Monotone convergence, Fatou’s lemma and dominated convergence

Pointwise convergence $f_n \to f$ does not imply $\int f_n \, \dd\mu \to \int f \, \dd\mu$. On $(0, 1]$ with Lebesgue measure, $f_n = n\mathbf 1_{(0, 1/n]}$ converges to zero at every point while $\int f_n = 1$ for all $n$; on $\R$, $f_n = \mathbf 1_{[n, n+1)}$ does the same. In the first case the mass escapes upwards into a vanishing set, in the second it escapes to infinity (Figure 1). The convergence theorems of Lebesgue integration say exactly when such escape is impossible: monotonicity prevents it, nonnegativity limits it to a loss, and domination by a single integrable function rules it out. In probability they are the standard tools for interchanging a limit and an expectation.

Setting

Throughout, $(X, \mathcal A, \mu)$ is a measure space. A nonnegative simple function is $s = \sum_{i=1}^m a_i\mathbf 1_{A_i}$ with $a_i \in [0, \infty)$ and $A_1, \dots, A_m \in \mathcal A$ forming a partition of $X$ (some $a_i$ may be zero); its integral is $\int s \, \dd\mu = \sum_i a_i\mu(A_i)$. For a measurable $f : X \to [0, \infty]$,

\[\int f \, \dd\mu = \sup\Bigl\{\int s \, \dd\mu : s \text{ simple},\ 0 \le s \le f\Bigr\} \in [0, \infty].\]

The integral is monotone: $0 \le f \le g$ implies $\int f \le \int g$. A measurable $f : X \to \R$ is integrable if $\int\abs f < \infty$, and then $\int f = \int f^+ - \int f^-$. One more fact is needed: for a nonnegative simple $s$, the set function $\nu(A) = \int s\mathbf 1_A \, \dd\mu = \sum_i a_i\mu(A_i \cap A)$ is a measure on $\mathcal A$, as a finite nonnegative combination of measures.

A property holds $\mu$-almost everywhere (a.e.) if it fails only on a set contained in a measurable set of measure zero.

Lemma 1. If $f, g : X \to [0, \infty]$ are measurable and $f = g$ a.e., then $\int f \, \dd\mu = \int g \, \dd\mu$. The same holds for integrable real-valued $f, g$.

Proof. Let $N \in \mathcal A$ with $\mu(N) = 0$ and $f = g$ off $N$. If $s = \sum_i a_i\mathbf 1_{A_i}$ is simple with $0 \le s \le f$, then $s\mathbf 1_{N^c}$ is simple, satisfies $s\mathbf 1_{N^c} \le g$, and $\int s\mathbf 1_{N^c} = \sum_i a_i\mu(A_i \cap N^c) = \int s$. Taking the supremum over $s$ gives $\int f \le \int g$, and symmetry gives equality. The real-valued case follows by applying this to positive and negative parts.

Monotone convergence

Theorem 2 (Monotone convergence). Let $f_n : X \to [0, \infty]$ be measurable with $f_1 \le f_2 \le \cdots$ pointwise, and let $f = \lim_n f_n$. Then $f$ is measurable and

\[\lim_{n\to\infty}\int f_n \, \dd\mu = \int f \, \dd\mu .\]

Proof. The limit exists in $[0, \infty]$ at each point because the sequence is nondecreasing, and $f = \sup_n f_n$ is measurable since $\{f > a\} = \bigcup_n\{f_n > a\}$ for every $a \in \R$. By monotonicity of the integral, $\int f_n$ is nondecreasing in $n$ and bounded by $\int f$, so $\alpha = \lim_n\int f_n$ exists and $\alpha \le \int f$.

