$ for all $n$ .

For any $\epsilon > 0$ , Egorov's theorem guarantees a set $E$ with $m(E) < \epsilon$ such that $f_n \to 0$ uniformly on $[0,1] \setminus E$ .

In this case, we can take $E = [1-\delta, 1]$ for small $\delta$ . On $[0, 1-\delta]$ , we have $|f_n(x)| = x^n \leq (1-\delta)^n$ , which tends to 0 uniformly.

This demonstrates how Egorov's theorem works: we exclude a small neighborhood of the problematic point (where convergence fails) to get uniform convergence on the remainder.

Example 2.10: Lusin's Theorem: Step Function Example

Problem: Consider the step function $f(x) = \sum_{i=1}^n c_i \chi_{[a_i, b_i)}(x)$ on $[0,1]$ . Show how Lusin's theorem applies.

Solution:

The function $f$ is measurable (as a simple function) but has jump discontinuities at the points $a_i$ and $b_i$ .

By Lusin's theorem, for any $\epsilon > 0$ , there exists a closed set $F$ with $m([0,1] \setminus F) < \epsilon$ such that $f$ is continuous on $F$ .

In this case, we can take $F$ to be $[0,1]$ minus small open neighborhoods around each discontinuity point. On the resulting closed set, $f$ is constant on each connected component, hence continuous.

This shows that even functions with many discontinuities can be made continuous on a set of nearly full measure.

Lemma 2.1: Approximation by Continuous Functions

Let $f$ be a measurable function on $\mathbb{R}^n$ that is finite a.e. on a measurable set $E$ of finite measure.

Then for any $\epsilon > 0$ , there exists a continuous function $g$ on $\mathbb{R}^n$ such that $m(\{x \in E : f(x) \neq g(x)\}) < \epsilon$ .

This is a consequence of Lusin's theorem combined with Tietze's extension theorem.

Proof of Lemma 2.1:

Proof:

By Lusin's theorem, there exists a closed set $F$ with $F \subseteq E$ , $m(E \setminus F) < \epsilon/2$ , and $f$ is continuous on $F$ .

By Tietze's extension theorem, the continuous function $f|_F$ can be extended to a continuous function $g$ on all of $\mathbb{R}^n$ .

Then $m(\{x \in E : f(x) \neq g(x)\}) \leq m(E \setminus F) < \epsilon$ . ∎

∎

Theorem 2.9: Density of Simple Functions in Lp

For $1 \leq p < \infty$ , the simple functions are dense in $L^p$ . That is, for any $f \in L^p$ and $\epsilon > 0$ , there exists a simple function $\varphi$ such that $\|f - \varphi\|_p < \epsilon$ .

Proof of Theorem 2.9:

Proof:

First approximate $f$ by a bounded function $f_N = f \chi_{\{|f| \leq N\}}$ . For large $N$ , $\|f - f_N\|_p < \epsilon/2$ .

Then approximate $f_N$ by a simple function using the construction in Theorem 2.7. The approximation can be made arbitrarily close in $L^p$ norm. ∎

∎

Corollary 2.4: Density of Continuous Functions

For $1 \leq p < \infty$ , continuous functions with compact support are dense in $L^p(\mathbb{R}^n)$ .

This follows from the density of simple functions and Lemma 2.1, which allows approximation of simple functions by continuous functions.

Example 2.11: Measurable Function That Is Not Borel Measurable

Problem: Show that there exist measurable functions that are not Borel measurable.

Solution:

Let $C$ be the Cantor set and $V$ be a non-Borel measurable subset of $C$ (such sets exist by the Axiom of Choice).

Define $f = \chi_V$ . Since $V$ is measurable (as a subset of a null set), $f$ is measurable.

However, $f^{-1}(\{1\}) = V$ is not Borel, so $f$ is not Borel measurable.

This shows that the class of measurable functions is strictly larger than the class of Borel measurable functions.

