11 Convergence Theorems for Integrals
This section will give rise to three most important theorems in analysis. Throughout this section, let \((X, \mathcal S, \mu)\) be a measure space.
11.1 The Notion of Almost Everywhere
Definition 11.1 A statement about \(x \in X\) will be said to hold almost everywhere, or a.e. iff it holds for all \(x \not\in A\) for some \(A\) having \(\mu(A)=0\). The set of all \(x\) for which the statement holds will thus be measurable for the completion of \(\mu\), as in Theorem 8.2, but will not necessarily in \(\mathcal S\).
Proposition 11.1 If \(f\) and \(g\) are two measurable functions from \(X\) into \([-\infty, \infty]\) such that \(f(x) = g(x)\) a.e., then \(\int f\,d\mu\) is defined if and only if \(\int g\,d\mu\) is defined. When defined, the integrals are equal.
Let \(f=g\) on \(X \setminus A\) where \(\mu(A)=0\).
Let us show that \(\int h\,d\mu = \int_{X\setminus A} h\,d\mu\), where \(h\) is any measurable function, and equality holds in the sense that the integrals are defined and equal if and only if either of them is defined.
This is clearly true if \(h\) is an indicator function of a set in \(\mathcal S\); then it is true if \(h\) is any nonnegative simple function; then it is true, by Proposition 10.3, if \(h\) is any nonnegative measurable function; and thus for a general \(h\) by definition of the integral.
Letting \(h=f\) and \(h=g\) finishes the proof.
Let a function \(f\) be defined on \(B \in \mathcal S\) with \(\mu(X \setminus B)=0\), where \(f\) has values in \([-\infty, \infty]\) and is measurable for \(\mathcal S_B\). Then \(f\) can be extended to a measurable function for \(\mathcal S\) on \(X\). For any two extensions \(g\) and \(h\) of \(f\), \(g=h\) a.e. Thus we can define \(\int f\,d\mu\) as \(\int g\,d\mu\) if this is defined. Then \(\int f\,d\mu\) is well-defined by Proposition 11.1.
If \(f_n = g_n\) a.e. for \(n=1,2,...,\) then \(\mu^*(\bigcup_n\{x: f_n(x) \ne g_n(x)\}) = 0\). Outside this set, \(f_n = g_n\) for all \(n\). Thus in theorem about integrals, even for sequences of functions, the hypotheses need only hold almost everywhere.
11.2 Monotone Convergence Theorem (MCT)
Theorem 11.1 (Monotone Convergence Theorem) Let \(f_n\) be measurable functions from \(X\) into \([-\infty, \infty]\), such that \(f_n \uparrow f\) and \(\int f_1\,d\mu > -\infty\). Then \(\int f_n\,d\mu \uparrow \int f\,d\mu\).
First, let us make sure \(f\) is measurable. It is easily seen that the set of all open half-lines \((c, \infty]\) generates the Borel \(\sigma\)-algebra of \([-\infty,\infty]\). Then \(f\) is measurable by Theorem 10.1.
Next, suppose \(f_1 \ge 0\). Then by Proposition 10.3, take simple functions \(f_{nm} \uparrow f_n\) as \(m\to\infty\) for each \(n\). Let \(g_n := max(f_{1n}, ..., f_{nn})\). Then each \(g_n\) is simple and \(0 \le g_n \uparrow f\). So by Proposition 10.3, \(\int g_n\,d\mu \uparrow \int f\,d\mu\). Since \(g_n\le f_n \uparrow f\), we get \(\int f_n\,d\mu \uparrow \int f\,d\mu\).
Next suppose \(f \ge 0\). Let \(g_n := -f_n \downarrow -f := g\). Then \(0 \le \int g\,d\mu \le \int g_n\,d\mu < \infty\) for all \(n\), by Lemma 10.1. Now \(0 \le g_1 - g_n \uparrow g_1 - g\), so \(\int g_1 - g_n\,d\mu \uparrow g_1 - g\,d\mu < \infty\). These integrals all being finite, we can subtract them from \(\int g_1\,d\mu\), by Theorem 10.2 and get \(\int g_n\,d\mu \downarrow \int g\,d\mu\), so \(\int f_n\,d\mu \uparrow \int f\,d\mu\), as desired.
