Lebesgue Integration

For a non-negative measurable simple function $s = \sum_{i=1}^n a_i \chi_{A_i}$ with $a_i \geq 0$ and $\{A_i\}$ disjoint, define:

$\int_X s\, d\mu = \sum_{i=1}^n a_i \mu(A_i)$

For a non-negative measurable function $f$, define:

$\int_X f\, d\mu = \sup\left\{\int_X s\, d\mu : 0 \leq s \leq f,\ s \text{ simple}\right\}$

This definition is consistent with Theorem 5.4: by monotone convergence, we also have

$\int_X f\, d\mu = \lim_{n \to \infty} \int_X s_n\, d\mu$

for any increasing sequence of simple functions $s_n \nearrow f$.

For a measurable function $f : X \to \mathbb{R}$, define $f^+ = \max(f, 0)$ and $f^- = \max(-f, 0)$, so $f = f^+ - f^-$ and $|f| = f^+ + f^-$. If $\int f^+\, d\mu < \infty$ and $\int f^-\, d\mu < \infty$ (i.e., $\int |f|\, d\mu < \infty$), define:

$\int_X f\, d\mu = \int_X f^+\, d\mu - \int_X f^-\, d\mu$

The function $f$ is called integrable (or $f \in L^1(\mu)$) if $\int |f|\, d\mu < \infty$.

Proposition 6.1 (Linearity). If $f, g \in L^1(\mu)$ and $a, b \in \mathbb{R}$, then $af + bg \in L^1(\mu)$ and $\int(af + bg)\, d\mu = a\int f\, d\mu + b\int g\, d\mu$.

Proposition 6.2 (Monotonicity). If $f \leq g$ a.e., then $\int f\, d\mu \leq \int g\, d\mu$.

Proposition 6.3 (Markov”s Inequality). If $f \geq 0$ is measurable, then for any $a > 0$:

$\mu(\{x : |f(x)| \geq a\}) \leq \frac{1}{a}\int |f|\, d\mu$

Theorem 6.4 (Chebyshev’s Inequality). If $f \in L^2(\mu)$, then for any $a > 0$:

$\mu(\{|f - \int f\, d\mu| \geq a\}) \leq \frac{1}{a^2}\mathrm{Var}(f)$

Theorem 6.5 (Monotone Convergence Theorem — Levi). If $0 \leq f_1 \leq f_2 \leq \cdots$ are measurable and $f_n \to f$ pointwise, then:

$\lim_{n \to \infty} \int f_n\, d\mu = \int f\, d\mu$

Proof sketch. Let $s$ be a simple function with $s \leq f$. Define $E_n = \{x : f_n(x) \geq (1 - \varepsilon)s(x)\}$. Then $E_n \nearrow X$ and $\int f_n\, d\mu \geq (1 - \varepsilon)\int s\, d\mu$ for large $n$. Take $\sup$ over $s$ and let $\varepsilon \to 0$. $\blacksquare$

Theorem 6.6 (Fatou’s Lemma). If $f_n \geq 0$ are measurable, then:

$\int \liminf_{n\to\infty} f_n\, d\mu \leq \liminf_{n\to\infty} \int f_n\, d\mu$

Proof. Define $g_n = \inf_{k \geq n} f_k$. Then $0 \leq g_1 \leq g_2 \leq \cdots$ and $g_n \to \liminf f_n$. By monotone convergence:

$\int \liminf f_n\, d\mu = \lim_{n\to\infty} \int g_n\, d\mu \leq \liminf_{n\to\infty} \int f_n\, d\mu$

$\blacksquare$

Theorem 6.7 (Dominated Convergence Theorem). If $f_n \to f$ a.e. and there exists $g \in L^1(\mu)$ with $|f_n| \leq g$ a.e. for all $n$, then:

$\lim_{n\to\infty} \int f_n\, d\mu = \int f\, d\mu$

Proof sketch. Apply Fatou’s lemma to $g + f_n$ and $g - f_n$:

$\int f\, d\mu \leq \liminf \int f_n\, d\mu \quad \text{and} \quad \int f\, d\mu \geq \limsup \int f_n\, d\mu$

Combining gives the result. $\blacksquare$

Problem. Compute $\lim_{n \to \infty} \int_0^1 (1 - x^2/n)^n\, dx$.

Solution. For each $x \in [0, 1]$, $(1 - x^2/n)^n \to e^{-x^2}$ as $n \to \infty$. Since $0 \leq (1 - x^2/n)^n \leq 1$ for all $n$ and $x$, we can apply the dominated convergence theorem with $g(x) = 1 \in L^1([0, 1])$:

$\lim_{n\to\infty} \int_0^1 (1 - x^2/n)^n\, dx = \int_0^1 e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2}\,\mathrm{erf}(1) \approx 0.7468$

$\blacksquare$

The three convergence theorems are related: DCT follows from Fatou, and Fatou follows from MCT. Together they form the backbone of Lebesgue integration theory.

• Applying DCT without a dominating function. If $|f_n| \leq g$ fails for some $n$, the limit may not pass through the integral. Fix: Always verify existence of $g \in L^1$ dominating all $f_n$ almost everywhere.
• Confusing pointwise and uniform convergence. DCT only requires a.e. pointwise convergence, not uniform. Fix: The theorem is powerful precisely because it relaxes the uniform-convergence requirement of Riemann integration.
• Assuming monotone convergence needs boundedness. MCT requires only monotonicity and non-negativity; the limit may be infinite. Fix: If $\int f_n$ diverges, the theorem correctly gives $\int f = \infty$.
• Forgetting the non-negativity in Fatou. Without $f_n \geq 0$, the inequality $\int \liminf f_n \leq \liminf \int f_n$ can fail. Fix: Apply Fatou to $f_n + g$ with $g$ integrable, then subtract.

• Fourier series: DCT justifies term-by-term integration of Fourier series, allowing computation of coefficients by integrating the series.
• Probability theory: Markov’s and Chebyshev’s inequalities are essential for proving laws of large numbers and concentration bounds.
• $L^p$ spaces: MCT and DCT are used to prove completeness of $L^p$ spaces and to exchange limits with norms.
• Fubini’s theorem: Tonelli’s theorem (MCT for non-negative functions) and Fubini’s theorem (DCT for integrable functions) justify swapping the order of integration.

Problem. Evaluate $\lim_{n\to\infty} \int_0^\pi \frac{\sin(nx)}{n}\, dx$ using the dominated convergence theorem.

Solution. Let $f_n(x) = \sin(nx)/n$. For each $x \in [0,\pi]$, $|f_n(x)| \leq 1/n \to 0$, so $f_n \to 0$ pointwise. Also $|f_n(x)| \leq 1$ for all $n$ and $x$, and $g(x) = 1$ is integrable on $[0,\pi]$. By DCT:

$\lim_{n\to\infty} \int_0^\pi \frac{\sin(nx)}{n}\, dx = \int_0^\pi 0\, dx = 0$

We can verify directly: $\int_0^\pi \sin(nx)\, dx = [-\cos(nx)/n]_0^\pi = (1 - (-1)^n)/n$, so the integral is $0$ for even $n$ and $2/n^2$ for odd $n$, both vanishing as $n\to\infty$.

$\blacksquare$