Fatou's lemma, monotone and dominated convergence, and the $L^p$ spaces

The lecture gives three theorems on the integrals of sequences of measurable functions and defines the spaces ${L^p.}$

Three most important theorems#

Theorem 1 (Fatou’s lemma). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let $(f_n)$ be a sequence of measurable functions ${f_n \colon X \to [0, +\infty].}$ Let ${f \colon X \to [0, +\infty]}$ be given by ${f(x) = \varliminf_{n \to \infty} f_n(x),}$ written ${f = \varliminf_n f_n.}$ Then

$$\varliminf_n \int_X f_n \, d\mu \ge \int_X f \, d\mu, \qquad \text{that is,} \qquad \varliminf_n \int_X f_n \, d\mu \ge \int_X \varliminf_n f_n \, d\mu.$$

Proof (sketch). Let

$$s = \sum_{k=1}^{m} a_k \chi_{A_k}$$

be a simple function such that ${0 \le s \le f,}$ written as in Definition 16 of the fourth post. Only the case ${0 < a_k < +\infty}$ for every ${k = 1, \dots, m}$ is treated. Let ${t \in (0, 1).}$ For every ${k = 1, \dots, m}$ and every ${n \in \mathbb{N}}$ define

$$B_{k,n} := \{ x \in A_k : f_j(x) \ge t a_k \ \text{for every } j \ge n \} \subset A_k,$$

as in Figure 1. These sets satisfy ${B_{k,n} \subset B_{k,n+1}}$ for every $n$ and every ${k.}$

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Figure 1. The drawing of the proof, with an instance on an interval $X$ and ${\mu = \lambda{:}}$ the function ${f,}$ the level $a_k$ of $s$ over the set ${A_k,}$ the dotted level ${t a_k,}$ the function $f_n$ and, faint, the next functions $f_j$ with ${j > n.}$ On the axis, the set ${B_{k,n},}$ under hatched rectangles of height ${t a_k;}$ the sliders set $n$ and ${t.}$

Let ${x \in A_k.}$ Then ${f(x) = \varliminf_n f_n(x) \ge s(x) = a_k > t a_k.}$ Hence there exists ${n_x \in \mathbb{N}}$ such that ${f_j(x) \ge t a_k}$ for every ${j \ge n_x.}$ In other words, for every ${x \in A_k}$ there exists $n_x$ such that ${x \in B_{k, n_x},}$ so that

$$\bigcup_n B_{k,n} = A_k$$

and, by the continuity of measures, Proposition 6 (i) of the second post,

$$\mu(A_k) = \lim_{n \to +\infty} \mu(B_{k,n}).$$

For every ${n,}$ by monotonicity, and since ${f_n \ge t a_k}$ on ${B_{k,n},}$

$$\begin{aligned} \int_X f_n \, d\mu &\ge \sum_{k=1}^{m} \int_{B_{k,n}} f_n \, d\mu \\ &\ge \sum_{k=1}^{m} \int_{B_{k,n}} t a_k \, d\mu \\ &= \sum_{k=1}^{m} t a_k \, \mu(B_{k,n}). \end{aligned}$$

Taking the lower limit as ${n \to \infty}$ and using the limit above,

$$\begin{aligned} \varliminf_{n \to \infty} \int_X f_n \, d\mu &\ge \sum_{k=1}^{m} t a_k \varliminf_n \mu(B_{k,n}) \\ &= \sum_{k=1}^{m} t a_k \, \mu(A_k) \\ &= t \sum_{k=1}^{m} a_k \, \mu(A_k) \\ &= t \int_X s \, d\mu. \end{aligned}$$

Taking the supremum over the simple functions $s$ with ${0 \le s \le f}$ gives, by Definition 3 of the previous post,

$$\varliminf_n \int_X f_n \, d\mu \ge t \int_X f \, d\mu,$$

and letting ${t \to 1^-}$ gives

$$\varliminf_n \int_X f_n \, d\mu \ge \int_X f \, d\mu. \tag*{$\blacksquare$}$$

Remark 2. In Theorem 1 the assumption ${f_n \colon X \to [0, +\infty]}$ is needed. Otherwise, on $\mathbb{R}$ with the Lebesgue measure ${\lambda,}$ let ${f_n = -n \chi_{[0, 1/n]}}$ for every ${n \ge 1,}$ as in Figure 2. Then ${\int_{\mathbb{R}} f_n \, d\lambda = -1}$ for every ${n,}$ and ${f_n \to 0}$ almost everywhere, so that, with ${f = 0,}$

$$\int_{\mathbb{R}} f \, d\lambda = 0 \not\le \varliminf_n \int_{\mathbb{R}} f_n \, d\lambda = -1.$$

