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RA-7

7-11 hours

Advanced

Bounded Variation & Absolute Continuity

Explore functions of bounded variation, Jordan decomposition, absolute continuity, and the Lebesgue fundamental theorem of calculus. Understand the relationship between function regularity and integration.

Learning Objectives

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

Define bounded variation and understand its geometric meaning

Master the Jordan decomposition of functions of bounded variation

Understand absolute continuity and its relationship to integration

Prove the Lebesgue fundamental theorem of calculus

Characterize absolutely continuous functions as integrals

Understand the relationship between BV, AC, and differentiability

Prerequisites

Before starting this course, you should be familiar with:

Lebesgue differentiation theorem
Lebesgue integration and L1 space
Basic properties of functions on intervals
Fundamental theorem of calculus (Riemann version)

Core Concepts

Bounded variation and absolute continuity

Definition 7.1: Total Variation

Let $f$ be a function on $[a,b]$ . For a partition $P = \{a = x_0 < x_1 < \cdots < x_n = b\}$ , define:

V(f, P) = \sum_{i=1}^n |f(x_i) - f(x_{i-1})|

The total variation of $f$ on $[a,b]$ is:

V_a^b(f) = \sup_P V(f, P)

where the supremum is over all partitions of $[a,b]$ .

Definition 7.2: Bounded Variation

A function $f$ on $[a,b]$ is said to be of bounded variation (BV) if $V_a^b(f) < \infty$ .

Definition 7.3: Absolute Continuity

A function $f$ on $[a,b]$ is absolutely continuous (AC) if for every $\epsilon > 0$ , there exists $\delta > 0$ such that for any finite collection of disjoint intervals $\{(a_i, b_i)\}$ in $[a,b]$ with $\sum (b_i - a_i) < \delta$ , we have:

\sum |f(b_i) - f(a_i)| < \epsilon

Theorem 7.1: Jordan Decomposition

A function $f$ on $[a,b]$ is of bounded variation if and only if it can be written as:

f = f_1 - f_2

where $f_1$ and $f_2$ are increasing functions.

Proof of Theorem 7.1:

Proof:

Define $f_1(x) = \frac{V_a^x(f) + f(x)}{2}$ and $f_2(x) = \frac{V_a^x(f) - f(x)}{2}$ .

Then $f = f_1 - f_2$ , and both $f_1$ and $f_2$ are increasing because the total variation function is increasing. ∎

∎

Theorem 7.2: BV Functions are Differentiable a.e.

If $f$ is of bounded variation on $[a,b]$ , then $f$ is differentiable almost everywhere.

Proof of Theorem 7.2:

Proof:

By the Jordan decomposition, $f = f_1 - f_2$ where $f_1$ and $f_2$ are increasing.

Increasing functions are differentiable almost everywhere (this follows from the Lebesgue differentiation theorem applied to the derivative measure).

Therefore, $f$ is differentiable almost everywhere. ∎

∎

Theorem 7.3: Characterization of Absolute Continuity

A function $f$ on $[a,b]$ is absolutely continuous if and only if there exists $g \in L^1([a,b])$ such that:

f(x) = f(a) + \int_a^x g(t) \, dt

for all $x \in [a,b]$ .

Proof of Theorem 7.3:

Proof:

If $f$ is absolutely continuous, then it is of bounded variation and differentiable a.e. The derivative $f'$ is in $L^1$ (this requires additional argument), and $f(x) = f(a) + \int_a^x f'(t) \, dt$ .

Conversely, if $f(x) = f(a) + \int_a^x g(t) \, dt$ for $g \in L^1$ , then absolute continuity follows from the absolute continuity of the integral. ∎

∎

Theorem 7.4: Lebesgue Fundamental Theorem of Calculus

Let $f \in L^1([a,b])$ and define $F(x) = \int_a^x f(t) \, dt$ . Then:

$F$ is absolutely continuous
$F'(x) = f(x)$ almost everywhere

Conversely, if $F$ is absolutely continuous, then $F' \in L^1$ and $F(x) = F(a) + \int_a^x F'(t) \, dt$ .

