Lebesgue Measure and Integral on R

Integrals of Bounded Measurable Functions

In real analysis, the Lebesgue integral is a powerful generalization of the Riemann integral. It provides a more robust framework for integration, particularly for functions that are not well-behaved in the Riemann sense. The foundation of the Lebesgue integral lies in the concept of measure theory.

Before defining the integral, we need to understand the concept of a measurable set and a measurable function. A set $E \subseteq \mathbb{R}$ is called Lebesgue measurable if it belongs to the Lebesgue $\sigma$-algebra, denoted by $\mathcal{M}$. The Lebesgue measure, denoted by $m(E)$, assigns a non-negative value to each measurable set, representing its "size" or "length" in $\mathbb{R}$. For intervals, the Lebesgue measure coincides with their length. For example, $m((a, b)) = m([a, b]) = m((a, b]) = m([a, b)) = b - a$.

A function $f: \mathbb{R} \to \mathbb{R}$ is called measurable if for every real number $c$, the set $\{x \in \mathbb{R} : f(x) > c\}$ is a measurable set. This condition ensures that we can meaningfully assign a measure to the "level sets" of the function.

Now, let's consider the integral of a bounded measurable function $f$ over a measurable set $E$ with finite measure, i.e., $m(E) < \infty$. We first define the Lebesgue integral for simple functions. A simple function $\phi$ is a finite linear combination of characteristic functions of disjoint measurable sets: $\phi(x) = \sum_{i=1}^n a_i \chi_{A_i}(x)$, where $a_i \in \mathbb{R}$ and $A_i$ are disjoint measurable sets. The Lebesgue integral of a simple function $\phi$ over $E$ is defined as: $$ \int_E \phi \, dm = \sum_{i=1}^n a_i m(A_i \cap E) $$

For a non-negative measurable function $f: E \to [0, \infty)$, its Lebesgue integral is defined as the supremum of the integrals of all simple functions $\phi$ such that $0 \le \phi \le f$: $$ \int_E f \, dm = \sup \left\{ \int_E \phi \, dm : \phi \text{ is simple}, 0 \le \phi \le f \right\} $$

If $f$ is any measurable function, we can decompose it into its positive and negative parts: $f = f^+ - f^-$, where $f^+(x) = \max(f(x), 0)$ and $f^-(x) = \max(-f(x), 0)$. Both $f^+$ and $f^-$ are non-negative measurable functions. The Lebesgue integral of $f$ over $E$ is defined as: $$ \int_E f \, dm = \int_E f^+ \, dm - \int_E f^- \, dm $$ provided that at least one of the integrals on the right-hand side is finite. If both are infinite, the integral is undefined.

For a bounded measurable function $f$ on a measurable set $E$ with finite measure, $m(E) < \infty$, and $|f(x)| \le M$ for some constant $M$, the integral is well-defined. In this case, we can also define the integral using partitions, similar to the Riemann integral. Let $P = \{x_0, x_1, \dots, x_n\}$ be a partition of an interval $[a, b]$ containing $E$. Let $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x)$ and $M_i = \sup_{x \in [x_{i-1}, x_i]} f(x)$. The lower Riemann sum is $L(f, P) = \sum_{i=1}^n m_i (x_i - x_{i-1})$ and the upper Riemann sum is $U(f, P) = \sum_{i=1}^n M_i (x_i - x_{i-1})$. The function $f$ is Riemann integrable if $\sup_P L(f, P) = \inf_P U(f, P)$. This common value is the Riemann integral $\int_a^b f(x) \, dx$.

For a bounded measurable function $f$ on a measurable set $E$ with finite measure, the Lebesgue integral $\int_E f \, dm$ is defined by considering partitions of the range of $f$. Let $f(E) = \{y : y = f(x) \text{ for some } x \in E\}$. Let $y_0 < y_1 < \dots < y_n$ be a partition of the range of $f$. Let $E_i = \{x \in E : y_{i-1} < f(x) \le y_i\}$. The Lebesgue integral can be approximated by $\sum_{i=1}^n y_i^* m(E_i)$, where $y_i^*$ is some value in $(y_{i-1}, y_i]$. For a bounded measurable function, this sum converges to the Lebesgue integral as the partition of the range becomes finer.