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Figure 2. The drawing of Remark 2: the function ${f_n,}$ equal to $-n$ on $[0, 1/n]$ and to $0$ elsewhere, and the region of area $1$ between its graph and the axis. The slider sets ${n.}$

Remark 3 (Reverse Fatou’s lemma). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let $(f_n)$ be a sequence of measurable functions ${f_n \colon X \to [0, +\infty),}$ and let ${f = \varlimsup_n f_n.}$ Assume that there exists a measurable ${g \colon X \to [0, +\infty]}$ such that ${g \in \mathcal{L}^1}$ and ${g \ge f_n}$ almost everywhere for every ${n.}$ Then

$$\varlimsup_n \int_X f_n \, d\mu \le \int_X f \, d\mu.$$

Proof (sketch). Let ${h_n = g - f_n,}$ so that ${h_n \ge 0}$ almost everywhere, and

$$\varliminf_n h_n = g - \varlimsup_n f_n = g - f.$$

Fatou’s lemma, Theorem 1, applied to $(h_n)$ gives

$$\varliminf_n \int_X h_n \, d\mu \ge \int_X (g - f) \, d\mu,$$

where

$$\varliminf_n \int_X h_n \, d\mu = \int_X g \, d\mu - \varlimsup_n \int_X f_n \, d\mu. \tag*{$\blacksquare$}$$

Theorem 4 (Monotone convergence, Beppo Levi). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let $(f_n)$ be a sequence of measurable functions ${f_n \colon X \to [0, +\infty].}$ Assume that ${f_{n+1} \ge f_n}$ almost everywhere for every ${n.}$ Let ${f \colon X \to [0, +\infty]}$ be such that ${f = \lim_{n \to \infty} f_n}$ almost everywhere, that is, ${f(x) = \sup_n f_n(x)}$ for almost every ${x.}$ Then

$$\lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu, \qquad \text{that is,} \qquad \sup_n \int_X f_n \, d\mu = \int_X \sup_n f_n \, d\mu.$$

Proof (complete). We have

$$\begin{aligned} \int_X f \, d\mu &\le \varliminf_n \int_X f_n \, d\mu \le \varlimsup_n \int_X f_n \, d\mu \\ &\le \varlimsup_n \int_X f \, d\mu = \int_X f \, d\mu, \end{aligned}$$

where the first inequality is Fatou’s lemma, Theorem 1, and the third is monotonicity, as ${f_n \le f}$ almost everywhere. $\blacksquare$

Remark 5. In Theorem 4 the assumption ${f_{n+1} \ge f_n}$ almost everywhere for every $n$ is needed. Otherwise, on $\mathbb{R}$ with the Lebesgue measure ${\lambda,}$ let ${f_n = \chi_{[n, n+1]}}$ for every ${n \in \mathbb{N},}$ as in Figure 3. Then ${f_n \to 0}$ pointwise and ${\int_{\mathbb{R}} f_n \, d\lambda = 1}$ for every ${n.}$

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Figure 3. The drawing of Remark 5: the function ${f_n,}$ equal to $1$ on ${[n, n + 1]}$ and to $0$ elsewhere, and the region of area $1$ under its graph. The slider sets ${n.}$

Remark 6 (Reverse monotone convergence). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let $(f_n)$ be a sequence of measurable functions ${f_n \colon X \to [0, +\infty).}$ Assume that ${f_{n+1} \le f_n}$ almost everywhere for every ${n.}$ Let ${f \colon X \to [0, +\infty]}$ be such that ${f = \lim_n f_n}$ almost everywhere, that is, ${f(x) = \inf_n f_n(x)}$ for almost every ${x.}$ Suppose that ${\int_X f_0 \, d\mu < +\infty.}$ Then

$$\lim_n \int_X f_n \, d\mu = \int_X f \, d\mu, \qquad \text{that is,} \qquad \inf_n \int_X f_n \, d\mu = \int_X \inf_n f_n \, d\mu.$$

