The Daniell integral#

Let \(\mu\) be a measure on a set \(X\) and let \(L\) be a set of real valued \(\mu\)-measurable functions on \(X\). The formula

\[T(f) := \int_X f\, \mathrm{d}\mu\]

defines an operator on \(L\). Moreover, it has the following two properties:

  • \(T\) is monotonic: for all \(f,g\in L\) with \(g\leq f\), \(T(g)\leq f\);

  • \(T\) is continuous with respect to monotone convergence: if \(f_i\in C(X)\) monotonically increase to \(f\) then \(T(f_i)\to T(f)\).

In the next theorem we see that these two properties completely characterise the Lebesgue integral. That is, we could equivalently develop a theory of integration (and hence measure) by beginning with operators on sets of functions.

Definition 13 (Lattice of functions)

Let \(X\) be a set. A lattice of functions on \(X\) is a non-empty set \(L\) of functions \(X\to\mathbb R\) which satisfies the following conditions: for any \(c\in \mathbb R^+\) and any \(f,g\in L\), \(f+g\), \(cf\), \(\inf\{f,g\}\) and \(\inf\{f,c\}\) all belong to \(L\) and if \(g\leq f\) then \(f-g\in L\) too. Note that any vector space of functions closed under \(\inf\) is a lattice.

If \(L\) is a lattice we let

\[L^+=\{f\in L: f\geq 0\}.\]

We say a \(T\colon L\to \mathbb R\) is

  1. Linear if, for all \(f,g\in L\) and \(a,b\in\mathbb R^+\),

  2. Monotonic if, for all \(f,g\in L\) with \(g\leq f\),

    \[T(g)\leq T(f);\]
  3. Continuous with respect to monotone convergence if, for all \(f_i\in L\) that monotonically increase to \(f\),

    \[T(f_i)\to T(f).\]
  4. Bounded if for every \(f\in L\),

    \[\sup\{T(g) : 0\leq g\leq f\}<\infty.\]

A \(T\) satisfying the first three items is called a monotone Daniell integral (it necessarily satisfies the fourth item. A \(T\) satisfying items 1, 3 and 4 is called a Daniell integral.

Theorem 6

Let \(L\) be a lattice on \(X\) and let \(T\colon L\to\mathbb R\) be a monotone Daniell integral. Then there exists a measure \(\mu\) on \(X\) for which each \(f\in L^+\) is \(\mu\)-measurable such that

\[\label{mu_rep_T}T(f) = \int f\, \mathrm{d}\mu\]

for all \(f\in L\).

Proof. First note that, for any \(f\in L^+\),

\[T(f) \geq T(0\cdot f)=0.\]

For \(A\subset X\) we say that a sequence \(f_i\in L^+\) suits \(A\) if the \(f_i\) monotonically increase and

\[\lim_{i\to\infty} f_i(x) \geq 1 \quad \forall x\in A.\]


\[\mu(A)=\inf\left\{\lim_{i\to\infty} T(f_i) : f_i \text{ suits } A\right\}.\]

Then \(\mu\) is a measure on \(X\). Indeed, \(\mu(\emptyset)=0\) and \(\mu\) is monotonic. If \(A_j\subset X\) for each \(j\in\mathbb N\) and \(f_i^j\in L^+\) suit \(A_j\) then

\[g_i := \sum_{j=1}^i f_i^j\]

suits \(\cup_j A_j\) and

\[T(g_i) = \sum_{j=1}^i T(f_i^j) \leq \sum_{j\in \mathbb N} \lim_{i\to\infty} T(f_i^j)= \sum_{j\in\mathbb N} \mu(A_j).\]

Next we show that each \(f\in L^+\) is \(\mu\)-measurable. By Exercise 15, it suffices to show, for every \(E\subset X\) and \(0\leq a<b\in \mathbb R\) that, for

\[A := f^{-1}((-\infty,a)), \quad B:= f^{-1}((b,\infty)),\]

we have

\[\mu(E) \geq \mu(E\cap A) + \mu(E\cap B).\]