Consider a function $f(x) = x^2$ on the interval $[0, 1]$. This function is continuous and hence Riemann integrable. Its Riemann integral is $\int_0^1 x^2 \, dx = [x^3/3]_0^1 = 1/3$. To compute the Lebesgue integral, we consider partitions of $[0, 1]$. For a partition $P = \{0=x_0, x_1, \dots, x_n=1\}$, let $m_i = \inf_{x \in [x_{i-1}, x_i]} x^2 = x_{i-1}^2$ and $M_i = \sup_{x \in [x_{i-1}, x_i]} x^2 = x_i^2$. The Lebesgue integral of $f$ over $[0, 1]$ is defined as the supremum of integrals of simple functions less than or equal to $f$. For a bounded measurable function, if it is Riemann integrable, its Lebesgue integral is equal to its Riemann integral.

Comparison of Riemann and Lebesgue Integrals

The Lebesgue integral is a significant improvement over the Riemann integral due to its broader applicability and stronger convergence theorems. The fundamental difference lies in how they "partition" the domain and range of the function.

The Riemann integral partitions the domain of the function into small intervals and approximates the area using rectangles whose heights are determined by the function's values (supremum or infimum) within those intervals. This approach works well for continuous or piecewise continuous functions. However, it struggles with functions that have many discontinuities.

The Lebesgue integral, on the other hand, partitions the range of the function. It considers the measure of the sets where the function takes values within certain ranges. For a function $f$, it asks: "What is the total measure of the set of points $x$ for which $f(x)$ lies in a particular interval?" This perspective allows it to handle a much wider class of functions, including those with "pathological" discontinuities.

Key Differences:

• Class of Integrable Functions: Every Riemann integrable function on a closed interval $[a, b]$ is Lebesgue integrable on $[a, b]$, and their integrals are equal. However, there exist Lebesgue integrable functions that are not Riemann integrable. A classic example is the Dirichlet function, $f(x) = 1$ if $x \in \mathbb{Q}$ and $f(x) = 0$ if $x \notin \mathbb{Q}$. This function is not Riemann integrable on any interval because any subinterval contains both rational and irrational numbers, making the upper Riemann sums always 1 and the lower Riemann sums always 0. However, it is Lebesgue integrable, and its integral over $[0, 1]$ is 0, because the set of rational numbers has Lebesgue measure zero.
• Convergence Theorems: The Lebesgue integral boasts powerful convergence theorems like the Monotone Convergence Theorem and the Dominated Convergence Theorem. These theorems provide conditions under which the limit of a sequence of functions can be interchanged with the integral. Such theorems are crucial in advanced analysis and probability theory, and their counterparts for the Riemann integral are much weaker or do not exist.
• Measurability Requirement: The Lebesgue integral requires the function to be measurable and the domain of integration to be a measurable set. The Riemann integral implicitly assumes the domain is an interval and does not explicitly require measurability in the same way.

Consider the Dirichlet function $f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}$ on $[0, 1]$. For Riemann integration: Let $P = \{x_0, x_1, \dots, x_n\}$ be any partition of $[0, 1]$. In each subinterval $[x_{i-1}, x_i]$, there are both rational and irrational numbers. The supremum of $f(x)$ in $[x_{i-1}, x_i]$ is $M_i = 1$ (since there's a rational number). The infimum of $f(x)$ in $[x_{i-1}, x_i]$ is $m_i = 0$ (since there's an irrational number). Upper Riemann sum: $U(f, P) = \sum_{i=1}^n M_i (x_i - x_{i-1}) = \sum_{i=1}^n 1 \cdot (x_i - x_{i-1}) = 1$. Lower Riemann sum: $L(f, P) = \sum_{i=1}^n m_i (x_i - x_{i-1}) = \sum_{i=1}^n 0 \cdot (x_i - x_{i-1}) = 0$. Since $\inf_P U(f, P) = 1$ and $\sup_P L(f, P) = 0$, the function is not Riemann integrable.

For Lebesgue integration: The function $f(x)$ is measurable. The set where $f(x) = 1$ is $\mathbb{Q} \cap [0, 1]$, which has Lebesgue measure $m(\mathbb{Q} \cap [0, 1]) = 0$. The set where $f(x) = 0$ is $[0, 1] \setminus \mathbb{Q}$, which has Lebesgue measure $m([0, 1] \setminus \mathbb{Q}) = 1 - 0 = 1$. We can represent $f$ as $f(x) = 1 \cdot \chi_{\mathbb{Q} \cap [0, 1]}(x) + 0 \cdot \chi_{([0, 1] \setminus \mathbb{Q})}(x)$. The Lebesgue integral is $\int_{[0, 1]} f \, dm = 1 \cdot m(\mathbb{Q} \cap [0, 1]) + 0 \cdot m([0, 1] \setminus \mathbb{Q}) = 1 \cdot 0 + 0 \cdot 1 = 0$. Thus, the Dirichlet function is Lebesgue integrable with integral 0.

