The proof proceeds in stages via the definition of the integral.

1. Suppose that \( g \) is a nonnegative simple function with \( g = 0 \) on \( A^c \). Then \( g \) has the representation \( g = \sum_{i \in I} a_i \bs{1}_{A_i} \) where \( a_i \in (0, \infty) \) and \( A_i \subseteq A \) for for \( i \in I \). But \( \mu(A_i) = 0 \) for each \( i \in I\) and so \( \int_S g \, d\mu = \sum_{i \in I} a_i \mu(A_i) = 0 \)
2. Suppose that \( f: S \to [0, \infty) \) is measurable. If \( g \) is a nonnegative simple function with \( g \le \bs{1}_A f \), then \( g = 0 \) on \( A^c \) so by (a), \( \int_S g \, d\mu = 0 \). Hence by part (b) of , \( \int_A f \, d\mu = \int_S \bs{1}_A f \, d\mu = 0 \).
3. Finally, suppose that \( f: S \to \R \) is measurable. Then \( \int_A f \, d\mu = \int_A f^+ \, d\mu - \int_A f^- \, d\mu \). But both integrals on the right are 0 by part (b).

Note that \( g = f \) if and only if \( g^+ = f^+ \) and \( g^- = f^- \). Let \( A = \{x \in S: g^+(x) = f^+(x)\} \). Then \( A \in \ms S \) and \( \mu(A^c) = 0 \). Hence by the additivity property and , \[ \int_S g^+ \, d\mu = \int_A g^+ \, d\mu + \int_{A^c} g^+ \, d\mu = \int_A f^+ \, d\mu + 0 = \int_A f^+ \, d\mu + \int_{A^c} f^+ \, d\mu = \int_S f^+ \, d\mu \] Similarly \( \int_S g^- \, d\mu = \int_S f^- \, d\mu\). Hence the integral of \( g \) exists and \( \int_S g \, d\mu = \int_S f \, d\mu \)

1. Let \( A = \{x \in S: f(x) \ge 0\} \). Then \( A \in \ms S \) and \( \mu(A^c) = 0 \). By the additivity of the integral over disjoint sets we have \[ \int_S f \, d\mu = \int_A f \, d\mu + \int_{A^c} f \, d\mu \] But \( \int_A f \, d\mu \ge 0 \) by the positive property and \( \int_{A^c} f \, d\mu = 0 \) by the null property , so \( \int_S f \, d\mu \ge 0 \).
2. Note first that if \( \mu(A) = 0 \) then both integrals in the displayed equation are 0 so \( \int_S f \, d\mu = 0 \). For the converse, let \( B_n = \left\{x \in S: f(x) \ge \frac{1}{n}\right\} \) for \( n \in \N_+ \) and \( B = \{x \in S: f(x) \gt 0\} \). Then \( B_n \) is increasing in \( n \) and \( \bigcup_{n=1}^\infty B_n = B \). If \( \mu(B) \gt 0 \) then \( \mu(B_n) \gt 0 \) for some \( n \in \N_+ \). But \( f \ge \frac{1}{n} \bs{1}_{B_n} \) on \( A \), so by the increasing property, \( \int_S f \, d\mu = \int_A f \, d\mu \ge \int_A \frac{1}{n} \bs{1}_{B_n} \, d\mu = \frac{1}{n} \mu(B_n) \gt 0 \).

1. Note that \( g = f + (g - f) \) and \( g - f \ge 0 \) almost everywhere on \( S \). If \( \int_S f \, d\mu = -\infty \) then trivially \( \int_S f \, d\mu \le \int_S g \, d\mu \). Otherwise, by the additive property, \[ \int_S g \, d\mu = \int_S f \, d\mu + \int_S (g - f) \, d\mu \] By the positive property , \(\int_S (g - f) \, d\mu \ge 0 \) so \( \int_S g \, d\mu \ge \int_S f \, d\mu \).
2. Except in the case that both integrals are \( \infty \) or both are \( -\infty \) we have \[ \int_S g \, d\mu - \int_S f \, d\mu = \int_S (g - f) \, d\mu \] By assumption \( g - f \ge 0 \) almost everywhere on \( S \), and hence by the positive property , the integral on the right is 0 if and only if \( g - f = 0 \) almost everywhere on \( S \).