Suppose \(g_i\) suit \(E\) and let

\[h=\frac{\inf\{f,b\}-\inf\{f,a\}}{b-a}, \quad k_i=\inf\{g_i,h\}.\]

Then \(0\leq k_{i+1}-k_i \leq g_{i+1}- g_i\) for all \(i\) and

\[h(x)=1 \text{ whenever } f(x)\geq b, \quad h(x)=0 \text{ whenever }f(x)\leq a.\]

Then \(k_i\) suit \(B\) and \(g_i-k_i\) suit \(A\). Therefore

\[\lim_{i\to\infty} T(g_i) = \lim_{i\to\infty} T(k_i) + T(g_i+k_i) \geq \mu(B) +\mu(A).\]

Finally we show that

(4)#\[T(f) = \int f\, \mathrm{d}\mu \quad f\in L^+.\]

First suppose that \(A\subset X\), \(f_i\) suit \(A\) and \(g\in L^+\) satisfies \(g\leq \chi_A\). Then \(h_i = \inf\{f_i,g\}\) monotonically increase to \(g\) and so

\[T(g) = \lim_{i\to\infty} T(h_i) \leq \lim_{i\to\infty} T(f_i).\]


(5)#\[T(g) \leq \mu(A).\]

Now fix \(f\in L^+\) and for each \(t\in \mathbb R^+\) let \(f_t=\inf\{f,t\}\). For a moment fix \(\epsilon>0\). Then, for each \(k\in\mathbb N\),

(6)#\[0\leq f_{k\epsilon}(x)-f_{(k-1)\epsilon}(x) \leq \epsilon \quad \forall x\in X,\]
(7)#\[f_{k\epsilon}(x) -f_{(k-1)\epsilon}(x) = \epsilon \quad \text{whenever }f(x)\geq k\epsilon\]


(8)#\[f_{k\epsilon}(x) -f_{(k-1)\epsilon}(x)=0 \quad \text{whenever }f(x)\leq (k-1)\epsilon.\]

Note that, for any \(k\in\mathbb N\), \((f_{k\epsilon}-f_{(k-1)\epsilon})/\epsilon\) suits

\[\{x: f(x)\geq k\epsilon\}\]

and so

(9)#\[T(f_{k\epsilon}-f_{(k-1)\epsilon}) \geq \epsilon \mu(\{x: f(x)\geq k\epsilon\}).\]

By (6) and (7) respectively,

(10)#\[\begin{split} \epsilon\mu(\{x: f(x)\geq k\epsilon\}) &\geq \int f_{(k+1)\epsilon} -f_{k\epsilon} \, \mathrm{d}\mu\\ &\geq \epsilon \mu(\{x: f(x)\geq (k+1)\epsilon\}).\end{split}\]

Finally by (5),

(11)#\[\epsilon \mu(\{x: f(x)\geq (k+1)\epsilon\}) \geq T(f_{(k+2)\epsilon}-f_{(k+1)\epsilon})\]

Combining (9), (11) and (10) gives

\[T(f_{k\epsilon}-f_{(k-1)\epsilon}) \geq \int f_{(k+1)\epsilon} -f_{k\epsilon} \, \mathrm{d}\mu \geq T(f_{(k+2)\epsilon}-f_{(k+1)\epsilon}).\]

Summing from 1 to \(i\) gives

\[T(f_{i\epsilon}) \geq \int f_{(i+1)\epsilon} -f_{\epsilon} \, \mathrm{d}\mu \geq T(f_{(i+2)\epsilon}-f_{2\epsilon}).\]

Since \(f_{i\epsilon}\) monotonically increases to \(f\),

\[T(f) \geq \int f-f_\epsilon \, \mathrm{d}\mu \geq T(f-f_\epsilon).\]

Therefore, since \(f_\epsilon\) monotonically decreases to 0, \(f-f_\epsilon\) increases to \(f\) and so this gives (4).