Proof (sketch). Let ${h_n = f_0 - f_n.}$ Then ${h_{n+1} \ge h_n}$ almost everywhere, and the monotone convergence theorem, Theorem 4, applies to ${(h_n).}$ $\blacksquare$

Remark 7. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let $(f_n)$ be a sequence of measurable functions ${f_n \colon X \to [0, +\infty].}$ Then

$$\int_X \Big( \sum_{n=0}^{\infty} f_n \Big) \, d\mu = \sum_{n=0}^{\infty} \int_X f_n \, d\mu,$$

where the series inside the integral is the function ${f \colon X \to [0, +\infty]}$ defined by

$$f(x) = \sum_{n=0}^{\infty} f_n(x).$$

Proof (complete). The partial sums

$$S_N = \sum_{n=0}^{N} f_n$$

satisfy ${S_N \ge S_{N-1}}$ for every ${N \ge 1}$ and converge to $f$ pointwise. Hence

$$\begin{aligned} \int_X \lim_{N \to \infty} \Big( \sum_{n=0}^{N} f_n \Big) \, d\mu &= \lim_N \int_X \sum_{n=0}^{N} f_n \, d\mu \\ &= \lim_N \sum_{n=0}^{N} \int_X f_n \, d\mu \\ &= \sum_{n=0}^{\infty} \int_X f_n \, d\mu, \end{aligned}$$

where the first equality is the monotone convergence theorem, Theorem 4. $\blacksquare$

Proposition 8 (Measure induced by a nonnegative function). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${\varphi \colon X \to [0, +\infty]}$ be measurable, and define ${\nu \colon \mathcal{M} \to [0, +\infty]}$ by

$$\nu(E) := \int_E \varphi \, d\mu.$$

Then $\nu$ is a measure on ${\mathcal{M}.}$

Remark 9. Let $\varphi$ and $\nu$ be as in Proposition 8. For every measurable function ${f \colon X \to [0, +\infty),}$

$$\int_X f \, d\nu = \int_X f \varphi \, d\mu.$$

For example, let ${X = \mathbb{R},}$ ${\mu = \lambda}$ and ${\varphi \colon \mathbb{R} \to [0, +\infty]}$ be measurable: Figure 4 draws $\lambda$ and the measure ${\nu,}$ written ${\varphi \lambda.}$

Figure 4. The drawing of Remark 9: on the left the Lebesgue measure ${\lambda,}$ drawn as a horizontal line, and on the right, after the arrow, the measure ${\varphi \lambda,}$ drawn as the graph of ${\varphi,}$ with the region under it over a small interval hatched.

Proof (sketch). Only the $\sigma$-additivity of the function $\nu$ of Proposition 8 is proved. Let ${(E_n) \subset \mathcal{M}}$ be pairwise disjoint and let ${E = \bigcup_n E_n.}$ The goal is

$$\nu(E) = \sum_{n=0}^{\infty} \nu(E_n), \qquad \text{that is,} \qquad \int_E \varphi \, d\mu = \sum_{n=0}^{\infty} \int_{E_n} \varphi \, d\mu.$$

We compute, as in Figure 5,

$$\begin{aligned} \sum_{n=0}^{\infty} \int_{E_n} \varphi \, d\mu &= \sum_{n=0}^{\infty} \int_X \varphi \chi_{E_n} \, d\mu \\ &= \int_X \sum_{n=0}^{\infty} \varphi \chi_{E_n} \, d\mu \\ &= \int_X \varphi \chi_E \, d\mu \\ &= \int_E \varphi \, d\mu, \end{aligned}$$

where the second equality is Remark 7 and the third holds because the series inside the integral is ${\varphi \chi_E.}$ $\blacksquare$

Figure 5. The drawing of the proof: three of the pairwise disjoint sets ${E_n,}$ here ${E_1,}$ $E_2$ and ${E_3,}$ in a plane, and above each of them, at height ${1,}$ the graph of its characteristic function, the first one marked ${\chi_{E_1}.}$