First note that \( -\left|f\right| \le f \le \left|f\right| \) on \( S \). The integrals of all three functions exist, so the increasing property and scaling properties give \[ -\int_S \left|f\right| \, d\mu \le \int_S f \, d\mu \le \int_S \left|f \right| \, d\mu \] which is equivalent to the inequality above. If \( f \) is integrable, then by the increasing property , equality holds if and only if \( f = -\left|f\right| \) almost everywhere on \( S \) or \( f = \left|f\right| \) almost everywhere on \( S \). In the first case, \( f \le 0 \) almost everywhere on \( S \) and in the second case, \( f \ge 0 \) almost everywhere on \( S \).

We will show that if either of the integrals exist then they both do, and are equal. The proof is a classical bootstrapping argument that parallels the definition of the integral.

1. Suppose first that \( f \) is a nonnegative simple function on \( T \) with the representation \( f = \sum_{i \in I} b_i \bs{1}_{B_i} \) where \( I \) is a finite index set, \( \{B_i: i \in I\} \) is a measurable partition of \( T \), and \( b_i \in [0, \infty) \) for \( i \in I \). Recall that \( f \circ u \) is a nonnegative simple function on \( S \), with representation \( f \circ u = \sum_{i \in I} b_i \bs{1}_{u^{-1}(B_i)} \). Hence \[ \int_T f \, d\nu = \sum_{i \in I} b_i \nu(B_i) = \sum_{i \in I} b_i \mu\left[u^{-1}(B_i)\right] = \int_S (f \circ u) \, d\mu \]
2. Next suppose that \( f: T \to [0, \infty) \) is measurable, so that \( f \circ u: S \to [0, \infty) \) is also measurable. There exists an increasing sequence \( (f_1, f_2, \ldots) \) of nonnegative simple functions on \( T \) with \( f_n \to f \) as \( n \to \infty \). Then \((f_1 \circ u, f_2 \circ u, \ldots)\) is an increasing sequence of simple functions on \( S \) with \( f_n \circ u \to f \circ u\) as \( n \to \infty \). By step (a), \( \int_T f_n \, d\nu = \int_S (f_n \circ u) \, d\mu \) for each \( n \in \N_+ \). But by the monotone convergence theorem, \( \int_T f_n \, d\nu \to \int_T f \, d\nu \) as \( n \to \infty \) and \( \int_S (f_n \circ u) \, d\mu \to \int_S (f \circ u) \, d\mu \) so we conclude that \( \int_T f \, d\nu = \int_S (f \circ u) \, d\mu \)
3. Finally, suppose that \( f: T \to \R \) is measurable, so that \( f \circ u: S \to \R \) is also measurable. Note that \( (f \circ u)^+ = f^+ \circ u \) and \( (f \circ u)^- = f^- \circ u \). By part (b), \begin{align} \int_T f^+ \, d\nu & = \int_S (f^+ \circ u) \, d\mu = \int_S (f \circ u)^+ \, d\mu \\ \int_T f^- \, d\nu & = \int_S (f^- \circ u) \, d\mu = \int_S (f \circ u)^- \, d\mu \end{align} Assuming that at least one of the integrals in the displayed equations is finite, we have \[ \int_T f \, d\nu = \int_T f^+ \, d\nu - \int_T f^- \, d\nu = \int_S (f \circ u)^+ \, d\mu - \int_S (f \circ u)^- \, d\mu = \int_S (f \circ u) \, d\mu\]

