The strategy is a four-stage bootstrap: prove the iterated integral identity for indicators of measurable rectangles (direct computation), extend to indicators of all product-measurable sets (via the [Dynkin $\pi$-system lemma](/theorems/505)), pass to non-negative measurable functions (via simple-function approximation and the [Monotone Convergence Theorem](/theorems/509)), and finally handle integrable functions by splitting $f = f^+ - f^-$. At each stage, the measurability of sections and of the inner integral must be verified.

**Step 1 (Measurable rectangles).** For $A_1 \in \mathcal{E}_1$ and $A_2 \in \mathcal{E}_2$, define $f = \mathbb{1}_{A_1 \times A_2}$. For fixed $x_1 \in E_1$, the section $x_2 \mapsto \mathbb{1}_{A_1 \times A_2}(x_1, x_2) = \mathbb{1}_{A_1}(x_1)\,\mathbb{1}_{A_2}(x_2)$ is $\mathcal{E}_2$-measurable, and \begin{align*} \int_{E_2} \mathbb{1}_{A_1 \times A_2}(x_1, x_2)\,\mu_2(dx_2) = \mathbb{1}_{A_1}(x_1)\,\mu_2(A_2). \end{align*} The function $x_1 \mapsto \mathbb{1}_{A_1}(x_1)\,\mu_2(A_2)$ is $\mathcal{E}_1$-measurable, and integrating against $\mu_1$: \begin{align*} \int_{E_1}\!\left(\int_{E_2} \mathbb{1}_{A_1 \times A_2}(x_1, x_2)\,\mu_2(dx_2)\right)\mu_1(dx_1) = \mu_1(A_1)\,\mu_2(A_2) = \mu(A_1 \times A_2) = \int_{E_1 \times E_2} \mathbb{1}_{A_1 \times A_2}\,d\mu. \end{align*} The same computation with the order of integration reversed gives the identity for both iterated integrals.

**Step 2 (All product-measurable sets).**

[claim:Iterated Integral For All Measurable Sets] For every $C \in \mathcal{E}_1 \otimes \mathcal{E}_2$, the section $x_2 \mapsto \mathbb{1}_C(x_1, x_2)$ is $\mathcal{E}_2$-measurable for each $x_1$, the function $x_1 \mapsto \mu_2(C_{x_1})$ (where $C_{x_1} = \{x_2 : (x_1, x_2) \in C\}$) is $\mathcal{E}_1$-measurable, and \begin{align*} \mu(C) = \int_{E_1} \mu_2(C_{x_1})\,\mu_1(dx_1). \end{align*} [/claim]

[proof] We first treat the case of finite measures, then extend to $\sigma$-finite.

*Finite case.* Assume $\mu_1(E_1) < \infty$ and $\mu_2(E_2) < \infty$. Define \begin{align*} \mathcal{D} = \left\{ C \in \mathcal{E}_1 \otimes \mathcal{E}_2 : \text{the three conclusions hold for } C \right\}. \end{align*} Step 1 shows that $\mathcal{D}$ contains the $\pi$-system $\mathcal{R} = \{A_1 \times A_2 : A_1 \in \mathcal{E}_1,\, A_2 \in \mathcal{E}_2\}$ of measurable rectangles. We show $\mathcal{D}$ is a $d$-system.

First, $E_1 \times E_2 \in \mathcal{D}$ (take $A_1 = E_1$, $A_2 = E_2$ in Step 1). Second, if $C, D \in \mathcal{D}$ with $C \subseteq D$, then $(D \setminus C)_{x_1} = D_{x_1} \setminus C_{x_1}$, so $x_2 \mapsto \mathbb{1}_{D \setminus C}(x_1, x_2)$ is measurable (as a difference of measurable functions) and $\mu_2((D \setminus C)_{x_1}) = \mu_2(D_{x_1}) - \mu_2(C_{x_1})$ (both finite since $\mu_2(E_2) < \infty$). The function $x_1 \mapsto \mu_2(D_{x_1}) - \mu_2(C_{x_1})$ is measurable, and \begin{align*} \int_{E_1} \mu_2((D \setminus C)_{x_1})\,\mu_1(dx_1) = \int_{E_1} \mu_2(D_{x_1})\,\mu_1(dx_1) - \int_{E_1} \mu_2(C_{x_1})\,\mu_1(dx_1) = \mu(D) - \mu(C) = \mu(D \setminus C). \end{align*} Third, if $(C_n)$ is increasing in $\mathcal{D}$ with $C = \bigcup_n C_n$, then $C_{x_1} = \bigcup_n (C_n)_{x_1}$ is a countable union of measurable sets (hence measurable), $\mu_2(C_{x_1}) = \lim_n \mu_2((C_n)_{x_1})$ by continuity from below, and $x_1 \mapsto \mu_2(C_{x_1})$ is measurable as a pointwise limit of measurable functions. By the [Monotone Convergence Theorem](/theorems/509), \begin{align*} \int_{E_1} \mu_2(C_{x_1})\,\mu_1(dx_1) = \lim_n \int_{E_1} \mu_2((C_n)_{x_1})\,\mu_1(dx_1) = \lim_n \mu(C_n) = \mu(C). \end{align*}

So $\mathcal{D}$ is a $d$-system containing the $\pi$-system $\mathcal{R}$. By the [Dynkin $\pi$-system lemma](/theorems/505), $\mathcal{E}_1 \otimes \mathcal{E}_2 = \sigma(\mathcal{R}) \subseteq \mathcal{D}$.