Remark 2.5: Convergence in Measure

In addition to pointwise and uniform convergence, there is another important mode of convergence:

Convergence in measure: A sequence $\{f_n\}$ converges in measure to $f$ if for every $\epsilon > 0$ ,

\lim_{n \to \infty} m(\{x : |f_n(x) - f(x)| \geq \epsilon\}) = 0

Convergence in measure is weaker than pointwise convergence a.e. but stronger than convergence in Lp. It plays an important role in probability theory (where it corresponds to convergence in probability).

Egorov's theorem shows that on finite measure spaces, pointwise convergence a.e. implies convergence in measure, and the converse holds along a subsequence.

Remark 2.6: Measurable Functions and Integration

The theory of measurable functions is essential for Lebesgue integration:

Definition of integral: The Lebesgue integral is first defined for simple functions, then extended to nonnegative measurable functions via approximation, and finally to general measurable functions.
Convergence theorems: The fact that limits of measurable functions are measurable ensures that convergence theorems (monotone, dominated) can be applied.
Lp spaces: The space Lp consists of measurable functions (modulo functions that are zero a.e.) with finite Lp norm.
Almost everywhere equality: Two functions that are equal a.e. have the same integral, which is why we work with equivalence classes in Lp spaces.

Without the theory of measurable functions, we could not have a satisfactory theory of integration for the wide class of functions that Lebesgue integration handles.

Remark 2.1: Key Insights

Key takeaways:

Measurable functions generalize continuous functions and are closed under limits
Simple functions provide a bridge between measure theory and integration
Egorov's theorem shows that pointwise convergence can be made uniform on large subsets
Lusin's theorem demonstrates that measurable functions are "almost continuous"
Almost-everywhere properties are sufficient for most purposes in analysis

Practice Quiz

Measurable Functions & Approximations

Questions

Correct

Accuracy

A function

f: \mathbb{R}^n \to \mathbb{R}

is measurable if and only if:

Easy

Not attempted

A simple function is:

Easy

Not attempted

Egorov's theorem states that:

Medium

Not attempted

Lusin's theorem states that:

Medium

Not attempted

f_n \to f

pointwise a.e. and each

f_n

is measurable, then:

Medium

Not attempted

A function

f

is measurable if

f^{-1}(B)

is measurable for all:

Hard

Not attempted

The simple function approximation theorem states that:

Medium

Not attempted

Almost everywhere (a.e.) means:

Easy

Not attempted

f

and

g

are measurable functions, then which of the following is NOT necessarily measurable?

Hard

Not attempted

Littlewood's three principles state that:

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Frequently Asked Questions

Why do we need measurable functions instead of just continuous functions?

Measurable functions form a much larger class than continuous functions and are closed under pointwise limits. Many important functions in analysis (like step functions, indicator functions, and limits of sequences) are measurable but not continuous. The theory of Lebesgue integration requires measurable functions.

What is the relationship between measurable functions and simple functions?

Simple functions are a special class of measurable functions that take only finitely many values. They serve as building blocks: any nonnegative measurable function can be approximated by an increasing sequence of simple functions, which is crucial for defining the Lebesgue integral.

How does Egorov's theorem relate to uniform convergence?

Egorov's theorem shows that on a set of finite measure, pointwise convergence almost everywhere can be 'upgraded' to uniform convergence on a large subset (all but a set of arbitrarily small measure). This bridges the gap between pointwise and uniform convergence in measure theory.

What does Lusin's theorem tell us about measurable functions?

Lusin's theorem shows that measurable functions are 'almost continuous' in a precise sense: given any measurable function on a set of finite measure, we can find a subset of nearly full measure where the function is continuous. This is remarkable since measurable functions can be very irregular.

Are all continuous functions measurable?

Yes, all continuous functions are measurable. This follows because the preimage of an open set under a continuous function is open, and open sets are measurable. However, not all measurable functions are continuous.

What does 'almost everywhere' mean in practice?

A property holds almost everywhere if it fails only on a set of measure zero. In practice, this means the property holds 'almost always' and exceptions are negligible from a measure-theoretic perspective. For example, two functions that differ only on a countable set are equal almost everywhere.

Why is the measurability of limits important?

The fact that pointwise limits of measurable functions are measurable ensures that the class of measurable functions is closed under limit operations. This is essential for integration theory, as we often need to take limits of sequences of functions.