Formal Statement: If $f: [a, b] \to \mathbb{R}$ is Riemann integrable, then $f$ is Lebesgue integrable on $[a, b]$, and $\int_a^b f(x) \, dx = \int_{[a, b]} f \, dm$. The converse is not true.

Monotone Convergence Theorem (MCT)

The Monotone Convergence Theorem is one of the most fundamental and powerful results in Lebesgue integration theory. It provides a condition under which the integral of a limit of a sequence of functions can be interchanged with the limit operation.

Statement of the Theorem: Let $\{f_n\}_{n=1}^\infty$ be a sequence of non-negative measurable functions defined on a measurable set $E$. Suppose that the sequence is pointwise non-decreasing, meaning $f_n(x) \le f_{n+1}(x)$ for all $n \in \mathbb{N}$ and for all $x \in E$. Let $f(x) = \lim_{n \to \infty} f_n(x)$ be the pointwise limit of the sequence. Then $f$ is a measurable function, and its Lebesgue integral is equal to the limit of the integrals of the sequence: $$ \int_E f \, dm = \lim_{n \to \infty} \int_E f_n \, dm $$ Furthermore, if the sequence is non-decreasing and bounded above by some integrable function $g$ (i.e., $f_n(x) \le g(x)$ for all $n$ and $x \in E$, and $\int_E g \, dm < \infty$), then the limit $f$ is also integrable.

Explanation and Significance: The theorem essentially states that if you have a sequence of non-negative measurable functions that are "growing" towards a limit function, then the integral of the limit is the limit of the integrals. This is incredibly useful because it often allows us to compute the integral of a complicated limit function by computing the integrals of simpler functions in the sequence.

The condition that the functions must be non-negative is crucial. The theorem does not hold for arbitrary sequences of functions. The non-decreasing nature ensures that the sequence of integrals $\int_E f_n \, dm$ is itself a non-decreasing sequence of real numbers (or possibly $\infty$), which is bounded below by 0. A non-decreasing sequence of real numbers always converges (possibly to $\infty$).

Example: Calculating the integral of a step function using MCT. Consider the function $f(x) = 1$ for $x \in [0, 1]$ and $f(x) = 0$ otherwise. We want to calculate $\int_{\mathbb{R}} f \, dm$. We can construct a sequence of non-negative measurable functions that converges to $f$ and is non-decreasing. Let $f_n(x)$ be the characteristic function of the interval $[0, 1 - 1/n]$. That is, $f_n(x) = 1$ if $x \in [0, 1 - 1/n]$ and $f_n(x) = 0$ otherwise. For $n \ge 1$, we have $1 - 1/(n+1) < 1 - 1/n$. So, if $x \in [0, 1 - 1/(n+1)]$, then $x \in [0, 1 - 1/n]$, which means $f_{n+1}(x) = 1$ implies $f_n(x) = 1$. Thus, $f_n(x) \le f_{n+1}(x)$ for all $x$. The sequence $\{f_n\}$ is pointwise non-decreasing. The pointwise limit is $f(x) = \lim_{n \to \infty} f_n(x)$. For $x \in [0, 1)$, $f_n(x) = 1$ for sufficiently large $n$, so $f(x) = 1$. For $x \ge 1$, $f_n(x) = 0$ for all $n$, so $f(x) = 0$. For $x < 0$, $f_n(x) = 0$ for all $n$, so $f(x) = 0$. The limit function is $f(x) = \chi_{[0, 1)}(x)$. The integral of each $f_n$ is $\int_{\mathbb{R}} f_n \, dm = m([0, 1 - 1/n]) = 1 - 1/n$. By the Monotone Convergence Theorem: $$ \int_{\mathbb{R}} f \, dm = \lim_{n \to \infty} \int_{\mathbb{R}} f_n \, dm = \lim_{n \to \infty} (1 - 1/n) = 1 $$ The limit function we constructed was $f(x) = \chi_{[0, 1)}(x)$, whose integral is $m([0, 1)) = 1$. This matches the result from MCT. Note that if we wanted the integral of $\chi_{[0, 1]}$, we could use a sequence like $f_n(x) = \chi_{[0, 1 + 1/n]}(x)$.

