Measure Theory and Lebesgue Integration: An Introduction

The Riemann integral you learned in calculus works beautifully for continuous functions, but it breaks down for wildly discontinuous ones and behaves badly with limits. Measure theory, developed by Henri Lebesgue around 1902, fixes these problems and underpins modern probability, functional analysis and Fourier analysis. This introduction explains the key ideas and shows them in action through worked examples.

Why Riemann Integration Is Not Enough

Consider the Dirichlet function on \([0, 1]\):

$$ D(x) = \begin{cases} 1, & x \text{ rational}, \\ 0, & x \text{ irrational}. \end{cases} $$

Every subinterval contains both rationals and irrationals, so every upper Riemann sum equals 1 and every lower sum equals 0. The Riemann integral does not exist. Yet intuitively the rationals are "tiny", so the integral should be 0. Lebesgue's idea delivers exactly that.

There is also a limit problem: a pointwise limit of Riemann integrable functions need not be Riemann integrable. Lebesgue integration comes with powerful convergence theorems that make swapping limits and integrals safe.

Sigma-Algebras and Measures

To measure sets we first decide which sets are allowed. A σ-algebra \(\mathcal{F}\) on a set \(X\) is a collection of subsets that:

1. contains \(X\),
2. is closed under complements,
3. is closed under countable unions.

A measure is a function \(\mu: \mathcal{F} \to [0, \infty]\) with \(\mu(\emptyset) = 0\) and countable additivity: for pairwise disjoint sets \(A_1, A_2, \dots\),

$$ \mu\Big(\bigcup_{n=1}^\infty A_n\Big) = \sum_{n=1}^\infty \mu(A_n). $$

The triple \((X, \mathcal{F}, \mu)\) is a measure space. A probability space is simply a measure space with \(\mu(X) = 1\).

Lebesgue Measure

Lebesgue measure \(\lambda\) on \(\mathbb{R}\) extends length: \(\lambda([a, b]) = b - a\). It is built from the outer measure

$$ \lambda^*(A) = \inf\Big\{\sum_{n} (b_n - a_n) : A \subseteq \bigcup_n (a_n, b_n)\Big\}, $$

the smallest total length of a countable cover by open intervals. Not every subset of \(\mathbb{R}\) can be measured consistently (the Vitali set is the classic counterexample), which is why we restrict to the Lebesgue measurable sets.

Worked Example 1: Null Sets

Part (a). Show that \(\mathbb{Q} \cap [0, 1]\) has Lebesgue measure zero.

Step 1. The rationals in \([0,1]\) are countable, so list them as \(q_1, q_2, q_3, \dots\).

Step 2. Fix \(\varepsilon > 0\). Cover \(q_n\) by an interval of length \(\varepsilon / 2^n\).

Step 3. The total length of the cover is

$$ \sum_{n=1}^\infty \frac{\varepsilon}{2^n} = \varepsilon \sum_{n=1}^\infty \frac{1}{2^n} = \varepsilon. $$

Step 4. Since \(\varepsilon\) is arbitrary, \(\lambda^*(\mathbb{Q}\cap[0,1]) = 0\). The same argument shows every countable set is a null set.

Part (b). Find the measure of the Cantor set \(C\).

Step 1. At stage 1 we remove the middle third \((1/3, 2/3)\), of length \(1/3\). At stage 2 we remove 2 intervals of length \(1/9\). At stage \(n\) we remove \(2^{n-1}\) intervals of length \(1/3^n\).

Step 2. The total removed length is a geometric series with first term \(1/3\) and ratio \(2/3\):

$$ \sum_{n=1}^{\infty} \frac{2^{n-1}}{3^n} = \frac{1/3}{1 - 2/3} = 1. $$

Step 3. Therefore \(\lambda(C) = 1 - 1 = 0\).

The surprise: \(C\) is uncountable (it has as many points as \([0,1]\)), yet its measure is zero. Size in the sense of counting and size in the sense of measure are different things.

Measurable Functions and Simple Functions

A function \(f: X \to \mathbb{R}\) is measurable if \(\{x : f(x) > a\} \in \mathcal{F}\) for every real \(a\). Continuous functions, monotone functions and pointwise limits of measurable functions are all measurable.

