Technopedia

The technologist’s encyclopedia

Integration

Integration measures accumulated quantity — area, volume, total mass, expected value — and, in its Lebesgue form, provides the rigorous setting in which limits and integrals may be interchanged. It underlies probability, physics, and the solution of differential equations. This page covers the Riemann integral and its limitations, then the Lebesgue integral and its convergence theorems (monotone and dominated convergence).

Authoritative sources

The Riemann integral

Definition. A bounded function ff on [a,b][a, b] is Riemann integrable if the upper and lower sums agree in the limit: for every partition PP of [a,b][a, b] with mesh →0\to 0,

∫abf(x) dx=lim⁡∥P∥→0∑if(xi∗)Δxi.\int_a^b f(x)\, dx = \lim_{\|P\| \to 0} \sum_i f(x_i^*) \Delta x_i.

Details

The idea

Partition [a,b][a, b] into small subintervals, approximate ff on each by a constant (its value at a sample point xi∗x_i^*), and sum the areas of the resulting rectangles. As the mesh shrinks, the sum approaches the integral — the signed area under the curve.

Worked example

Compute ∫01x dx\int_0^1 x\, dx by Riemann sums. Take the right endpoints xi∗=i/nx_i^* = i/n on [i−1,i]/n[i-1, i]/n. Then

∑i=1nin⋅1n=1n2∑i=1ni=1n2⋅n(n+1)2→12.\sum_{i=1}^{n} \frac{i}{n} \cdot \frac{1}{n} = \frac{1}{n^2} \sum_{i=1}^{n} i = \frac{1}{n^2} \cdot \frac{n(n+1)}{2} \to \frac{1}{2}.

So ∫01x dx=1/2\int_0^1 x\, dx = 1/2.

The fundamental theorem of calculus

If FF is an antiderivative of ff (i.e. F′=fF' = f), then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\, dx = F(b) - F(a). This is the practical engine of integration: it reduces computing an area to finding an antiderivative.

Example problem

Problem. Compute ∫01x2 dx\int_0^1 x^2\, dx using the fundamental theorem of calculus.

Solution

An antiderivative of x2x^2 is F(x)=x3/3F(x) = x^3/3. Hence ∫01x2 dx=F(1)−F(0)=1/3−0=1/3\int_0^1 x^2\, dx = F(1) - F(0) = 1/3 - 0 = 1/3. ∎

Limitations of the Riemann integral

The Riemann integral fails on some natural limits. The classic example is the sequence fn(x)=1{x=1/n}f_n(x) = \mathbf{1}_{\{x = 1/n\}} on [0,1][0,1] (a single spike of height 1 at x=1/nx = 1/n): each fnf_n is Riemann integrable with integral 0, and fn→0f_n \to 0 pointwise, yet the Riemann framework gives no clean way to pass to the limit for more general pointwise limits. Worse, the indicator of the rationals 1Q\mathbf{1}_{\mathbb{Q}} is not Riemann integrable at all (upper sum 1, lower sum 0 on every interval), even though it is a perfectly reasonable pointwise limit of integrable functions.

The Riemann integral is also awkward for functions with “too many” discontinuities and for interchanging limits and integrals. The Lebesgue integral fixes both problems.

The Lebesgue integral

Definition (sketch). The Lebesgue integral integrates by slicing the range rather than the domain: it measures the set of points where ff takes values in a given interval, and sums value×measure\text{value} \times \text{measure}. It is defined first for non-negative measurable functions as a supremum of integrals of simple functions, then extended to general measurable functions.

Details

Why slice the range?

The Riemann integral slices the domain into intervals and asks “what is ff like here?” — which fails when ff oscillates wildly (as 1Q\mathbf{1}_{\mathbb{Q}} does). The Lebesgue integral instead asks “on how much of the domain is ff near the value vv?” — a question that has a clean answer via measure. For 1Q\mathbf{1}_{\mathbb{Q}}, the set where it equals 1 (the rationals) has measure 0, so the integral is 0.

Measurability

The Lebesgue integral is defined for measurable functions, a class that includes every Riemann integrable function and is closed under pointwise limits. This closure is exactly what makes limit theorems possible. The measure-theoretic foundation (σ-algebras, measure, measurable functions) is developed under Measure theory.

Example problem

Problem. Compute the Lebesgue integral of f(x)=1Q(x)f(x) = \mathbf{1}_{\mathbb{Q}}(x) on [0,1][0, 1].