Theorem 10 (Dominated convergence). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let $(f_n)$ be a sequence of measurable functions ${f_n \colon X \to \mathbb{R},}$ and let ${f \colon X \to \mathbb{R}.}$ Suppose that ${f_n \to f}$ almost everywhere and that there exists ${g \in \mathcal{L}^1(X)}$ such that ${|f_n| \le g}$ almost everywhere for every ${n.}$ Then ${f \in \mathcal{L}^1(X)}$ and

$$\lim_{n \to \infty} \int_X |f_n - f| \, d\mu = 0.$$

Proof (complete). At almost every ${x \in X,}$

$$\begin{aligned} |f(x)| &= \lim_{n \to \infty} |f_n(x)| \\ &\le g(x), \end{aligned}$$

because ${|f_n(x)| \le g(x)}$ for every $n$ at almost every ${x.}$ Hence

$$\int_X |f| \, d\mu \le \int_X g \, d\mu < +\infty,$$

and ${f \in \mathcal{L}^1.}$

(i) For every ${n,}$ almost everywhere,

$$|f_n - f| \le |f_n| + |f| \le 2g \in \mathcal{L}^1,$$

where ${|f_n| \le g}$ almost everywhere by assumption and ${|f| \le g}$ almost everywhere by the first step.

(ii) By Fatou’s lemma, Theorem 1, and by the reverse Fatou’s lemma, Remark 3, applied to the sequence ${(|f_n - f|)}$ with $2g$ in place of ${g,}$

$$\begin{aligned} 0 &= \int_X \varliminf_n |f_n - f| \, d\mu \\ &\le \varliminf_n \int_X |f_n - f| \, d\mu \\ &\le \varlimsup_n \int_X |f_n - f| \, d\mu \\ &\le \int_X \varlimsup_n |f_n - f| \, d\mu \\ &= 0, \end{aligned}$$

where ${\varliminf_n |f_n - f| = \varlimsup_n |f_n - f| = 0}$ almost everywhere. Hence the limit of ${\int_X |f_n - f| \, d\mu}$ exists and is ${0.}$ $\blacksquare$

Remark 11. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${(f_n) \subset \mathcal{L}^1(X).}$ Suppose that

$$\sum_{n=0}^{\infty} \int_X |f_n| \, d\mu < +\infty.$$

Then

$$\sum_{n=0}^{\infty} \int_X f_n \, d\mu = \int_X \sum_{n=0}^{\infty} f_n \, d\mu.$$

The $L^p$ spaces#

Definition 12. Let ${(X, \mathcal{M}, \mu)}$ be a measure space. For every ${p \in [1, +\infty)}$ we define

$$\mathcal{L}^p(X, \mathcal{M}, \mu) := \Big\{ f \colon X \to \mathbb{R} : f \text{ measurable}, \ \int_X |f|^p \, d\mu < +\infty \Big\}.$$

Remark 13. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${p \in [1, +\infty).}$
(i) If ${f, g \colon X \to \mathbb{R}}$ are measurable and ${f = g}$ almost everywhere, then ${\int_X |f|^p \, d\mu = \int_X |g|^p \, d\mu.}$

(ii) For functions ${f, g \colon X \to \mathbb{R},}$ the relation given by ${f \sim g}$ if and only if ${f = g}$ almost everywhere is an equivalence relation. Hence we can consider the equivalence class of a function ${f,}$

$$[f] = \{ g : g \sim f \}.$$

Definition 14 ($L^p$ spaces). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${p \in [1, +\infty).}$ We define

$$L^p(X, \mathcal{M}, \mu) := \{ [f] : f \in \mathcal{L}^p(X, \mathcal{M}, \mu) \},$$

also written ${L^p(X)}$ or ${L^p.}$

Notation. The elements of $L^p$ are indicated simply by ${f,}$ thinking of $f$ as a representative of ${[f].}$

Remark 15. Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${p \in [1, +\infty)}$ and let ${f, g \in L^p.}$
(1) It makes sense to write ${\int_X |f|^p \, d\mu,}$ because it does not depend on the choice of the representative. It makes sense to write ${f + g}$ and ${f \cdot g,}$ since ${f + g}$ is a representative of the class ${[f + g]}$ for any choice of representatives $f$ and $g$ of $[f]$ and ${[g],}$ and the same holds for ${f \cdot g.}$ It makes sense to write ${f \ge g}$ almost everywhere.
(2) It does not make sense to write ${f(x_0)}$ for a fixed ${x_0 \in X,}$ if ${\mu(\{x_0\}) = 0.}$ It does not make sense to write ${f \ge g}$ for every ${x \in X.}$