Proof Sketch of MCT: 1. Measurability of $f$: Since each $f_n$ is measurable, the set $\{x : f(x) > c\} = \bigcup_{n=1}^\infty \{x : f_n(x) > c\}$ is a union of measurable sets, hence measurable. 2. Inequality $\int_E f \, dm \ge \lim_{n \to \infty} \int_E f_n \, dm$: Since $f_n \le f$ and the sequence is non-decreasing, we have $\int_E f_n \, dm \le \int_E f \, dm$ for all $n$. Taking the limit, $\lim_{n \to \infty} \int_E f_n \, dm \le \int_E f \, dm$. 3. Inequality $\int_E f \, dm \le \lim_{n \to \infty} \int_E f_n \, dm$: This is the harder part. Let $\phi$ be any simple function such that $0 \le \phi \le f$. We need to show that $\int_E \phi \, dm \le \lim_{n \to \infty} \int_E f_n \, dm$. For any $\epsilon > 0$, consider the sets $E_n = \{x \in E : f_n(x) \ge \phi(x)\}$. Since $f_n$ is non-decreasing and converges to $f$, $E_n$ is an increasing sequence of sets, and $\bigcup_{n=1}^\infty E_n = E$. We have $\int_E f_n \, dm \ge \int_{E_n} f_n \, dm \ge \int_{E_n} \phi \, dm$. Also, $\int_E \phi \, dm = \lim_{n \to \infty} \int_{E_n} \phi \, dm$ because $\bigcup E_n = E$. So, $\lim_{n \to \infty} \int_E f_n \, dm = \lim_{n \to \infty} \int_{E_n} f_n \, dm \ge \lim_{n \to \infty} \int_{E_n} \phi \, dm = \int_E \phi \, dm$. Since this holds for any simple function $\phi \le f$, by taking the supremum over all such $\phi$, we get $\int_E f \, dm \le \lim_{n \to \infty} \int_E f_n \, dm$. 4. Combining both inequalities yields the equality.

Repeated Integrals (Iterated Integrals)

Repeated integrals, also known as iterated integrals, are integrals computed sequentially. In the context of Lebesgue integration, particularly over $\mathbb{R}^n$, this concept leads to Fubini's Theorem, which relates multiple integrals to iterated integrals. However, the question specifically mentions "repeated integrals" in the context of Lebesgue measure on $\mathbb{R}$, which usually implies integrating a function $f(x, y)$ over a region in $\mathbb{R}^2$ by integrating with respect to one variable at a time. If we are strictly on $\mathbb{R}$, it might refer to integrating a function $f(x)$ multiple times, or perhaps integrating a function of two variables $f(x, y)$ where the domain is a subset of $\mathbb{R}^2$. Let's assume the context is integration over $\mathbb{R}^2$.

Consider a function $f(x, y)$ defined on a rectangle $R = [a, b] \times [c, d]$ in $\mathbb{R}^2$. The double Lebesgue integral of $f$ over $R$ is denoted by $\int_R f \, dm$, where $m$ is the Lebesgue measure on $\mathbb{R}^2$. Fubini's Theorem provides conditions under which the double integral can be calculated by iterated integration: $$ \iint_R f \, dm = \int_a^b \left( \int_c^d f(x, y) \, dy \right) \, dx = \int_c^d \left( \int_a^b f(x, y) \, dx \right) \, dy $$

Fubini's Theorem Statement: Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a function such that its absolute value $|f|$ is integrable with respect to the Lebesgue measure on $\mathbb{R}^2$ over a measurable set $E \subseteq \mathbb{R}^2$. That is, $\iint_E |f(x, y)| \, dm(x, y) < \infty$. Then, the iterated integrals exist almost everywhere and are equal to the double integral: $$ \iint_E f \, dm = \int_{E_x} \left( \int_{E_y(x)} f(x, y) \, dm(y) \right) \, dm(x) = \int_{E_y} \left( \int_{E_x(y)} f(x, y) \, dm(x) \right) \, dm(y) $$ where $E_x = \{x : \exists y \text{ such that } (x, y) \in E\}$ and $E_y(x) = \{y : (x, y) \in E\}$, and similarly for $E_y$ and $E_x(y)$.

Tonelli's Theorem: If $f(x, y)$ is a non-negative measurable function on $E$, then the iterated integrals are equal, regardless of whether the double integral is finite: $$ \int_{E_x} \left( \int_{E_y(x)} f(x, y) \, dm(y) \right) \, dm(x) = \int_{E_y} \left( \int_{E_x(y)} f(x, y) \, dm(x) \right) \, dm(y) $$ and this common value is equal to $\iint_E f \, dm$ (which may be $\infty$).

Practical Use: Tonelli's theorem is often used first. If $f$ is non-negative, we can compute the iterated integrals. If they are finite, then by Fubini's theorem, $f$ is integrable, and the double integral equals the iterated integral. If the iterated integrals are infinite, the double integral is also infinite. If $f$ is not necessarily non-negative, we first apply Tonelli's theorem to $|f|$. If $\iint_E |f| \, dm < \infty$, then Fubini's theorem applies, and we can compute the iterated integrals of $f$ (not $|f|$).