Proof of Theorem 7.4:

Proof:

(1) Absolute continuity of $F$ follows from the absolute continuity of the integral with respect to the measure of the integration domain.

(2) The equality $F'(x) = f(x)$ a.e. is the Lebesgue differentiation theorem.

The converse follows from the characterization theorem. ∎

∎

Example 7.1: Function of Bounded Variation

Problem: Show that $f(x) = x^2$ on $[0,1]$ is of bounded variation.

Solution:

Since $f'(x) = 2x$ is bounded on $[0,1]$ , $f$ is Lipschitz continuous, hence absolutely continuous, and therefore of bounded variation.

More directly, for any partition, the variation is bounded by the total change, which is finite. ∎

Corollary 7.1: AC Functions are BV

Every absolutely continuous function is of bounded variation.

Theorem 7.5: Jordan Decomposition Construction

Every function $f$ of bounded variation on $[a,b]$ can be written as $f = g - h$ where $g$ and $h$ are increasing functions.

One such decomposition is given by $g(x) = V_a^x(f)$ (the total variation) and $h(x) = V_a^x(f) - f(x)$ .

Proof of Theorem 7.5:

Proof:

Define $g(x) = V_a^x(f)$ and $h(x) = V_a^x(f) - f(x)$ .

Since the total variation is increasing, $g$ is increasing.

For $x < y$ , we have $V_a^y(f) = V_a^x(f) + V_x^y(f)$ , so

h(y) - h(x) = (V_a^y(f) - f(y)) - (V_a^x(f) - f(x)) = V_x^y(f) - (f(y) - f(x)) \geq 0

since $V_x^y(f) \geq |f(y) - f(x)|$ . Therefore, $h$ is also increasing.

Clearly, $f = g - h$ . ∎

∎

Theorem 7.6: Equivalent Conditions for Absolute Continuity

For a function $f$ on $[a,b]$ , the following are equivalent:

$f$ is absolutely continuous
$f$ is of bounded variation and $f' \in L^1$ with $f(x) = f(a) + \int_a^x f'(t) \, dt$
$f$ maps null sets to null sets
$f$ is an indefinite integral: $f(x) = f(a) + \int_a^x g(t) \, dt$ for some $g \in L^1$

Proof of Theorem 7.6:

Proof:

(1) ⇒ (2): Absolute continuity implies BV. The derivative exists a.e. and is in L1. The integral representation follows from the fundamental theorem.

(2) ⇒ (3): If $E$ is a null set, then $f(E) = \{f(x) : x \in E\}$ has measure zero because $f$ is Lipschitz on sets where $|f'|$ is bounded, and we can approximate.

(3) ⇒ (4): This is the most technical step, requiring the construction of $g$ from the behavior of $f$ .

(4) ⇒ (1): This follows from the absolute continuity of the integral. ∎

∎

Example 7.4: Cantor Function: Detailed Analysis

Problem: Analyze the Cantor function (devil's staircase) in terms of bounded variation and absolute continuity.

Solution:

The Cantor function $C$ is continuous, increasing, and constant on each removed interval in the Cantor set construction.

Bounded variation: Since $C$ is increasing, it is of bounded variation with $V_0^1(C) = C(1) - C(0) = 1$ .

Absolute continuity: The Cantor function is not absolutely continuous. To see this, note that $C' = 0$ almost everywhere (on the complement of the Cantor set), but $C(1) - C(0) = 1 \neq 0 = \int_0^1 C'(t) \, dt$ .

This violates the fundamental theorem of calculus for absolutely continuous functions, showing that $C$ is not absolutely continuous.

The Cantor function demonstrates that BV does not imply absolute continuity.

Example 7.5: Constructing a BV Function

Problem: Construct a function of bounded variation that is not absolutely continuous.

Solution:

The Cantor function (Example 7.4) is one example.

Another example: Let $f(x) = \begin{cases} 0 & x = 0 \\ x \sin(1/x) & 0 < x \leq 1 \end{cases}$ .

This function oscillates infinitely often near 0, but is still of bounded variation (with variation approximately 2). However, it is not absolutely continuous because it has unbounded derivative near 0.