Note that, if \(f\in L\), then \(f^+,f^-\in L^+\) and so

\[T(f)=T(f^+)-T(f^-)=\int f^+\, \mathrm{d}\mu -\int f^-\, \mathrm{d}\mu = \int f\, \mathrm{d}\mu.\]

Observation 1

For any measure \(\mu\) satisfying the conclusion of Theorem 6, the value of \(\mu(\{f>t\})\), for \(f\in L^+\) and \(t>0\), is uniquely determined by the values of \(T\) on \(L^+\).

Proof. For any \(t>0\) and \(0<h<t\), observe that the functions

\[g_h := \frac{\inf\{f,t+h\}-\inf\{f,t\}}{h}\]

converge pointwise to the characteristic function of \(f^{-1}((t,\infty))\) and are bounded above by \(2f\). Therefore, by [mu_rep_T] and the dominated convergence theorem,

\[\mu(\{x:f(x)>t\}) = \lim_{h\to 0} h^{-1} T(\inf\{f,t+h\} -\inf\{f,t\}).\]

The Banach–Alaoglu theorem (see Example 26) motivates us to consider representations of Borel measures by elements of the dual of a Banach space, namely of \(C(X)\). To use Theorem 6, we must upgrade the pointwise convergence in the hypotheses to uniform convergence in \(C(X)\). Recall that \(C_c(X)\) is the set of all continuous functions on \(X\) with compact support, and that pointwise monotonic convergence in \(C_c(X)\) implies uniform convergence (see Example 22). Therefore, for any monotonic \(T\in C_c(X)'\), Theorem 6 produces a measure \(\mu\) that represents \(T\). If \(X\) is a metric space, then all compact sets are measurable with respect to \(\mu\). We next show, on locally compact spaces, how to obtain a Borel measure.

Lemma 3

Let \(X\) be a locally compact metric space and suppose that \(\tau\) is a finite measure on \(X\). There exists a unique Radon measure \(\mu\) on \(X\) that agrees with \(\tau\) on \(\mathcal K(X)\).

Proof. Let \(\mathcal U\) be the set of open subsets of \(X\). For each \(U \in \mathcal U\) define

\[\nu(U) = \sup\{\tau(K): K\subset U \text{ compact}\}.\]

Since \(\tau\) is monotone, for any \(K\in\mathcal K(X)\), \(\nu(K^o) \leq \tau(K)\), for \(K^o\) the interior of \(K\).

Since \(X\) is locally compact, for any \(U\in \mathcal U\) and \(K\in\mathcal K(X)\) with \(K\subset U\), there exists \(V\in\mathcal U\) with \(\overline V\in\mathcal K(X)\) and

\[K\subset V \subset \overline V \subset U.\]


(12)#\[\nu(U) = \sup\{\nu(V) : V\in \mathcal U,\ V\subset U,\ \overline V\in \mathcal K(X)\}.\]

Also, since \(\tau\) is a measure and \(X\) is a metric space,

(13)#\[\tau(K) =\inf\{\tau(S) : K\subset S^o,\ S\in\mathcal K(X)\}.\]

For each \(A\subset X\) define

\[\mu(A) = \inf\{\nu(U): U\supset A \text{ open}\}.\]

Note that both \(\nu\) and \(\mu\) are monotone and give value 0 to the empty set. To see that \(\mu\) is a measure, let \(A_i\subset X\) and let \(U_i\supset A_i\) be open. We must show that

\[\mu\left(\bigcup_{i\in\mathbb N}A_i\right) \leq \sum_{i\in\mathbb N} \mu(A_i).\]

It suffices to show that

\[\nu\left(\bigcup_{i\in\mathbb N}U_i\right) \leq \sum_{i\in\mathbb N} \nu(U_i).\]

By (12), for any \(\epsilon>0\), there exists \(V\subset \cup_i U_i\) with compact closure such that

\[\nu(V)\geq \nu\left(\bigcup_{i\in\mathbb N} U_i\right)-\epsilon\]

Since \(V\) has compact closure, it is contained in the union of finitely many \(U_i\). Therefore, it suffices to show that \(\nu\) is finitely sub-additive.

