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Lebesgue integral

The Lebesgue integral generalizes the Riemann integral by integrating with respect to a measure rather than with respect to the length of subintervals. It handles a much larger class of functions and provides powerful convergence theorems. It answers the question how do we sum (integrate) a measurable function? The integral is the central object of Lp spaces, where the convergence theorems below are the key tools.

Authoritative sources

Construction from simple functions

For a nonnegative measurable function f:X→[0,∞]f: X \to [0,\infty]: ∫Xf dμ=sup⁡{∫Xs dμ:s simple,  0≤s≤f}\int_X f \, d\mu = \sup \left\{ \int_X s \, d\mu : s \text{ simple}, \; 0 \leq s \leq f \right\}

where the integral of a simple function is defined algebraically as ∫Xs dμ=∑iai μ(Ai)\int_X s \, d\mu = \sum_i a_i \, \mu(A_i) (see Measurable functions for the definition of simple functions and this formula).

For a general measurable f:X→Rf: X \to \mathbb{R} (with ∫f− dμ<∞\int f^- \, d\mu < \infty or ∫f+ dμ<∞\int f^+ \, d\mu < \infty): ∫Xf dμ=∫Xf+ dμ−∫Xf− dμ\int_X f \, d\mu = \int_X f^+ \, d\mu - \int_X f^- \, d\mu

where f+=max⁡(f,0)f^+ = \max(f, 0) and f−=max⁡(−f,0)f^- = \max(-f, 0).”

Details

Why this works. The Riemann integral slices the domain into subintervals and sums f(xi)Δxif(x_i) \Delta x_i. The Lebesgue integral slices the range: it asks “what is the measure of the set where ff is between aa and bb?” and sums a⋅μ({x:a≤f(x)<b})a \cdot \mu(\{x : a \leq f(x) < b\}).

Key properties:

Comparison with Riemann. Every Riemann-integrable function on a bounded interval is Lebesgue-integrable, and the two integrals agree. The Lebesgue integral additionally handles: functions with dense discontinuities (e.g., 1Q\mathbf{1}_{\mathbb{Q}}), unbounded functions (e.g., 1/x1/x on (0,1)(0,1)), and functions on general measure spaces.

Monotone convergence theorem

Theorem (MCT). If f1≤f2≤⋯f_1 \leq f_2 \leq \cdots are nonnegative measurable functions with fn→ff_n \to f pointwise, then ∫f dμ=lim⁡n→∞∫fn dμ\int f \, d\mu = \lim_{n \to \infty} \int f_n \, d\mu

(in particular, ∫f dμ\int f \, d\mu may be infinite).

Details

Intuition. You can swap the limit and the integral when the sequence is increasing. The “mass” cannot escape to infinity because the functions are growing.

Example. Let fn=1[0,n]f_n = \mathbf{1}_{[0,n]} on (R,L,λ)(\mathbb{R}, \mathcal{L}, \lambda). Then fn↑1Rf_n \uparrow \mathbf{1}_{\mathbb{R}} and ∫1R dλ=lim⁡n∫1[0,n] dλ=lim⁡nn=∞\int \mathbf{1}_{\mathbb{R}} \, d\lambda = \lim_n \int \mathbf{1}_{[0,n]} \, d\lambda = \lim_n n = \infty

Limitation. MCT requires monotonicity. For general sequences, one needs the dominated convergence theorem.

Fatou’s lemma

Theorem (Fatou). If fn≥0f_n \geq 0 are measurable functions, then ∫lim inf⁡n→∞fn dμ≤lim inf⁡n→∞∫fn dμ\int \liminf_{n \to \infty} f_n \, d\mu \leq \liminf_{n \to \infty} \int f_n \, d\mu

Details

Intuition. Fatou’s lemma is the weakest of the three fundamental convergence theorems: it gives an inequality (≤) rather than an equality. It requires only nonnegativity, no monotonicity, and no dominating function. It is typically proved first and serves as a stepping stone to the dominated convergence theorem.

Relation to MCT and DCT.

Example. Let fn=1{1/n}f_n = \mathbf{1}_{\{1/n\}} on (0,1)(0,1). Then fn→0f_n \to 0 pointwise, so ∫lim inf⁡fn dλ=0\int \liminf f_n \, d\lambda = 0. But ∫fn dλ=0\int f_n \, d\lambda = 0 for every nn (a single point has measure zero), so lim inf⁡∫fn=0\liminf \int f_n = 0. The inequality is an equality here.

A sharper example: fn=n⋅1(0,1/n)f_n = n \cdot \mathbf{1}_{(0, 1/n)}. Then fn→0f_n \to 0 a.e., but ∫fn dλ=1\int f_n \, d\lambda = 1 for all nn. Fatou gives 0≤10 \leq 1 (trivially true), while DCT does not apply (no integrable dominator exists).

Dominated convergence theorem

Theorem (DCT). If fn→ff_n \to f pointwise a.e., and there exists a measurable gg with ∣fn∣≤g|f_n| \leq g for all nn and ∫g dμ<∞\int g \, d\mu < \infty, then ∫f dμ=lim⁡n→∞∫fn dμ\int f \, d\mu = \lim_{n \to \infty} \int f_n \, d\mu

Details

Intuition. If the sequence is “trapped” by an integrable function, you can pass the limit through the integral. The dominating function gg prevents mass from escaping.