For the reverse inequality fix a simple $s = \sum_i a_i\mathbf 1_{A_i}$ with $0 \le s \le f$ and a number $c \in (0, 1)$, and let

\[E_n = \{x : f_n(x) \ge c\,s(x)\} = \bigcup_i \bigl(A_i \cap \{f_n \ge c\,a_i\}\bigr) \in \mathcal A .\]

The sets $E_n$ increase because the $f_n$ do, and their union is $X$: if $s(x) = 0$ then $x \in E_1$, and if $s(x) > 0$ then $c\,s(x) < s(x) \le f(x) = \lim_n f_n(x)$, so $f_n(x) \ge c\,s(x)$ for large $n$. Since $f_n \ge f_n\mathbf 1_{E_n} \ge c\,s\mathbf 1_{E_n}$,

\[\int f_n \, \dd\mu \ge c\int s\mathbf 1_{E_n} \, \dd\mu = c\,\nu(E_n),\]

where $\nu(A) = \int s\mathbf 1_A \, \dd\mu$ is the measure introduced above. Continuity of $\nu$ from below gives $\nu(E_n) \to \nu(X) = \int s$, hence $\alpha \ge c\int s$. Letting $c \uparrow 1$ and taking the supremum over $s$ gives $\alpha \ge \int f$.

The factor $c < 1$ is what makes the sets $E_n$ exhaust $X$: with $c = 1$, a point where $f_n(x) \uparrow f(x) = s(x)$ strictly from below would belong to no $E_n$.

Since null sets do not affect integrals, the hypotheses need only hold almost everywhere.

Corollary 3. Let $f_n, f : X \to [0, \infty]$ be measurable, with $f_n \le f_{n+1}$ a.e. for each $n$ and $f_n \to f$ a.e. Then $\int f_n \to \int f$.

Proof. The set $S = \bigcup_n\{f_n > f_{n+1}\} \cup \{f_n \not\to f\}$ is measurable, because $\{f_n \to f\} = \{\liminf_n f_n = f\} \cap \{\limsup_n f_n = f\}$ and $\liminf_n f_n$, $\limsup_n f_n$ are measurable. It is a countable union of null sets, so $\mu(S) = 0$. The functions $f_n\mathbf 1_{S^c}$ increase everywhere to $f\mathbf 1_{S^c}$, and Theorem 2 together with Lemma 1 gives $\int f_n = \int f_n\mathbf 1_{S^c} \to \int f\mathbf 1_{S^c} = \int f$.

The measurability of $f$ is assumed here and not deduced: an a.e. limit of measurable functions need not be measurable unless the measure space is complete.

Fatou’s lemma

Without monotonicity, nonnegative functions can lose mass in the limit but cannot gain it.

Theorem 4 (Fatou’s lemma). If $f_n : X \to [0, \infty]$ are measurable, then

\[\int \liminf_{n\to\infty} f_n \, \dd\mu \le \liminf_{n\to\infty}\int f_n \, \dd\mu .\]

Proof. Let $g_k = \inf_{n \ge k} f_n$. Each $g_k$ is measurable, since $\{g_k \ge a\} = \bigcap_{n \ge k}\{f_n \ge a\}$, and $g_k \uparrow \liminf_n f_n$ pointwise. Because $g_k \le f_n$ for every $n \ge k$, monotonicity gives $\int g_k \le \inf_{n \ge k}\int f_n$. By monotone convergence,

\[\int\liminf_n f_n \, \dd\mu = \lim_{k\to\infty}\int g_k \, \dd\mu \le \lim_{k\to\infty}\inf_{n \ge k}\int f_n \, \dd\mu = \liminf_n\int f_n \, \dd\mu .\]

Both examples of the introduction make the inequality strict: $\liminf_n f_n = 0$ while $\int f_n = 1$. So does $f_n = \mathbf 1_{\{n\}}$ on $\N$ with counting measure. Nonnegativity cannot be dropped: $-f_n$ with $f_n = n\mathbf 1_{(0, 1/n]}$ has $\int\liminf_n(-f_n) = 0 > -1 = \liminf_n\int(-f_n)$.

Dominated convergence

Theorem 5 (Dominated convergence). Let $f_n, f : X \to \R$ be measurable with $f_n \to f$ a.e., and suppose there is an integrable $g : X \to [0, \infty)$ with $\abs{f_n} \le g$ a.e. for every $n$. Then $f$ is integrable and

\[\lim_{n\to\infty}\int\abs{f_n - f} \, \dd\mu = 0, \qquad\text{in particular}\qquad \lim_{n\to\infty}\int f_n \, \dd\mu = \int f \, \dd\mu .\]