Solution

The set Q∩[0,1]\mathbb{Q} \cap [0,1] is countable, hence has Lebesgue measure 0. The function equals 1 on a set of measure 0 and 0 elsewhere, so ∫01f dx=1⋅0+0⋅1=0\int_0^1 f\, dx = 1 \cdot 0 + 0 \cdot 1 = 0. (Contrast: this function is not Riemann integrable.) ∎

Convergence theorems

Theorem (monotone convergence). If 0≤f1≤f2≤⋯0 \leq f_1 \leq f_2 \leq \cdots are measurable and fn→ff_n \to f pointwise, then

lim⁡n→∞∫fn=∫f.\lim_{n \to \infty} \int f_n = \int f.

Theorem (dominated convergence). If fn→ff_n \to f pointwise and there is an integrable gg with ∣fn∣≤g|f_n| \leq g for all nn, then

lim⁡n→∞∫fn=∫f.\lim_{n \to \infty} \int f_n = \int f.

Details

Why these matter

These theorems justify interchanging limits and integrals — the single most common operation in analysis, probability, and PDE theory. Monotone convergence handles increasing sequences (no domination needed, but the functions must be non-negative and increasing). Dominated convergence is the workhorse: a single integrable “envelope” gg controls the whole sequence, allowing the limit to pass through the integral.

Worked example: dominated convergence

Let fn(x)=nx1+n2x2f_n(x) = \frac{n x}{1 + n^2 x^2} on [0,1][0, 1]. Pointwise, fn(x)→0f_n(x) \to 0 for x>0x > 0 and fn(0)=0f_n(0) = 0, so fn→0f_n \to 0 pointwise. By AM–GM, 1+n2x2≥2nx1 + n^2 x^2 \geq 2 n x, so fn(x)≤12f_n(x) \leq \frac{1}{2} for all xx and nn — an integrable envelope. Hence dominated convergence applies and lim⁡n→∞∫01fn dx=∫010 dx=0\lim_{n \to \infty} \int_0^1 f_n\, dx = \int_0^1 0\, dx = 0. (Indeed, substituting u=nxu = n x gives ∫01fn dx=12nln⁡(1+n2)→0\int_0^1 f_n\, dx = \frac{1}{2n} \ln(1 + n^2) \to 0.)

Example problem

Problem. Let fn(x)=n1(0,1/n)(x)f_n(x) = n \mathbf{1}_{(0, 1/n)}(x) on [0,1][0, 1]. Compute ∫01fn dx\int_0^1 f_n\, dx and lim⁡n→∞fn(x)\lim_{n \to \infty} f_n(x) pointwise. Explain why dominated convergence does not apply.

Solution

∫01fn dx=n⋅(1/n)=1\int_0^1 f_n\, dx = n \cdot (1/n) = 1 for every nn. Pointwise, for each fixed x>0x > 0, eventually x>1/nx > 1/n, so fn(x)=0f_n(x) = 0; and fn(0)=0f_n(0) = 0. Thus fn→0f_n \to 0 pointwise, so ∫f=0≠1=lim⁡∫fn\int f = 0 \neq 1 = \lim \int f_n. Dominated convergence fails because no single integrable gg dominates all fnf_n: near 0, fnf_n has a spike of height nn on a set of measure 1/n1/n, and any dominating gg would need g(x)≥ng(x) \geq n on (0,1/n)(0, 1/n) for all nn, forcing gg to be non-integrable. ∎

Quiz

Example problem

Problem. Let fn(x)=11+n2x2f_n(x) = \frac{1}{1 + n^2 x^2} on [0,1][0, 1].

(a) Find the pointwise limit f(x)f(x).

(b) Use dominated convergence to compute lim⁡n→∞∫01fn(x) dx\lim_{n \to \infty} \int_0^1 f_n(x)\, dx.

Solution

(a) For x=0x = 0, fn(0)=1f_n(0) = 1. For x>0x > 0, fn(x)→0f_n(x) \to 0. So f(x)=1{0}(x)f(x) = \mathbf{1}_{\{0\}}(x).

(b) ∣fn(x)∣≤1|f_n(x)| \leq 1 and 11 is integrable on [0,1][0, 1], so dominated convergence applies. Hence lim⁡∫01fn=∫01f=∫011{0}=0\lim \int_0^1 f_n = \int_0^1 f = \int_0^1 \mathbf{1}_{\{0\}} = 0 (the point 00 has measure 0). ∎