Let \( g_n = \sum_{i=1}^n f_i \) for \( n \in \N_+ \). Then \( g_n: S \to [0, \infty) \) is measurable and \( g_n \) is increasing in \( n \). Moreover, by definition, \( g_n \to \sum_{i=1}^\infty f_i \) as \( n \to \infty \). Hence by the monotone convergence theorem, \( \int_S g_n \, d\mu \to \int_S \sum_{i=1}^\infty f_i \, d\mu \) as \( n \to \infty \). But we know the additivity property holds for finite sums, so \(\int_S g_n \, d\mu = \sum_{i=1}^n \int_S f_i \, d\mu\) and again, by definition, this sum converges to \(\sum_{i=1}^\infty \int_S f_i \, d\mu\) as \( n \to \infty \).

Suppose first that \( f \) is nonnegative. Note that \( \bs{1}_A = \sum_{n=1}^\infty \bs{1}_{A_n} \) and hence \( \bs{1}_A f = \sum_{n=1}^\infty \bs{1}_{A_n} f \). Thus from , \[ \int_A f \, d\mu = \int_S \bs{1}_A f \, d\mu = \int_S \sum_{n=1}^\infty \bs{1}_{A_n} f \, d\mu = \sum_{n=1}^\infty \int_S \bs{1}_{A_n} f \, d\mu = \sum_{n=1}^\infty \int_{A_n} f \, d\mu \] Suppose now that \( f: S \to \R \) is measurable and \( \int_S f \, d\mu \) exists. Note that for \( B \in \ms S \), \( \left(\bs{1}_B f\right)^+ = \bs{1}_B f^+ \) and \( \left(\bs{1}_B f\right)^- = \bs{1}_B f^- \). Hence from the previous argument, \[ \int_A f^+ \, d\mu = \sum_{n=1}^\infty \int_{A_n} f^+ \, d\mu, \quad \int_A f^- \, d\mu = \sum_{n=1}^\infty \int_{A_n} f^- \, d\mu \] Both of these are sums of nonnegative terms, and one of the sums, at least, is finite. Hence we can group the terms to get \[ \int_A f \, d\mu = \int_A f^+ \, d\mu - \int_A f^- \, d\mu = \sum_{n=1}^\infty \int_{A_n} (f^+ - f^-) \, d\mu = \sum_{n=1}^\infty \int_{A_n} f \, d\mu \]

Let \( f(x) = \lim_{n \to \infty} f_n(x) \) for \( x \in S \) which exists in \( \R \cup \{\infty\} \) since \( f_n(x) \) is increasing in \( n \in \N_+ \). If \( \int_S f_1 \, d\mu = \infty \), then by the increasing property, \( \int_S f_n \, d\mu = \infty \) for all \( n \in \N_+ \) and \( \int_S f \, d\mu = \infty \), so the conclusion of the monotone convergence theorem trivially holds. Thus suppose that \( f_1 \) is integrable. Let \( g_n = f_n - f_1 \) for \( n \in \N \) and let \( g = f - f_1 \). Then \( g_n \) is nonnegative and increasing in \( n \) on \( S \), and \( g_n \to g \) as \( n \to \infty \) on \( S \). By the ordinary monotone convergence theorem, \( \int_S g_n \, d\mu \to \int_S g \, d\mu \) as \( n \to \infty \). But since \( \int_S f_1 \, d\mu \) is finite, \( \int_S g_n \, d\mu = \int_S f_n \, d\mu - \int_S f_1 \, d\mu \) and \( \int_S g \, d\mu = \int f \, d\mu - \int_S f_1 \, d\mu \). Again since \( \int_S f_1 \, d\mu \) is finite, it follows that \( \int_S f_n \, d\mu \to \int_S f \, d\mu \) as \( n \to \infty \).