More precisely, $f$ is differentiable a.e., but $f' \notin L^1$ near 0, so the fundamental theorem fails.

Example 7.6: Verifying Absolute Continuity

Problem: Show that $f(x) = x^2$ on $[0,1]$ is absolutely continuous.

Solution:

Method 1: Since $f'(x) = 2x$ is bounded on $[0,1]$ , $f$ is Lipschitz continuous, hence absolutely continuous.

Method 2: For any $\epsilon > 0$ , choose $\delta = \epsilon/2$ . For any collection of disjoint intervals $\{(a_i, b_i)\}$ with $\sum (b_i - a_i) < \delta$ , we have

\sum |f(b_i) - f(a_i)| = \sum |b_i^2 - a_i^2| = \sum (b_i + a_i)(b_i - a_i) \leq 2 \sum (b_i - a_i) < 2\delta = \epsilon

Therefore, $f$ is absolutely continuous.

Example 7.7: Non-Absolutely Continuous BV Function

Problem: Show that a step function with a jump discontinuity is of bounded variation but not absolutely continuous.

Solution:

Let $f(x) = \begin{cases} 0 & x < 1/2 \\ 1 & x \geq 1/2 \end{cases}$ on $[0,1]$ .

The total variation is $V_0^1(f) = 1$ (the size of the jump), so $f$ is of bounded variation.

However, $f$ is not absolutely continuous. To see this, take a single interval $(1/2 - \delta, 1/2 + \delta)$ with $\delta < \epsilon$ . The variation over this interval is 1, which does not tend to 0 as $\delta \to 0$ .

More formally, absolute continuity requires that the function can be made arbitrarily close to constant on small intervals, which fails at the jump.

Corollary 7.3: Properties of BV Functions

Functions of bounded variation have the following properties:

They are differentiable almost everywhere
They have at most countably many discontinuities (all jump discontinuities)
They can be approximated uniformly by step functions
The derivative (where it exists) is in L1 if and only if the function is absolutely continuous

Remark 7.3: Historical Significance of the Cantor Function

The Cantor function (devil's staircase) is one of the most important counterexamples in analysis:

Continuous but not absolutely continuous: It shows that continuity and bounded variation do not imply absolute continuity.
Constant derivative almost everywhere: It has derivative 0 almost everywhere, yet increases from 0 to 1, showing that the fundamental theorem of calculus can fail for BV functions.
Singular functions: It is an example of a singular function—continuous, increasing, but with derivative zero almost everywhere.
Measure theory: It demonstrates the distinction between topological and measure-theoretic properties.

The Cantor function was constructed by Georg Cantor in the late 19th century and has been fundamental in understanding the relationship between differentiation and integration.

Remark 7.4: Applications in Functional Analysis

BV and AC functions are important in functional analysis:

BV spaces: The space of BV functions forms a Banach space under appropriate norms, important in the study of functions of one variable.
AC and Sobolev spaces: Absolutely continuous functions are closely related to Sobolev spaces W1,1, which are fundamental in PDE theory.
Riesz representation: BV functions appear in the Riesz representation theorem for the dual of C([a,b]).
Stieltjes integrals: BV functions are exactly those for which the Riemann-Stieltjes integral is well-defined for all continuous functions.

These connections show that BV and AC are not just technical conditions, but fundamental concepts in analysis.

Example 7.8: BV Function with Non-Integrable Derivative

Problem: Construct a function of bounded variation whose derivative exists almost everywhere but is not integrable.

Solution:

The Cantor function $C$ is of bounded variation (it's increasing) and differentiable almost everywhere with $C' = 0$ a.e.

However, $\int_0^1 C'(x) \, dx = 0$ while $C(1) - C(0) = 1$ , so the fundamental theorem fails.

More precisely, $C' \in L^1$ (it's identically 0), but $C(x) \neq C(0) + \int_0^x C'(t) \, dt$ , showing that $C$ is not absolutely continuous.

This demonstrates that BV does not imply absolute continuity, and that the integrability of the derivative is not automatic for BV functions.