Proof. Let $N$ be the union of $\{f_n \not\to f\}$ and the sets $\{\abs{f_n} > g\}$; it is a countable union of null sets. Replacing $f_n$, $f$ and $g$ by $0$ on $N$ changes no integral (Lemma 1), so we may assume convergence and domination everywhere. Then $\abs f \le g$, so $f$ is integrable, and $h_n = 2g - \abs{f_n - f} \ge 0$ with $h_n \to 2g$. Fatou’s lemma gives

\[\int 2g \, \dd\mu \le \liminf_n\int\bigl(2g - \abs{f_n - f}\bigr)\dd\mu = \int 2g \, \dd\mu - \limsup_n\int\abs{f_n - f} \, \dd\mu ,\]

where splitting the integral is legitimate because $2g$ and $\abs{f_n - f}$ are integrable. Since $\int 2g < \infty$, it can be cancelled, leaving $\limsup_n\int\abs{f_n - f} \le 0$. Finally $\abs{\int f_n - \int f} \le \int\abs{f_n - f}$.

The dominating function must be integrable and must not depend on $n$. In the first example, $\sup_n f_n(x) = \lfloor 1/x \rfloor$ for $x \in (0, 1]$, which behaves like $1/x$ and is not integrable; in the second, $\sup_n f_n = \mathbf 1_{[1, \infty)}$ is not integrable on $\R$. Panel (c) of Figure 1 shows a sequence where domination holds: for $x \ge 0$ the sequence $(1 + x/n)^n$ increases to $e^x$, so for $n \ge 2$

\[0 \le \Bigl(1 + \frac xn\Bigr)^{-n} \le \Bigl(1 + \frac x2\Bigr)^{-2}, \qquad \int_0^\infty\Bigl(1 + \frac x2\Bigr)^{-2}\dd x = 2 < \infty,\]

and dominated convergence gives $\int_0^\infty(1 + x/n)^{-n}\,\dd x \to \int_0^\infty e^{-x}\,\dd x = 1$. Here the integrals can be computed exactly, $\int_0^\infty(1 + x/n)^{-n}\,\dd x = n/(n-1)$, which confirms the limit.

Three panels: rectangles of height n on (0, 1/n], unit rectangles moving to the right, and decreasing curves (1 + x/n) to the power minus n under a shaded dominating function. — Figure 1. (a) $f_n = n\mathbf{1}_{(0,1/n]}$ and (b) $f_n = \mathbf{1}_{[n,n+1)}$ converge to zero pointwise, yet every $f_n$ has integral one; no integrable function dominates them. (c) $f_n(x) = (1 + x/n)^{-n}$ for $n = 2, 4, 8$ (colours as in (a)) lies under the integrable envelope $g(x) = (1 + x/2)^{-2}$ (shaded), which is $f_2$ itself, and converges to $e^{-x}$ (dashed); the integrals $n/(n-1)$ converge to $1$.

When $\mu$ is finite, a constant bound is an integrable dominating function, which gives the bounded convergence theorem: if $\mu(X) < \infty$, $\abs{f_n} \le M$ a.e. and $f_n \to f$ a.e., then $\int f_n \to \int f$. On a probability space this says $\E X_n \to \E X$ for uniformly bounded random variables converging almost surely.

Convergence in measure

Definition 6. Measurable functions $f_n : X \to \R$ converge in measure to a measurable $f : X \to \R$ if $\mu\bigl(\{\abs{f_n - f} \ge \varepsilon\}\bigr) \to 0$ for every $\varepsilon > 0$.

On a probability space this is convergence in probability. On a general measure space it is neither weaker nor stronger than a.e. convergence. With counting measure on $\N$, $f_n = \mathbf 1_{\{n\}}$ converges to $0$ everywhere, but $\mu(\{\abs{f_n} \ge 1\}) = 1$ for all $n$; by Proposition 8 below, any limit in measure would have to agree with the pointwise limit $0$ a.e., so $(f_n)$ does not converge in measure to anything. Conversely, the “typewriter” sequence of indicators of the dyadic intervals $[j2^{-k}, (j+1)2^{-k})$, $j = 0, \dots, 2^k - 1$, $k = 0, 1, \dots$, on $[0, 1)$ converges to $0$ in Lebesgue measure but converges at no point. When $\mu(X) < \infty$, a.e. convergence does imply convergence in measure: if $f_n \to f$ a.e., then $\mathbf 1\{\abs{f_n - f} \ge \varepsilon\} \to 0$ a.e., and bounded convergence gives $\mu(\abs{f_n - f} \ge \varepsilon) \to 0$.