The functions \( -f_n \) for \( n \in \N_+ \) satisfy the hypotheses of the monotone convergence theorem for increasing functions and hence \(\int_S \lim_{n \to \infty} -f_n \, d\mu = \lim_{n \to \infty} -\int_S f_n \, d\mu \). By the scaling property, \( \int_S \lim_{n \to \infty} f_n \, d\mu = \lim_{n \to \infty} \int_S f_n \, d\mu \).

Let \( g_n = \inf\left\{f_k: k \in \{n, n + 1, \ldots \}\right\} \) for \( n \in \N_+ \). Then \( g_n: S \to [0, \infty) \) is measurable for \( n \in \N_+ \), \( g_n \) is increasing in \( n \), and by definition, \( \lim_{n \to \infty} g_n = \liminf_{n \to \infty} f_n \). By the monotone convergence theorem, \[ \int_S \liminf_{n \to \infty} f_n \, d\mu = \lim_{n \to \infty} \int_S g_n \, d\mu \] But \( g_n \le f_k \) on \( S \) for \( n \in \N_+ \) and \( k \in \{n, n + 1, \ldots\} \) so by the increasing property, \( \int_S g_n \, d\mu \le \int_S f_k \, d\mu\) for \( n \in \N_+ \) and \( k \in \{n, n + 1, \ldots\} \). Hence \( \int_S g_n \, d\mu \le \inf\left\{\int_S f_k \, d\mu: k \in \{n, n+1, \ldots\}\right\} \) for \( n \in \N_+ \) and therefore \[ \lim_{n \to \infty} \int_S g_n \, d\mu \le \liminf_{n \to \infty} \int_S f_n \, d\mu \]

First note that by the increasing property, \( \int_S \left|f_n\right| \, d\mu \le \int_S g \, d\mu \lt \infty \) and hence \( f_n \) is integrable for \( n \in \N_+ \). Let \( f = \lim_{n \to \infty} f_n \). Then \( f \) is measurable, and by the increasing property again, \( \int_S \left| f \right| \, d\mu \lt \int_S g \, d\mu \lt \infty \), so \( f \) is integrable.

Now for \( n \in \N_+ \), let \( u_n = \inf\left\{f_k: k \in \{n, n + 1, \ldots\}\right\} \) and let \( v_n = \sup\left\{f_k: k \in \{n, n + 1, \ldots\}\right\} \). Then \( u_n \le f_n \le v_n \) for \( n \in \N_+ \), \( u_n \) is increasing in \( n \), \( v_n \) is decreasing in \( n \), and \( u_n \to f \) and \( v_n \to f \) as \( n \to \infty \). Moreover, \( \int_S u_1 \, d\mu \ge - \int_S g \, d\mu \gt -\infty \) so by the version of the monotone convergence theorem above, \( \int_S u_n \, d\mu \to \int_S f \, d\mu \) as \( n \to \infty \). Similarly, \( \int_S v_1 \, d\mu \lt \int_S g \, d\mu \lt \infty \), so by the monotone convergence theorem , \( \int_S v_n \, d\mu \to \int_S f \, d\mu \) as \( n \to \infty \). But by the increasing property, \( \int_S u_n \, d\mu \le \int_S f_n \, d\mu \le \int_S v_n \, d\mu \) for \( n \in \N_+ \) so by the squeeze theorem for limits, \( \int_S f_n \, d\mu \to \int_S f \, d\mu \) as \( n \to \infty \).