A simple function takes finitely many values:

$$ s = \sum_{k=1}^{m} c_k \, \mathbf{1}_{A_k}, $$

where \(\mathbf{1}_{A}\) is the indicator of a measurable set \(A\). Its integral is defined in the most natural way:

$$ \int s \, d\mu = \sum_{k=1}^m c_k \, \mu(A_k). $$

The Lebesgue Integral

For a measurable \(f \ge 0\),

$$ \int f\,d\mu = \sup\Big\{ \int s\,d\mu : 0 \le s \le f,\ s \text{ simple} \Big\}. $$

For a general \(f\), write \(f = f^+ - f^-\) and define \(\int f = \int f^+ - \int f^-\) whenever at least one part is finite. If \(\int |f| < \infty\), \(f\) is Lebesgue integrable.

The key mental shift: Riemann slices the domain into intervals; Lebesgue slices the range into levels and asks how large the set is where \(f\) lies at each level. Lebesgue himself compared it to counting money by sorting coins into piles of equal value first.

Now the Dirichlet function is easy. \(D = 1\cdot\mathbf{1}_{\mathbb{Q}\cap[0,1]}\) is simple, so

$$ \int_{[0,1]} D\,d\lambda = 1 \cdot \lambda(\mathbb{Q}\cap[0,1]) = 1 \cdot 0 = 0. $$

Whenever a function is Riemann integrable on \([a,b]\), its Lebesgue integral agrees with the Riemann value, so nothing you already know is lost.

The Big Convergence Theorems

• Monotone Convergence Theorem (MCT): if \(0 \le f_1 \le f_2 \le \dots\) and \(f_n \to f\) pointwise, then \(\int f_n \to \int f\).
• Fatou's Lemma: for \(f_n \ge 0\), \(\int \liminf f_n \le \liminf \int f_n\).
• Dominated Convergence Theorem (DCT): if \(f_n \to f\) pointwise and \(|f_n| \le g\) with \(g\) integrable, then \(\int f_n \to \int f\).

Worked Example 2: When Limits and Integrals Do and Do Not Commute

Part (a). Let \(f_n = n \cdot \mathbf{1}_{(0, 1/n)}\) on \([0,1]\).

Step 1. For each fixed \(x > 0\), once \(n > 1/x\) we have \(f_n(x) = 0\). Also \(f_n(0) = 0\). So \(f_n \to 0\) pointwise.

Step 2. Each integral is \(\int f_n = n \cdot \lambda((0, 1/n)) = n \cdot \tfrac{1}{n} = 1\).

Step 3. Hence \(\lim \int f_n = 1 \ne 0 = \int \lim f_n\). The DCT does not apply because no integrable \(g\) dominates all \(f_n\): the smallest candidate, \(\sup_n f_n\), behaves like \(1/x\) near 0, which has infinite integral. Fatou's inequality \(0 \le 1\) still holds.

Part (b). Compute \(\displaystyle \lim_{n\to\infty} \int_0^1 \frac{n x}{1 + n^2 x^2}\,dx\) using the DCT. Pointwise, for \(x > 0\), \(\dfrac{nx}{1 + n^2x^2} \to 0\). For domination, the AM-GM inequality gives \(1 + n^2x^2 \ge 2nx\), so the integrand is at most \(1/2\), an integrable constant on \([0,1]\). By the DCT,

$$ \lim_{n\to\infty}\int_0^1 \frac{nx}{1+n^2x^2}\,dx = \int_0^1 0\,dx = 0. $$

Check directly: the integral equals \(\dfrac{\ln(1 + n^2)}{2n}\), which indeed tends to 0 (for \(n = 100\) it is about \(0.046\)).

Did You Know?

• Lebesgue introduced his integral in his 1902 doctoral thesis, "Intégrale, longueur, aire".
• Andrey Kolmogorov used measure theory in 1933 to give probability its modern axioms: an expectation is just a Lebesgue integral.
• There exist sets of real numbers that are not Lebesgue measurable, but constructing one requires the Axiom of Choice.

Key Takeaways

• A measure assigns sizes to sets in a countably additive way; Lebesgue measure extends length.
• Countable sets, and even some uncountable sets like the Cantor set, have measure zero.
• The Lebesgue integral partitions the range, handles functions such as the Dirichlet function, and agrees with Riemann where both exist.
• The MCT and DCT tell you exactly when you may swap a limit and an integral.

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