Now in the general case, we have \(f_n^+ \uparrow f^+\) and \(f_n^- \downarrow f^-\) with \(\int f^-\,d\mu < \infty\), so by previous cases, \(\int f_n^+\,d\mu \uparrow \int f^+\,d\mu\) and \(+\infty > \int f_n^-\,d\mu \downarrow \int f^-\,d\mu \ge 0\). Thus \(\int f_n\,d\mu \uparrow \int f\,d\mu\).
Example 11.1 Let \(f_n := -1_{[n, \infty)}\). Then \(f_n\uparrow 0\) but \(\int f_n\,d\mu\equiv -\infty\), not converging to \(0\). This shows why a hypothesis such as \(\int f_1\,d\mu > -\infty\) is needed in the monotone convergence theorem.
There is a symmetric form of MCT with \(f_n \downarrow f\), \(\int f_1\,d\mu < +\infty\).
11.3 Fatou’s Lemma
For any real \(a_n\), \(\lim\inf_{n\to\infty} := \sup_m \inf_{n \ge m} a_n\) is defined (possibly infinite). The next fact, though a one-sided inequality, is rather general
Theorem 11.2 (Fatou’s Lemma) Let \(f_n\) be any nonnegative measurable functions on \(X\). Then \(\int \lim\inf f_n\,d\mu\le \lim\inf\int f_n\,d\mu\).
Let \(g_n(x) := \inf\{f_m(x): m\ge n\}\). Then \(g_n \uparrow \lim\inf f_n\). Thus by MCT, \(\int g_n\,d\mu \uparrow \int \lim\inf f_n\,d\mu\).
For all \(m\ge n\), \(g_n \le f_m\), so \(\int g_n\,d\mu \le \int f_m\,d\mu\). Hence \(\int g_n\,d\mu \le \inf\{\int f_m\,d\mu: m \ge n\}\). Taking the limit of both sides as \(n \to\infty\) finishes the proof.
The following corollary is an upper bound version of Fatou’s lemma.
Corollary 11.1 Suppose \(f_n\) are nonnegative and measurable, and that \(f_n(x) \to f(x)\) for all \(x\). Then \(\int f\,d\mu \le \sup_n \int f_n\,d\mu\).
Example 11.2 Let \(\mu = \lambda\), \(f_n = 1_{[n, n+1]}\). Then \(f_n(x) \to f(x) := 0\) for all \(x\), while \(\int f_n\,d\mu = 1\) for all \(n\).
This shows that the inequality in Fatou’s lemma may be strict; also this example should help in remembering which way the inequality goes.
11.4 Dominated Convergence Theorem (DCT)
Theorem 11.3 (Dominated Convergence Theorem): Let \(f_n\) and \(g\) be in \(\mathcal L^1(X, \mathcal S, \mu)\), \(|f_n(x)|\le g(x)\) and \(f_n(x) \to f(x)\) for all \(x\). Then \(f\in \mathcal L^1\) and \(\int f_n\,d\mu \to \int f\,d\mu\)
Let \(h_n(x) := \inf \{f_m(x): m\ge n\}\) and \(j_n(x):=\sup \{f_m(x): m \ge n\}\). Then \(h_n\le f_n \le j_n\).
Since \(h_n \uparrow f_n\), and \(\int h_1\,d \mu \ge -\int |g|\,d\mu > -\infty\), we have monotone convergence \(\int h_n\,d\mu \uparrow \int f\,d\mu\) (by MCT). Likewise considering \(-j_n\), we have monotone convergence \(\int j_n\,d\mu \downarrow \int f\,d\mu\).
Since \(\int h_n\,d\mu \le \int f_n\,d\mu \le \int j_n\,d\mu\), we get \(\int f_n\,d\mu \to \int f\,d\mu\).
Example 11.3 \(1_{[n, 2n]}\to 0\) but \(\int 1_{[n, 2n]}d\lambda = n\not\to 0\). This shows how the “domination” hypothesis \(|f_n| \le g\in \mathcal L^1\) is useful.
Remark 11.1. Sums can be considered as integrals for counting measures (which give measure 1 to each singleton), so the above convergence theorems can all be applied to sums.