The assumption that \( g = \sum_{i = 1}^\infty \left| f_i \right| \) is integrable implies that \( g \lt \infty \) almost everywhere on \( S \). In turn, this means that \( \sum_{i=1}^\infty f_i \) is absolutely convergent almost everywhere on \( S \). Let \( f(x) = \sum_{i=1}^\infty f_i(x) \) if \( g(x) \lt \infty \), and for completeness, let \( f(x) = 0 \) if \( g(x) = \infty \). Since only the integral of \( f \) appears in the theorem, it doesn't matter how we define \( f \) on the null set where \( g = \infty \). Now let \( g_n = \sum_{i=1}^n f_i \). Then \( g_n \to f \) as \( n \to \infty \) almost everywhere on \( S \) and \( \left| g_n \right| \le g \) on \( S \). Hence by the dominated convergence theorem , \( \int_S g_n \, d\mu \to \int_S f \, d\mu \) as \( n \to \infty \). But we know the additivity property holds for finite sums, so \( \int_S g_n \, d\mu = \sum_{i=1}^n \int_S f_i \, d\mu \), and in turn this converges to \( \sum_{i=1}^\infty \int_S f_i \, d\mu \) as \( n \to \infty \). Thus we have \( \sum_{i=1}^\infty \int_S f_i \, d\mu = \int_S f \, d\mu \).

Suppose that \( \left|f_n\right| \) is bounded in \( n \) on \( A \) by \( c \in (0, \infty) \). The constant \( c \) is integrable on \( A \) since \( \int_A c \, d\mu = c \mu(A) \lt \infty \), and \( \left|f_n\right| \le c \) on \( A \) for \( n \in \N_+ \). Thus the result follows from the dominated convergence theorem .

We will show that \[ \int_{S \times T} f(x, y) \, d(\mu \times \nu)(x, y) = \int_S \int_T f(x, y) \, d\nu(y) \, d\mu(x) \] The proof with the other iterated integral is symmetric. The proof is a bootstrapping argument that proceeds in stages, paralleling the definition of the integral.