Proposition 7. If $f_n \to f$ in measure, then $f_n \to g$ in measure if and only if $f = g$ a.e.

Proof. If $f_n$ converges in measure to both, then for $\varepsilon > 0$ the triangle inequality gives $\{\abs{f - g} \ge \varepsilon\} \subseteq \{\abs{f - f_n} \ge \varepsilon/2\} \cup \{\abs{f_n - g} \ge \varepsilon/2\}$, so $\mu(\abs{f - g} \ge \varepsilon)$ is bounded by a quantity tending to zero and equals zero. Taking $\varepsilon = 1/k$ and a union over $k$ gives $\mu(f \ne g) = 0$. Conversely, if $f = g$ a.e., then $\{\abs{f_n - g} \ge \varepsilon\}$ and $\{\abs{f_n - f} \ge \varepsilon\}$ differ by a null set.

Proposition 8. If $f_n \to f$ in measure, there is a subsequence $n_1 < n_2 < \cdots$ with $f_{n_k} \to f$ a.e.

Proof. Choose $n_k > n_{k-1}$ with $\mu(S_k) \le 2^{-k}$, where $S_k = \{\abs{f_{n_k} - f} > 2^{-k}\}$. Let $S = \bigcap_{m \ge 1}\bigcup_{k \ge m}S_k$, the set of points lying in infinitely many $S_k$. For every $m$, countable subadditivity gives

\[\mu(S) \le \mu\Bigl(\bigcup_{k \ge m}S_k\Bigr) \le \sum_{k \ge m}2^{-k} = 2^{1-m},\]

so $\mu(S) = 0$. If $x \notin S$, then $x \notin S_k$ for all large $k$, so $\abs{f_{n_k}(x) - f(x)} \le 2^{-k}$ eventually and $f_{n_k}(x) \to f(x)$.

The summability of the bounds $\mu(S_k)$ is essential; with bounds such as $1/k$, whose sum diverges, the argument gives nothing. Proposition 8 turns any statement about a.e. convergence into one about convergence in measure along subsequences, and in particular extends dominated convergence.

Theorem 9 (Dominated convergence in measure). Let $f_n \to f$ in measure, and suppose $\abs{f_n} \le g$ a.e. for every $n$, with $g$ integrable. Then $f$ is integrable and $\int\abs{f_n - f} \, \dd\mu \to 0$; in particular $\int f_n \to \int f$.

Proof. By Proposition 8 a subsequence converges to $f$ a.e., so $\abs f \le g$ a.e. and $f$ is integrable. Let $a_n = \int\abs{f_n - f}$. A sequence of real numbers converges to $0$ if every subsequence has a further subsequence converging to $0$. Given a subsequence $(f_{n_k})$, it still converges to $f$ in measure, so by Proposition 8 a further subsequence converges to $f$ a.e., and along it $a_n \to 0$ by Theorem 5. Hence $a_n \to 0$.

Remarks

On a probability space, Theorem 9 reads: if $X_n \to X$ in probability and $\abs{X_n} \le Y$ with $\E Y < \infty$, then $\E\abs{X_n - X} \to 0$. Domination is sufficient but not necessary. The sharp condition is uniform integrability, $\lim_{M\to\infty}\sup_n\E\bigl[\abs{X_n}\mathbf 1\{\abs{X_n} > M\}\bigr] = 0$: on a probability space, $X_n \to X$ in $L^1$ if and only if $X_n \to X$ in probability and $(X_n)$ is uniformly integrable (Vitali’s convergence theorem). Example (a) of Figure 1 fails uniform integrability, since its mass sits where $f_n$ is large. Example (b) does not: it is bounded by $1$, so the truncated integrals vanish for $M \ge 1$. On a space of infinite measure, Vitali’s theorem needs a second condition, tightness: for every $\eta > 0$ there is a set $E$ with $\mu(E) < \infty$ and $\sup_n\int_{E^c}\abs{f_n}\,\dd\mu < \eta$. Example (b) fails this condition, because its mass escapes every set of finite measure.

Standard references are Folland, Real Analysis: Modern Techniques and Their Applications (Chapter 2), Rudin, Real and Complex Analysis (Chapter 1), and, for the probabilistic versions, Durrett, Probability: Theory and Examples.