1. Concept Limits of Riemann Integration
In elementary calculus, we learn Riemann integration. The domain \([a, b]\) is divided into subintervals by a partition \(P = \{x_0, x_1, \dots, x_n\}\). We construct upper and lower Darboux sums based on the supremum (\(M_i\)) and infimum (\(m_i\)) of the function on each subinterval: \[U(P, f) = \sum_{i=1}^n M_i \Delta x_i \quad \text{and} \quad L(P, f) = \sum_{i=1}^n m_i \Delta x_i\] The function is Riemann integrable if the supremum of all lower sums equals the infimum of all upper sums.
While simple and intuitive, this formulation has two major weaknesses:
1. Fails on highly discontinuous functions: If a function oscillates too rapidly, the upper and lower sums never meet.
2. Poor limit interchange behavior: If a sequence of Riemann integrable functions \(f_n(x)\) converges pointwise to a function \(f(x)\), \(f(x)\) is not guaranteed to be Riemann integrable, making it difficult to write \(\lim_{n \rightarrow \infty} \int f_n dx = \int \lim_{n \rightarrow \infty} f_n dx\).
2. Lebesgue Integration: Partitioning the Range
Henri Lebesgue solved these limitations in 1902 by partitioning the range (y-axis) rather than the domain.
Suppose the range of a bounded function \(f(x)\) lies in \([m, M]\). We partition the range: \(m = y_0 < y_1 < \dots < y_r = M\).
Instead of looking at arbitrary intervals on the x-axis, we look at the set of all \(x\) values that map into a specific slice of the range:
\[E_i = \{x \in [a, b] \mid y_{i-1} \le f(x) < y_i\}\]
We then sum the values of the range weighted by the "size" of their preimages:
\[\text{Lebesgue Sum} \approx \sum_{i=1}^r y_i \mu(E_i)\]
where \(\mu(E_i)\) is the Lebesgue measure of the set \(E_i\). This measure is a rigorous generalization of length.
To integrate general functions, we start by integrating simple functions (functions that take a finite number of values \(a_k\) on measurable sets \(A_k\)): \[\int f d\mu = \sum_{k=1}^m a_k \mu(A_k)\] For any non-negative measurable function \(f\), its Lebesgue integral is defined as the supremum of the integrals of all simple functions \(s\) that lie below it: \[\int f d\mu = \sup \left\{ \int s d\mu \ \middle|\ 0 \le s \le f, \ s \text{ is simple} \right\}\]
3. The Classic Counterexample: The Dirichlet Function
The difference between the two integration methods is best demonstrated by the Dirichlet Function on the interval \([0, 1]\): \[f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}\]
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Under Riemann Integration: Since both rational and irrational numbers are dense in \(\mathbb{R}\), any subinterval \([x_{i-1}, x_i]\) contains both rational and irrational numbers.
Therefore, the supremum is \(M_i = 1\) and the infimum is \(m_i = 0\) for every subinterval.
This yields \(U(P, f) = 1 \cdot (1 - 0) = 1\) and \(L(P, f) = 0 \cdot (1 - 0) = 0\) for all partitions \(P\).
Since \(0 \neq 1\), the function is not Riemann integrable. -
Under Lebesgue Integration: The function \(f\) takes only two values, 1 and 0, so it is a simple function:
\[f = 1 \cdot \chi_{\mathbb{Q} \cap [0,1]} + 0 \cdot \chi_{[0,1] \setminus \mathbb{Q}}\]
The set of rational numbers in \([0, 1]\) is countable, so its Lebesgue measure is \(\mu(\mathbb{Q} \cap [0,1]) = 0\).
The set of irrationals has measure \(\mu([0,1] \setminus \mathbb{Q}) = 1 - 0 = 1\).
The Lebesgue integral is: \[\int_{[0,1]} f d\mu = 1 \cdot \mu(\mathbb{Q} \cap [0,1]) + 0 \cdot \mu([0,1] \setminus \mathbb{Q}) = 1 \cdot 0 + 0 \cdot 1 = 0\] The integral exists and is exactly 0.
4. Solved CSIR-NET & GATE Questions
Question 1 (CSIR-NET Mathematics)
Let \(f: [0, 1] \rightarrow \mathbb{R}\) be defined by \(f(x) = x^3\) for irrational \(x\), and \(f(x) = \cos(x)\) for rational \(x\). Find the Lebesgue integral \(\int_0^1 f(x) d\mu\).