Let \(U,V\in \mathcal U\) and \(K\subset U\cup V\) be compact. Let \(W=U\cap V\) and

\[K_U :=\{x\in K : d(x,U\setminus W) \leq d(x,V\setminus W)\}\]


\[K_V := \{x\in K : d(x,V\setminus W) \leq d(x,U\setminus W)\}.\]

Then \(K_U,K_V\) are closed subsets of \(K\) with \(K=K_U\cup K_V\) and \(K_U\subset U\) and \(K_V\subset V\). Since \(\tau\) is finitely sub-additive,

\[\tau(K) \leq \tau(K_U) +\tau(K_V)\]

and so

\[\nu(U\cup V) \leq \nu(U) + \nu(V).\]

Therefore, \(\nu\) is finitely sub-additive by induction. As shown above, this implies that \(\mu\) is a measure.

To see that \(\mu\) is a Borel measure, if \(A,B\subset X\) are separated, then there exist separated open sets \(U\supset A\), \(V\supset B\). Since \(\tau\) is finitely additive, \(\nu(U\cap V) = \nu(U) + \nu(V)\) and so \(\mu\) is additive on separated sets. By construction, \(\mu\) is Borel regular. Also, (12) shows that open sets are inner regular by compact sets. Combining this with outer regularity by open sets shows that \(\mu\) is a Radon measure.

To see that \(\mu\) agrees with \(\tau\) on \(\mathcal K(X)\), note that for any \(K\in\mathcal K(X)\) and \(U\in\mathcal U\) with \(K\subset U\),

\[\tau(K)\leq \nu(U) =\mu(U)\]

and so \(\tau(K) \leq \mu(K)\). For the other inequality,

\[\begin{split}\mu(K) &=\inf\{\nu(U): U\supset K \text{ open}\}\\ &\leq \inf\{\nu(S^o): S\in\mathcal K(X),\ K\subset S^o\}\\ &\leq \inf\{\tau(S) : S\in \mathcal K(X), K\subset S^o\} =\tau(K)\end{split}\]

by (13).

Finally, if \(\mu_1,\mu_2\) are two Radon measures that agree with \(\tau\) on \(\mathcal K(X)\), then for any open \(U\subset X\),

\[\begin{split}\mu_1(U)&=\sup\{\mu_1(K): K\subset U,\ K\in\mathcal K(X)\}\\ &=\sup\{\mu_2(K): K\subset U,\ K\in \mathcal K(X)\} = \mu_2(U).\end{split}\]

Since \(\mu_1,\mu_2\) are both Borel regular, they must agree.

Theorem 7 (Riesz representation theorem)

Let \(X\) be a locally compact metric space and let \(T\in C_c(X)'\) be monotone. Then there exists a unique Radon measure \(\mu\) such that

(14)#\[T(f)= \int f\, \mathrm{d}\mu \quad \forall f\in C_c(X).\]

Proof. Observe that \(T\) is a monotone Daniell integral on \(L=C_c(X)\). By Theorem 6, there exists a measure \(\tau\) on \(X\) for which (14) holds with \(\mu\) replaced by \(\tau\). By Lemma 3, there exists a Radon measure \(\mu\) that agrees with \(\tau\) on \(\mathcal K(X)\). For any \(f\in C_c(X)\), the value of

\[\int f \, \mathrm{d}\tau\]

is determined by the value of \(\tau\) on compact sets, and so

\[\int f\, \mathrm{d}\tau=\int f\, \mathrm{d}\mu,\]

so that (14) holds. Observation 1 implies that, on a locally compact space, the measure of any compact set is uniquely determined by (14). Thus any Radon measure satisfying (14) is uniquely determined.

The Riesz representation theorem allows us to identify the set of finite Radon measures on a locally compact space \(X\) with the set of monotonic elements of \(C_0(X)'\). For \(B\) a Banach space, a sequence \(T_n\in B'\) weak* converges to \(T\in B'\) if \(T_n(x)\to T(x)\) for every \(x\in B\). In \(C_0(X)'\), this translates to the following.