Example 7.9: Absolute Continuity: Verification Example

Problem: Show that $f(x) = x^3$ on $[0,1]$ is absolutely continuous.

Solution:

Since $f'(x) = 3x^2$ is bounded on $[0,1]$ , $f$ is Lipschitz continuous, hence absolutely continuous.

Alternatively, we can verify directly: for any collection of disjoint intervals $\{(a_i, b_i)\}$ with $\sum (b_i - a_i) < \delta$ ,

\sum |f(b_i) - f(a_i)| = \sum |b_i^3 - a_i^3| = \sum (b_i^2 + a_i b_i + a_i^2)(b_i - a_i) \leq 3 \sum (b_i - a_i) < 3\delta

So for $\epsilon > 0$ , choose $\delta = \epsilon/3$ to verify absolute continuity.

Theorem 7.7: AC Functions and L1 Derivatives

A function $f$ on $[a,b]$ is absolutely continuous if and only if:

$f$ is of bounded variation
$f$ is differentiable almost everywhere
$f' \in L^1([a,b])$
$f(x) = f(a) + \int_a^x f'(t) \, dt$ for all $x \in [a,b]$

Proof of Theorem 7.7:

Proof:

The "if" direction: If $f$ satisfies (1)-(4), then absolute continuity follows from the absolute continuity of the integral (Theorem 3.6).

The "only if" direction: If $f$ is absolutely continuous, then it is BV (Corollary 7.1), differentiable a.e. (as BV functions are), and the derivative is in L1 with the integral representation (Theorem 7.3). ∎

∎

Corollary 7.4: AC Functions are Differentiable a.e.

Every absolutely continuous function is differentiable almost everywhere, and its derivative is in L1.

This is stronger than the corresponding result for BV functions, where the derivative may not be integrable.

Remark 7.5: BV and AC in Multiple Variables

The concepts of bounded variation and absolute continuity extend to functions of several variables:

BV functions: Functions whose distributional derivatives are finite measures. Important in geometric measure theory and image processing.
AC functions: Functions whose distributional derivatives are absolutely continuous with respect to Lebesgue measure. These are functions in Sobolev spaces W1,1.
Applications: BV and AC functions in multiple variables are fundamental in PDE theory, especially for problems involving free boundaries and interfaces.

These extensions demonstrate the broad applicability of BV and AC concepts beyond one-dimensional analysis.

Example 7.10: BV Function: Computing Total Variation

Problem: Compute the total variation of $f(x) = \sin(x)$ on $[0, 2\\pi]$ .

Solution:

Since $f'(x) = \cos(x)$ is bounded, $f$ is Lipschitz and hence of bounded variation.

The total variation is the total "up and down" movement. For $\sin(x)$ on $[0, 2\\pi]$ , it goes from 0 to 1, down to -1, and back to 0.

More precisely, $V_0^{2\pi}(f) = \int_0^{2\pi} |f'(x)| \, dx = \int_0^{2\pi} |\cos(x)| \, dx = 4$ .

This equals the total "length" of the graph of the function, which for a smooth function equals $\int \sqrt{1 + (f')^2}$ , but for this calculation, $\int |f'|$ suffices.

Example 7.11: AC vs BV: A Concrete Comparison

Problem: Compare the Cantor function (BV but not AC) with $f(x) = x$ (both BV and AC) on $[0,1]$ .

Solution:

Cantor function $C$ :

BV: Yes (it's increasing, so $V_0^1(C) = C(1) - C(0) = 1$ )
AC: No ( $C' = 0$ a.e. but $C(1) - C(0) = 1 \neq 0 = \int_0^1 C'$ )
Differentiable: Yes a.e. ( $C' = 0$ a.e.)
Derivative integrable: Yes ( $\int C' = 0$ ), but fundamental theorem fails

Function $f(x) = x$ :

BV: Yes ( $V_0^1(f) = 1$ )
AC: Yes ( $f' = 1$ is bounded, so Lipschitz)
Differentiable: Yes everywhere ( $f' = 1$ )
Fundamental theorem: Holds ( $f(1) - f(0) = 1 = \int_0^1 f'$ )

Theorem 7.8: BV Functions and Discontinuities

A function of bounded variation has at most countably many discontinuities, and all discontinuities are jump discontinuities.