Example: Calculating a double integral using Fubini's Theorem. Calculate the integral of $f(x, y) = x^2 y$ over the rectangle $R = [0, 1] \times [0, 2]$. First, check the integrability of $|f(x, y)| = |x^2 y| = x^2 y$ over $R$ (since $x, y \ge 0$). Using Tonelli's theorem (since $f$ is non-negative): Inner integral with respect to $y$: $$ \int_0^2 x^2 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_0^2 = x^2 \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = x^2 \cdot 2 = 2x^2 $$ Outer integral with respect to $x$: $$ \int_0^1 2x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^1 = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 2 \cdot \frac{1}{3} = \frac{2}{3} $$ So, the iterated integral is $2/3$. Now, let's compute the other iterated integral: Inner integral with respect to $x$: $$ \int_0^1 x^2 y \, dx = y \left[ \frac{x^3}{3} \right]_0^1 = y \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = y \cdot \frac{1}{3} = \frac{y}{3} $$ Outer integral with respect to $y$: $$ \int_0^2 \frac{y}{3} \, dy = \frac{1}{3} \left[ \frac{y^2}{2} \right]_0^2 = \frac{1}{3} \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \frac{1}{3} \cdot 2 = \frac{2}{3} $$ Since both iterated integrals are equal to $2/3$, and $f(x, y) = x^2 y \ge 0$ on $R$, Tonelli's theorem guarantees that $f$ is integrable over $R$ and $\iint_R x^2 y \, dm = 2/3$.

When Fubini's Theorem Fails: Fubini's theorem requires the integral of $|f|$ to be finite. If $\iint_E |f| \, dm = \infty$, the iterated integrals may not be equal, or one or both may not exist. Consider $f(x, y) = \frac{x^2 - y^2}{(x^2 + y^2)^2}$ on the square $R = [0, 1] \times [0, 1]$. Let's compute the iterated integrals: $$ \int_0^1 \left( \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dx \right) \, dy $$ The inner integral $\int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dx$ can be evaluated. Consider the function $g(x) = \frac{x}{x^2+y^2}$. Then $g'(x) = \frac{(x^2+y^2)(2x) - x(2x)}{(x^2+y^2)^2} = \frac{2x^3 + 2xy^2 - 2x^3}{(x^2+y^2)^2} = \frac{2xy^2}{(x^2+y^2)^2}$. This is not quite what we have. Let's consider the derivative of $\frac{x}{x^2+y^2}$ with respect to $x$. It is $\frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$. So, $\int \frac{x^2-y^2}{(x^2+y^2)^2} dx = - \int \frac{y^2-x^2}{(x^2+y^2)^2} dx = - \frac{x}{x^2+y^2}$. Inner integral: $$ \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dx = \left[ -\frac{x}{x^2 + y^2} \right]_{x=0}^{x=1} = -\frac{1}{1 + y^2} - (-\frac{0}{0 + y^2}) = -\frac{1}{1 + y^2} $$ Outer integral: $$ \int_0^1 \left( -\frac{1}{1 + y^2} \right) \, dy = -[\arctan(y)]_0^1 = -(\arctan(1) - \arctan(0)) = -(\pi/4 - 0) = -\pi/4 $$ Now, let's compute the other iterated integral: $$ \int_0^1 \left( \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy \right) \, dx $$ By symmetry, replacing $x$ with $y$ and $y$ with $x$ in the previous calculation, we get: Inner integral: $\int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy = \left[ -\frac{y}{x^2 + y^2} \right]_{y=0}^{y=1} = -\frac{1}{x^2 + 1} - (-\frac{0}{x^2 + 0}) = -\frac{1}{x^2 + 1}$ Outer integral: $$ \int_0^1 \left( -\frac{1}{x^2 + 1} \right) \, dx = -[\arctan(x)]_0^1 = -(\arctan(1) - \arctan(0)) = -(\pi/4 - 0) = -\pi/4 $$ In this case, the iterated integrals are equal. However, the function $f(x, y)$ is not integrable over $R$ because $\iint_R |f(x, y)| \, dm = \infty$. Fubini's theorem does not apply, and the equality of iterated integrals is coincidental (related to the origin being a point of singularity).

A more standard example where Fubini fails is $f(x, y) = \frac{x-y}{(x+y)^3}$ on $(0, 1] \times (0, 1]$. $\int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dx dy = 1/2$ $\int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} dy dx = -1/2$ Here, $\iint |f| \, dm = \infty$, so Fubini's theorem does not apply, and the iterated integrals differ.