Example. 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 each fixed x>0x > 0. But ∣fn(x)∣≤12|f_n(x)| \leq \frac{1}{2} by AM-GM (since nx≤12(n2x2+1)nx \leq \frac{1}{2}(n^2x^2 + 1)). So g=1/2g = 1/2 dominates, and DCT applies. Explicitly, ∫01fn dx=ln⁡(1+n2)2n→0=∫010 dx\int_0^1 f_n \, dx = \frac{\ln(1+n^2)}{2n} \to 0 = \int_0^1 0 \, dx, confirming the convergence of the integrals.

Contrast with MCT. DCT does not require monotonicity, but requires a uniform integrable bound. MCT requires monotonicity but no bound (the integral may be infinite).

Fubini and Tonelli theorems

Tonelli’s theorem. If f≥0f \geq 0 is measurable on X×YX \times Y (with product measure μ×ν\mu \times \nu), then ∫X×Yf d(μ×ν)=∫X(∫Yf(x,y) dν(y))dμ(x)=∫Y(∫Xf(x,y) dμ(x))dν(y)\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left(\int_Y f(x,y) \, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x,y) \, d\mu(x)\right) d\nu(y)

(All three sides are equal, possibly ∞\infty.)

Fubini’s theorem. If ff is measurable on X×YX \times Y and ∫X×Y∣f∣ d(μ×ν)<∞\int_{X \times Y} |f| \, d(\mu \times \nu) < \infty, then the iterated integrals exist and are equal to the double integral.

Details

Key difference. Tonelli requires only nonnegativity (no integrability needed); Fubini requires integrability of ∣f∣|f|.

Why it matters. Without these theorems, iterated integrals can differ. For the integrand x2−y2(x2+y2)2\frac{x^2 - y^2}{(x^2 + y^2)^2} on (0,1)2(0,1)^2: ∫01(∫01x2−y2(x2+y2)2 dx)dy=−π4\int_0^1 \left(\int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dx\right) dy = -\frac{\pi}{4} ∫01(∫01x2−y2(x2+y2)2 dy)dx=π4\int_0^1 \left(\int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy\right) dx = \frac{\pi}{4}

The two iterated integrals are unequal (−π/4≠π/4-\pi/4 \neq \pi/4), and the integrand is not absolutely integrable, so Fubini does not apply.

Connection to probability. In probability, E[X]=∫X dP\mathbb{E}[X] = \int X \, dP, and Fubini/Tonelli justify exchanging expectation and summation/integration: E[∑Xn]=∑E[Xn]\mathbb{E}[\sum X_n] = \sum \mathbb{E}[X_n] (Tonelli for nonnegative, Fubini for integrable).

Example problem

Problem. Compute ∫0∞e−xsin⁡(x) dx\int_0^\infty e^{-x} \sin(x) \, dx using the Lebesgue integral, and justify interchanging the order of integration in the representation sin⁡(x)=x∫01cos⁡(tx) dt\sin(x) = x \int_0^1 \cos(tx) \, dt

Solution

Step 1: Set up the iterated integral.

Using the identity sin⁡(x)=x∫01cos⁡(tx) dt\sin(x) = x \int_0^1 \cos(tx) \, dt (obtained by integrating cos⁡(tx)\cos(tx) in tt and evaluating at the endpoints):

∫0∞e−xsin⁡(x) dx=∫0∞x e−x(∫01cos⁡(tx) dt)dx\int_0^\infty e^{-x} \sin(x) \, dx = \int_0^\infty x \, e^{-x} \left(\int_0^1 \cos(tx) \, dt\right) dx

Step 2: Justify Fubini.

The integrand x e−xcos⁡(tx)x \, e^{-x} \cos(tx) is measurable on [0,∞)×[0,1][0,\infty) \times [0,1]. Check absolute integrability: ∫0∞∫01∣x e−xcos⁡(tx)∣ dt dx≤∫0∞∫01x e−x dt dx=∫0∞x e−x dx=1<∞\int_0^\infty \int_0^1 |x \, e^{-x} \cos(tx)| \, dt \, dx \leq \int_0^\infty \int_0^1 x \, e^{-x} \, dt \, dx = \int_0^\infty x \, e^{-x} \, dx = 1 < \infty

So Fubini applies.

Step 3: Swap and compute the inner integral.

∫01(∫0∞x e−xcos⁡(tx) dx)dt\int_0^1 \left(\int_0^\infty x \, e^{-x} \cos(tx) \, dx\right) dt

The inner integral is the real part of ∫0∞x e−(1−it)x dx=1(1−it)2\int_0^\infty x \, e^{-(1-it)x} \, dx = \frac{1}{(1-it)^2}: ∫0∞x e−xcos⁡(tx) dx=1−t2(1+t2)2\int_0^\infty x \, e^{-x} \cos(tx) \, dx = \frac{1 - t^2}{(1+t^2)^2}

Step 4: Evaluate the outer integral.

∫011−t2(1+t2)2 dt\int_0^1 \frac{1 - t^2}{(1+t^2)^2} \, dt

Note that ddt(t1+t2)=1−t2(1+t2)2\frac{d}{dt}\left(\frac{t}{1+t^2}\right) = \frac{1 - t^2}{(1+t^2)^2}, so: ∫011−t2(1+t2)2 dt=[t1+t2]01=12−0=12\int_0^1 \frac{1 - t^2}{(1+t^2)^2} \, dt = \left[\frac{t}{1+t^2}\right]_0^1 = \frac{1}{2} - 0 = \frac{1}{2}

Answer: ∫0∞e−xsin⁡(x) dx=12\int_0^\infty e^{-x} \sin(x) \, dx = \frac{1}{2}.