1. Suppose that \( f = \bs{1}_{A \times B} \) where \( A \in \ms S \) and \( B \in \ms T \). The equation holds by definition of the product measure, since the double integral is \( (\mu \times \nu)(A \times B) \) and the iterated integral is \[ \int_S \int_T \bs{1}_{A \times B} (x, y) \, d\nu(y) \, d\nu(x) = \int_S \int_T \bs{1}_A(x) \bs{1}_B(y) \, d\nu(y) \, d\mu(x) \int_S \bs{1}_A(x) \nu(B) \, d\mu = \mu(A) \nu(B) \]
2. Consider \( f = \bs{1}_C \) where \( C \in \ms S \times \ms T \). The double integral is \( (\mu \times \nu)(C) \), and so as a function of \( C \in \ms S \times \ms T \) defines the measure \( \mu \times \nu \). On the other hand, the iterated integral is \[ \int_S \int_T \bs{1}_C(x, y) \, d\nu(y) \, d\mu(x) = \int_S \int_T \bs{1}_{C_x}(y) \, d\nu(y) \, d\mu(x) = \int_S \nu(C_x) \, d\mu(x) \] where \( C_x = \{y \in T: (x, y) \in C\} \) is the cross-section of \( C \) at \( x \in S \). Recall from the section on existence and uniqueness that \( x \mapsto \nu(C_x) \) is a nonnegative, measurable function of \( x \), so \( C \mapsto \int_S \nu(C_x) \, d\mu(x) \) makes sense. Moreover, as a function of \( C \in \ms S \times \ms T \), this integral also forms a measure: If \( \{C^i: i \in I\} \) is a countable, disjoint collection sets in \( \ms S \times \ms T \), then \( \{C_x^i: i \in I\} \) is a countable, disjoint collection of sets in \( \ms T \). Cross-sections preserve set operations, so if \( C = \bigcup_{i \in I} C^i \) then \( C_x = \bigcup_{i \in I} C_x^i \). By the additivity of the measure \( \nu \) and the integral we have \[ \int_S \nu(C_x) \, d\mu(x) = \int_S \nu\left(\bigcup_{i \in I} C_x^i \right) \, d\mu(x) = \int_S \sum_{i \in I} \nu\left(C_x^i\right) \, d\mu(x) = \sum_{i \in I} \int_S \nu\left(C_x^i\right) \, d\mu(x)\] To summarize, the double integral and the iterated integral define positive measures on \( \ms S \times \ms T \). By (a), these measure agree on the measurable rectangles. By the uniqueness theorem, they must be the same measure. Thus the double integral and the iterated integral agree with integrand \( f = \bs{1}_C \) for every \( C \in \ms S \times \ms T \).
3. Suppose \( f = \sum_{i \in I} c_i \bs{1}_{C_i} \) is a nonnegative simple function on \( S \times T \). Thus, \( I \) is a finite index set, \( c_i \in [0, \infty) \) for \( i \in I \), and \( \{C_i: i \in I\} \) is a disjoint collection of sets in \( \ms S \times \ms T \). The double integral and the iterated integral satisfy the linearity properties, and hence by (b), agree with integrand \( f \).
4. Suppose that \( f: S \to [0, \infty) \) is measurable. Then there exists a sequence of nonnegative simple functions \( g_n, \; n \in \N_+ \) such that \( g_n \) is increasing in \( n \in \N_+ \) on \( S \times T \), and \( g_n \to f \) as \( n \to \infty \) on \( S \times T \). By the monotone convergence theorem, \( \int_{S \times T} g_n \, d(\mu \times \nu) \to \int_{S \times T} f \, d(\mu \times \nu) \). But for fixed \( x \in S \), \( y \mapsto g_n(x, y) \) is increasing in \( n \) on \( T \) and has limit \( f(x, y) \) as \( n \to \infty \). By another application of the montone convergence theorem, \( \int_T g_n(x, y) \, d\nu(y) \to \int_T f(x, y) \, d\nu(y) \) as \( n \to \infty \). But \(x \mapsto \int_T g_n(x, y) \, d\nu(y) \) is measurable and is increasing in \( n \in \N_+ \) on \( S \), so by yet another application of the monotone convergence theorem, \( \int_S \int_T g_n(x, y) \, d\nu(y) \, d\mu(x) \to \int_S \int_T f(x, y) \, d\nu(y) \, d\mu(x) \) as \( n \to \infty \). But the double integral and the iterated integral agree with integrand \( g_n \) by (c) for each \( n \in \N_+ \), so it follows that the double integral and the iterated integral agree with integrand \( f \).
5. Suppose that \( f: S \times T \to \R \) is measurable. By (d), the double integral and the iterated integral agree with integrand functions \( f^+ \) and \( f^- \). Assuming that at least one of these is finite, then by the additivity property, they agree with integrand function \( f = f^+ - f^- \).

Define \(f: S \times T \to \R\) by \(f(x, y) = g(x) h(y)\) for \((x, y) \in S \times T\). Then \(f\) is measurable. If \(g\) and \(h\) are nonnegative, then so is \(f\) and hence \(f\) is integrable. If \(g\) and \(h\) are absolutely integrable then by Fubini's theorem applied to \(|f|\) we have \begin{align*} \int_{S \times T} |f(x, y)| \, d(\mu \times \nu)(x, y) & = \int_T\left(\int_S |g(x)| |h(y)| \, d\mu(x)\right) \, d\nu(y) = \int_T |h(y)| \left(\int_S |g(x)| \, d\mu(x)\right) \, d\nu(y) \\ & = \left(\int_S |g(x)| \, d\mu(x)\right) \left(\int_T |h(y)| \, d\nu(y)\right) \lt \infty \end{align*} Hence \(f\) is absolutely integrable. So in either case, applying Fubini's theorem to \(f\), with the same computations as above we have \begin{align*} \int_{S \times T} f(x, y) \, d(\mu \times \nu)(x, y) & = \int_T\left(\int_S g(x) h(y) \, d\mu(x)\right) \, d\nu(y) = \int_T h(y) \left(\int_S g(x) \, d\mu(x)\right) \, d\nu(y) \\ & = \left(\int_S g(x) \, d\mu(x)\right) \left(\int_T h(y) \, d\nu(y)\right) \end{align*}