Moreover, the sum of the jump sizes is at most the total variation.

Proof of Theorem 7.8:

Proof:

By the Jordan decomposition, $f = g - h$ where $g$ and $h$ are increasing.

Increasing functions have at most countably many discontinuities (all jumps), and the sum of jumps is bounded by the total variation.

Since $f$ is the difference of two such functions, it inherits these properties. ∎

∎

Corollary 7.5: AC Functions are Continuous

Every absolutely continuous function is continuous (and even uniformly continuous on compact intervals).

This is stronger than BV functions, which may have jump discontinuities.

Remark 7.6: BV and AC in Signal Processing

BV and AC functions have important applications in signal and image processing:

Total variation denoising: Minimizing the total variation of an image while fitting data is a powerful denoising technique.
Edge preservation: BV functions can have sharp edges (discontinuities), making them suitable for modeling images with clear boundaries.
Regularization: The total variation is used as a regularizer in inverse problems, promoting piecewise constant or smooth solutions.
Compressed sensing: BV and related concepts are used in compressed sensing and sparse recovery.

These applications show that BV and AC are not just abstract mathematical concepts, but have practical importance in modern technology.

Remark 7.1: Key Insights

Key takeaways:

Bounded variation captures the 'total oscillation' of a function
Jordan decomposition shows BV functions are differences of increasing functions
Absolute continuity is the precise condition for the fundamental theorem
AC functions are exactly indefinite integrals of L1 functions
BV functions are differentiable a.e., but derivatives may not be integrable

Practice Quiz

Bounded Variation & Absolute Continuity

Questions

Correct

Accuracy

A function

f

[a,b]

has bounded variation if:

Easy

Not attempted

The Jordan decomposition theorem states that a function of bounded variation can be written as:

Medium

Not attempted

A function

f

[a,b]

is absolutely continuous if:

Medium

Not attempted

The Lebesgue fundamental theorem of calculus states that if

f \in L^1([a,b])

and

F(x) = \int_a^x f

, then:

Hard

Not attempted

f

is absolutely continuous, then:

Medium

Not attempted

The total variation function

V_a^x(f)

is:

Easy

Not attempted

A function is absolutely continuous if and only if:

Hard

Not attempted

The Cantor function is:

Hard

Not attempted

f

is absolutely continuous and

f' = 0

a.e., then:

Medium

Not attempted

The relationship between BV, AC, and differentiability is:

Hard

Not attempted

Frequently Asked Questions

What is the geometric meaning of bounded variation?

A function has bounded variation if the total 'length' of its graph is finite. More precisely, if you trace the graph, the total vertical distance traveled (counting both up and down) is bounded.

What is the difference between continuity and absolute continuity?

Continuity controls the function at individual points. Absolute continuity is a uniform condition that controls the function's variation over collections of intervals. It's much stronger than continuity and is equivalent to being an indefinite integral.

Why is absolute continuity important?

Absolute continuity is exactly the condition needed for the fundamental theorem of calculus to hold in the Lebesgue setting. It characterizes functions that are integrals of L1 functions.

What is the Jordan decomposition?

The Jordan decomposition shows that any function of bounded variation can be written as the difference of two increasing functions. This is useful because increasing functions are well-behaved and easier to work with.

Can a function be of bounded variation but not absolutely continuous?

Yes. The Cantor function is a classic example: it's continuous and increasing (hence BV) but not absolutely continuous, as it has zero derivative almost everywhere yet is not constant.

What is the relationship between absolute continuity and the fundamental theorem of calculus?

A function is absolutely continuous if and only if it is an indefinite integral of an L1 function. This gives a complete characterization: $F$ is AC if and only if $F(x) = F(a) + \int_a^x f(t) \, dt$ for some $f \in L^1$.

Do all functions of bounded variation have derivatives?

Functions of bounded variation are differentiable almost everywhere, but the derivative may not be integrable. Absolutely continuous functions have derivatives that are in L1.