Definition 14 (Weak* convergence)

Let \(X\) be a locally compact metric space. A sequence \(\mu_n\) of finite Radon measures on \(X\) weak* converges to a finite Radon measure \(\mu\) if, for every \(f\in C_c(X)\),

\[\int_X f\, \mathrm{d}\mu_n \to \int_X f \, \mathrm{d}\mu.\]

By the Banach-Alaoglu theorem (see Example 26), the unit ball of \(B'\) is weak* compact. Since the weak* limit of a sequence of monotonic operators on \(C_0(X)\) is monotonic, we have the following.

Theorem 8

Let \(X\) be a locally compact metric space and \(\mu_n\) a sequence of Radon measures with uniformly bounded total measures. There exists a finite Radon measure \(\mu\) on \(X\) and subsequence \(\mu_{n_k}\) that weak* converges to \(\mu\).


Example 19

Let \(L\) be a lattice of functions on \(X\). For any \(f\in L\) show that \(f^+,f^-\in L^+\)

Example 20

Let \(L\) be a lattice on \(X\) and \(T\) a monotone Daniell integral on \(L\). Give an example to show that there may be more than one measure satisfying (14) (recall Example 9). Compare to Observation 1.

Example 21

Let \(L\) be a lattice on \(X\) and \(T\) a Daniell integral on \(L\). Define \(T^+,T^-\) on \(L^+\) by

\[T^+(f) = \sup\{T(g): g\in L^+,\ g \leq f\}\]


\[T^-(f) = -\inf\{T(g) : g\in L^+\ g\leq f\}.\]
  1. There show that \(T^+,T^-\) are monotone Daniell integrals on \(L\).

  2. If \(f,g\in L^+\) with \(g\leq f\) then \(f\geq f-g\in L^+\) and so

    \[T(g)-T^-(f) \leq T(g) +T(f-g) \leq T(g) + T^+(f).\]
  3. Deduce that \(T=T^+-T^-.\)

Example 22

Let \(K\) be a compact metric space and suppose that \(f_i \in C(K)\) monotonically increase to \(f\in C(K)\). Show that \(f_i\to f\) uniformly.

Example 23

Let \(c\) be the set of all \(f\colon \mathbb N\to \mathbb R\) such that \(\lim_{j} f(x_j)\) exists and define \(T\colon c\to \mathbb R\) by \(T(f)= \lim_{j} f(x_j)\). Let \(\mathcal C\) be the set of all bounded \(f\colon \mathbb N\to \mathbb R\). Equip \(c\) and \(\mathcal C\) with the supremum norm.

  1. Observe that \(T\) is linear and continuous on \(c\) and hence can be extended by the Hahn-Banach theorem to a linear and continuous element of \(\mathcal C\). (Such an extension is called a Banach limit.)

  2. Show that any finite Borel measure on \(\mathbb N\) is a convergent sum of Dirac masses.

  3. Hence show that there is no Borel measure \(\mu\) on \(\mathbb R\) such that \(T_\mu=T\).

Example 24

Adapt the previous exercise to show that the Riesz representation theorem is false in non locally compact metric spaces.

Example 25

Let \(X\) be a metric space. Show that any \(T\in C_c(X)'\) is a Daniell integral.

  1. Let let \(T^+,T^-\in C_c(X)'^+\) be obtained from Example 21.

  2. Show that \(\|T\|=\|T^+\|+\|T^-\|\).

  3. By Theorem 7, any \(T\in C_0(X)\) can be identified with two measures \(\mu^+\) and \(\mu^-\) and hence with a signed measure (recall Example 60). Show that this identification is an isometric isomorphism.

Example 26 (Banach–Alaoglu theorem)

Let \(B\) be a separable Banach space and \(\mathcal D\) a countable dense subset of \(B\). Suppose that \(T_n\in B'\) satisfy \(\|T_n\|\leq M\) for some \(M>0\).

  1. Show that there exists a subsequence \(T_{n_j}\) and a \(T\in B'\) such that \(T_{n_j}(d) \to T(d)\) for each \(d\in\mathcal D\).

  2. Deduce that, for any \(x\in B\), \(T_{n_j}(x) \to T(x)\).

That is, closed and bounded subsets of \(B'\) are weak* compact.