Remark 16. Let ${(f_n) \subset L^p}$ and ${f \in L^p.}$ Then it makes sense to write ${f_n \to f}$ almost everywhere. We also write

$$C([0, 1]) \subset L^p([0, 1], \mathcal{L}([0, 1]), \lambda),$$

where ${C([0, 1])}$ is the set of the continuous functions on ${[0, 1].}$

Definition 17 ($L^p$-norm). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${p \in [1, +\infty).}$ For ${f \in L^p(X)}$ we define the $L^p$-norm of $f$ as

$$\|f\|_p := \Big( \int_X |f|^p \, d\mu \Big)^{1/p}.$$

Notation. The norm ${\|f\|_p}$ is also written ${\|f\|_{L^p}}$ or ${\|f\|_{L^p(X)}.}$

Remark 18. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${p \in [1, +\infty).}$
(i) The set ${L^p(X, \mathcal{M}, \mu)}$ is a vector space.

(ii) If ${f, g \in L^p(X, \mathcal{M}, \mu),}$ then

$$\begin{aligned} \|f - g\|_p = 0 &\iff f = g \text{ in } L^p \\ &\iff f = g \text{ almost everywhere}. \end{aligned}$$

Proof (sketch). In (i) only sums are checked. For ${f, g \in L^p(X),}$

$$\begin{aligned} \int_X |f + g|^p \, d\mu &\le \int_X 2^{p-1} \big( |f|^p + |g|^p \big) \, d\mu \\ &= 2^{p-1} \Big( \int_X |f|^p \, d\mu + \int_X |g|^p \, d\mu \Big) \\ &< +\infty, \end{aligned}$$

where the inequality uses, for ${a, b \in \mathbb{R},}$

$$|a + b|^p \le 2^{p-1} \big( |a|^p + |b|^p \big).$$

In (ii), the equivalence of ${\|f - g\|_p = 0}$ and ${f = g}$ almost everywhere is the vanishing lemma for nonnegative functions, (1) of Corollary 10 of the previous post, applied to ${|f - g|^p.}$ $\blacksquare$

Definition 19 (Essentially bounded functions and $L^\infty$). Let ${(X, \mathcal{M}, \mu)}$ be a measure space.
(i) A measurable function ${f \colon X \to \mathbb{R}}$ is said to be essentially bounded if there exists ${M \ge 0}$ such that ${\mu(\{ |f| > M \}) = 0,}$ as in Figure 6.

(ii) For a measurable function ${f \colon X \to \mathbb{R}}$ we define

$$\begin{aligned} \operatorname*{ess\,sup}_X f &:= \inf \{ M \in \mathbb{R} : \mu(\{ f > M \}) = 0 \}, \\ \operatorname*{ess\,inf}_X f &:= \sup \{ M \in \mathbb{R} : \mu(\{ f < M \}) = 0 \}, \end{aligned}$$

and ${\operatorname*{ess\,sup}_X f}$ is called the essential supremum of ${f.}$

(iii) We define

$$\begin{aligned} \mathcal{L}^\infty(X, \mathcal{M}, \mu) &:= \{ f \colon X \to \mathbb{R} : f \text{ measurable and essentially bounded} \}, \\ L^\infty(X, \mathcal{M}, \mu) &:= \{ [f] : f \in \mathcal{L}^\infty(X, \mathcal{M}, \mu) \}. \end{aligned}$$

(iv) For ${f \in L^\infty(X)}$ we define the $L^\infty$-norm ${\|f\|_\infty := \operatorname*{ess\,sup}_X |f|,}$ also written ${\|f\|_{L^\infty}}$ or ${\|f\|_{L^\infty(X)}.}$

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Figure 6. The drawing of Definition 19 (i): a function ${f,}$ the dashed level $M$ and the set ${\{ |f| > M \},}$ hatched on the axis. The slider sets ${M.}$

Remark 20. If ${f \in L^\infty(X),}$ then ${|f| \le \|f\|_\infty}$ almost everywhere.

Lemma 21. Let ${(X, \mathcal{M}, \mu)}$ be a measure space with ${\mu(X) < +\infty,}$ and let ${f \colon X \to \mathbb{R}}$ be measurable. Then ${f \in L^\infty(X)}$ if and only if the limit ${\lim_{p \to +\infty} \|f\|_p}$ exists and is finite. Moreover, in this case,

$$\lim_{p \to \infty} \|f\|_p = \|f\|_\infty.$$