Dominated Convergence Theorem

The dominated convergence theorem is a fundamental result in measure theory concerning the interchange of limits and integration. It states that if a sequence of measurable functions converges almost everywhere and the absolute values of all its terms are bounded by one integrable function, then the limit is integrable and the integrals converge to its integral. The theorem belongs to the theory of the Lebesgue integral and applies to functions that may change sign or take complex values. (math.mit.edu)

Statement

Let $(X,\Sigma,\mu)$ be a measure space, and let $f_n:X\to\mathbb R$ or $\mathbb C$ be measurable functions. Suppose that:

$f_n(x)\to f(x)$ almost everywhere, where $f$ is measurable.
There is a nonnegative measurable function $g$ such that
$\int_X g\,d\mu<\infty, \qquad |f_n(x)|\le g(x)$
almost everywhere for every $n$ .

Then every $f_n$ and the limit $f$ are integrable, and

\lim_{n\to\infty}\int_X f_n\,d\mu = \int_X f\,d\mu.

The function $g$ is called an integrable dominating function, or integrable majorant. The same $g$ must control the entire sequence; separate bounds depending on $n$ are not sufficient. (math.mit.edu)

The almost-everywhere qualifications permit exceptional sets of measure zero. Because the sequence is countable, the exceptional sets for its individual bounds can be combined into a single null set. For complex-valued functions, the absolute value means the complex modulus; the integral identity also follows by applying the real-valued theorem to the real and imaginary parts. (ocw.mit.edu)

Meaning and strength of the conclusion

Pointwise convergence controls the values at each fixed point but does not, by itself, control the total integral. A sequence can develop increasingly narrow, high peaks, or move its contribution farther out in an unbounded domain. Domination supplies global control through a function whose total integral is finite. (maths.tcd.ie)

The theorem actually gives the stronger conclusion

\int_X |f_n-f|\,d\mu\longrightarrow0.

This is convergence in the $L^1$ space, rather than merely convergence of the signed or complex integrals. Indeed, $|f|\le g$ almost everywhere, so

|f_n-f|\le2g.

Applying dominated convergence to $|f_n-f|$ , which tends to zero almost everywhere, proves the assertion. The triangle inequality then gives

\left|\int_X f_n\,d\mu-\int_X f\,d\mu\right| \le\int_X|f_n-f|\,d\mu.

These are direct consequences of the theorem. (abel.math.harvard.edu)

Proof using Fatou’s lemma

For real-valued functions, a standard proof uses Fatou’s lemma twice. Since $|f|\le g$ , all relevant integrals are finite. The functions $g+f_n$ and $g-f_n$ are nonnegative, so Fatou’s lemma gives

\int_X(g+f)\,d\mu \le \liminf_n\int_X(g+f_n)\,d\mu

and

\int_X(g-f)\,d\mu \le \liminf_n\int_X(g-f_n)\,d\mu.

Cancelling the finite integral of $g$ yields

\int_X f\,d\mu \le\liminf_n\int_X f_n\,d\mu, \qquad \limsup_n\int_X f_n\,d\mu \le\int_X f\,d\mu.

Consequently, the lower and upper limits agree with $\int_X f\,d\mu$ . The dominating function thus turns the one-sided inequality of Fatou’s lemma into an equality of limits. (ocw.mit.edu)

Examples and failures without domination

On $[0,1]$ , equipped with Lebesgue measure, consider

f_n(x)=x^n.

The functions tend to zero except at $x=1$ , and $0\le f_n\le1$ . Because the constant function $1$ is integrable on this interval, the theorem gives

\int_0^1x^n\,dx\longrightarrow0,

consistent with the value $1/(n+1)$ . This illustrates the finite-interval application of a constant majorant. (www-users.cse.umn.edu)

For a contrasting construction on $(0,1)$ , let

f_n(x)=n\,\mathbf1_{(0,1/n)}(x),

where $\mathbf1_A$ denotes the indicator function of $A$ . For each fixed $x>0$ , eventually $f_n(x)=0$ , but

\int_0^1f_n(x)\,dx=1.

The limit function has integral zero. Thus pointwise convergence and even a uniform bound on the integrals of $|f_n|$ do not replace an integrable pointwise majorant. This is a narrow-peak example of the failure that domination excludes. (maths.tcd.ie)

The integrability of the majorant also matters on infinite-measure domains. For example, the one-dimensional heat kernel tends pointwise to zero as time tends to infinity, while its integral remains one. Its mass spreads over an increasingly large region, and no common integrable majorant exists for that family. (www-users.cse.umn.edu)

Applications

Expectations in probability

On a probability space, integration is expectation. If random variables $X_n$ converge almost surely to $X$ , and

|X_n|\le Y,\qquad \mathbb E[Y]<\infty,

then

\mathbb E[X_n]\to\mathbb E[X], \qquad \mathbb E[|X_n-X|]\to0.

No independence assumption is needed. This is the measure-theoretic theorem applied with the probability measure as $\mu$ . (math.mit.edu)

Parameter-dependent integrals

Suppose $h(t,x)$ is measurable in $x$ , continuous in the real parameter $t$ for almost every $x$ , and bounded in absolute value by a common integrable function for $t$ near $t_0$ . Applying the theorem along every sequence $t_n\to t_0$ proves that

H(t)=\int_Xh(t,x)\,d\mu(x)

is continuous at $t_0$ . (maths.tcd.ie)

The theorem also justifies differentiation under the integral sign. One sufficient set of conditions is that $h$ is real-valued, its sections are measurable, $h(t_0,\cdot)$ is integrable, and, outside a fixed null set, $h(\cdot,x)$ is differentiable on a neighborhood of $t_0$ , with

|\partial_t h(t,x)|\le g(x),\qquad g\in L^1(\mu).

The mean value theorem bounds the difference quotients by $g$ . Dominated convergence then gives

H'(t_0)=\int_X\partial_t h(t_0,x)\,d\mu(x).

The crucial domination is therefore of the difference quotients, not simply of the original integrand. (maths.tcd.ie)

Related convergence theorems

The bounded convergence theorem is a special case: on a finite-measure space, a common constant bound is integrable. A constant bound does not generally suffice when the space has infinite measure. (math.mit.edu)

The monotone convergence theorem instead assumes a nonnegative, increasing sequence. It requires no integrable majorant and permits the limiting integral to be infinite. Dominated convergence does not require monotonicity, but its integrable bound ensures finite integrals. (math.mit.edu)

There is also a dominated-convergence formulation in which convergence in measure replaces almost-everywhere convergence, while a common integrable majorant is retained. In probability, the corresponding convergence mode is convergence in probability. (math.mit.edu)

References

175: Lecture 4 — Expectation properties, law of large numbers statement, and Kolmogorov’s extension theoremmath.mit.edu
Lecture Notes for 18.102, Spring 2009 — Lecture 7ocw.mit.edu
Chapter 4. The dominated convergence theorem and applicationsmaths.tcd.ie
Real and Complex Analysis — Section 5.6: The Dominated Convergence Theoremabel.math.harvard.edu
125, Spring 2016math.mit.edu