Detailed Solution:
- Identify the set of points where \(f(x) \ne x^3\):
The function \(f(x)\) differs from \(x^3\) only when \(x\) is a rational number. \[D = \{x \in [0, 1] \mid f(x) \ne x^3\} = \mathbb{Q} \cap [0, 1]\] - Evaluate the measure of this set:
The set of all rational numbers is countable, and any countable set has a Lebesgue measure of 0. \[\mu(D) = 0\] - Apply Lebesgue integral properties:
If two functions \(f\) and \(g\) are equal almost everywhere (meaning they differ only on a set of measure zero), their Lebesgue integrals are identical. \[\text{Since } f(x) = x^3 \text{ almost everywhere (a.e.) on } [0, 1]\] \[\int_0^1 f(x) d\mu = \int_0^1 x^3 d\mu\] - Evaluate the standard Riemann integral, which matches the Lebesgue integral for continuous functions: \[\int_0^1 x^3 d\mu = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}\]
Answer: \(\int_0^1 f(x) d\mu = 1/4\).
Question 2 (GATE Mathematics)
State the Dominated Convergence Theorem (DCT) and use it to find the limit: \[\lim_{n \rightarrow \infty} \int_0^1 \frac{n \sin(x)}{1 + n^2 x^2} dx\]
Detailed Solution:
- State the Dominated Convergence Theorem:
Let \(\{f_n\}\) be a sequence of measurable functions converging pointwise almost everywhere to \(f\).
If there exists an integrable function \(g\) such that \(|f_n(x)| \le g(x)\) for all \(n\) and almost all \(x\), then \(f\) is integrable and: \[\lim_{n \rightarrow \infty} \int f_n d\mu = \int f d\mu\] - Identify the sequence of functions \(f_n(x)\) on \([0, 1]\): \[f_n(x) = \frac{n \sin(x)}{1 + n^2 x^2}\]
- Find the pointwise limit \(f(x)\) as \(n \rightarrow \infty\):
• For \(x = 0\): \(f_n(0) = 0 \implies \lim f_n(0) = 0\).
• For \(x \in (0, 1]\): Since the denominator contains \(n^2\) and the numerator contains \(n\): \[\lim_{n \rightarrow \infty} \frac{n \sin(x)}{1 + n^2 x^2} = \lim_{n \rightarrow \infty} \frac{\sin(x)}{\frac{1}{n} + n x^2} = 0\] Thus, \(f_n(x) \rightarrow 0\) pointwise for all \(x \in [0, 1]\). - Find a dominating integrable function \(g(x)\):
Using the inequality \(\sin(x) \le x\) for \(x \ge 0\), and the basic inequality \(2nx \le 1 + n^2 x^2 \implies \frac{n}{1+n^2 x^2} \le \frac{1}{2x}\):
Alternatively, we can write: \[|f_n(x)| = \left|\sin(x)\right| \frac{n}{1 + n^2 x^2} \le x \frac{n}{2nx} \text{ (not quite, let's use calculus to find maximum)}\] Let's find the maximum of \(h(u) = \frac{u}{1+u^2}\) where \(u = nx\):
The maximum value of \(\frac{u}{1+u^2}\) is \(1/2\) (achieved at \(u = 1\)).
Therefore: \[f_n(x) = \sin(x) \frac{n}{1 + n^2 x^2} = \frac{\sin(x)}{x} \frac{nx}{1 + n^2 x^2}\] Since \(0 \le \frac{\sin(x)}{x} \le 1\) on \([0, 1]\) and \(\frac{nx}{1 + n^2 x^2} \le \frac{1}{2}\): \[|f_n(x)| \le 1 \times \frac{1}{2} = \frac{1}{2}\] The constant function \(g(x) = 1/2\) is integrable on \([0, 1]\) (\(\int_0^1 \frac{1}{2} dx = \frac{1}{2} < \infty\)). - Apply DCT:
Since the conditions of DCT are satisfied, we can move the limit inside the integral: \[\lim_{n \rightarrow \infty} \int_0^1 f_n(x) dx = \int_0^1 \left( \lim_{n \rightarrow \infty} f_n(x) \right) dx = \int_0^1 0 dx = 0\]
Answer: The limit is 0.
Abdul Raheem K
Mathematics Stream Head | IIT Madras Alumni | GATE AIR-27Abdul Raheem K completed his MSc in Mathematics at IIT Madras and secured GATE AIR 27 and a CSIR JRF qualification. He directs the Real Analysis and Algebra teaching programs at Benzil Academy.