*$\sigma$-finite case.* Write $E_1 = \bigcup_j F_j$ and $E_2 = \bigcup_k G_k$ with $\mu_1(F_j) < \infty$ and $\mu_2(G_k) < \infty$. For any $C \in \mathcal{E}_1 \otimes \mathcal{E}_2$, the sets $C \cap (F_j \times G_k)$ increase to $C$ as $j, k \to \infty$. The finite-measure case applies to each $C \cap (F_j \times G_k)$, and the [Monotone Convergence Theorem](/theorems/509) extends the identity to $C$. [/proof]

**Step 3 (Non-negative measurable functions).** Let $f: E_1 \times E_2 \to [0, \infty]$ be $\mathcal{E}_1 \otimes \mathcal{E}_2$-measurable.

[claim:Fubini For Non Negative Functions] The section $x_2 \mapsto f(x_1, x_2)$ is $\mathcal{E}_2$-measurable for each $x_1 \in E_1$, the function \begin{align*} \varphi: E_1 &\to [0, \infty] \\ x_1 &\mapsto \int_{E_2} f(x_1, x_2)\,\mu_2(dx_2) \end{align*} is $\mathcal{E}_1$-measurable, and $\int_{E_1 \times E_2} f\,d\mu = \int_{E_1} \varphi\,d\mu_1$. [/claim]

[proof] By linearity and Step 2, the identity holds for all non-negative simple functions on $\mathcal{E}_1 \otimes \mathcal{E}_2$: any such function is a finite linear combination of indicators of product-measurable sets. Choose non-negative simple functions $\phi_n \uparrow f$ pointwise (by the standard dyadic approximation). For each $n$, \begin{align*} \int_{E_1 \times E_2} \phi_n\,d\mu = \int_{E_1}\!\left(\int_{E_2} \phi_n(x_1, x_2)\,\mu_2(dx_2)\right)\mu_1(dx_1). \end{align*} For fixed $x_1$, $\phi_n(x_1, \cdot) \uparrow f(x_1, \cdot)$ pointwise on $E_2$. By the [Monotone Convergence Theorem](/theorems/509) applied to $\mu_2$, \begin{align*} \int_{E_2} \phi_n(x_1, x_2)\,\mu_2(dx_2) \uparrow \int_{E_2} f(x_1, x_2)\,\mu_2(dx_2) = \varphi(x_1). \end{align*} The functions $x_1 \mapsto \int_{E_2} \phi_n(x_1, \cdot)\,d\mu_2$ are measurable (by Step 2 and linearity), so $\varphi$ is measurable as a pointwise limit of measurable functions. Applying the [Monotone Convergence Theorem](/theorems/509) to $\mu_1$: \begin{align*} \int_{E_1} \varphi\,d\mu_1 = \lim_n \int_{E_1}\!\left(\int_{E_2} \phi_n(x_1, x_2)\,\mu_2(dx_2)\right)\mu_1(dx_1) = \lim_n \int_{E_1 \times E_2} \phi_n\,d\mu = \int_{E_1 \times E_2} f\,d\mu, \end{align*} where the last equality is the [Monotone Convergence Theorem](/theorems/509) applied to $\mu$. [/proof]

The identity with the order of integration reversed follows by the same argument with the roles of $(E_1, \mu_1)$ and $(E_2, \mu_2)$ exchanged.

**Step 4 (Integrable functions).** Let $f: E_1 \times E_2 \to \mathbb{R}$ be integrable. Write $f = f^+ - f^-$. By Step 3 applied to $|f| = f^+ + f^-$, \begin{align*} \int_{E_1}\!\left(\int_{E_2} |f(x_1, x_2)|\,\mu_2(dx_2)\right)\mu_1(dx_1) = \int_{E_1 \times E_2} |f|\,d\mu < \infty. \end{align*} Since the outer integral is finite, the inner integral $\int_{E_2} |f(x_1, \cdot)|\,d\mu_2$ is finite for $\mu_1$-almost every $x_1$, meaning $x_2 \mapsto f(x_1, x_2)$ is $\mu_2$-integrable for $\mu_1$-a.e. $x_1$. Applying Step 3 to $f^+$ and $f^-$ separately and subtracting: \begin{align*} \int_{E_1 \times E_2} f\,d\mu &= \int_{E_1 \times E_2} f^+\,d\mu - \int_{E_1 \times E_2} f^-\,d\mu \\ &= \int_{E_1}\!\left(\int_{E_2} f^+(x_1, x_2)\,\mu_2(dx_2)\right)\mu_1(dx_1) - \int_{E_1}\!\left(\int_{E_2} f^-(x_1, x_2)\,\mu_2(dx_2)\right)\mu_1(dx_1) \\ &= \int_{E_1}\!\left(\int_{E_2} f(x_1, x_2)\,\mu_2(dx_2)\right)\mu_1(dx_1), \end{align*} where the last equality uses linearity of the inner integral (valid $\mu_1$-a.e. since both $\int f^+\,d\mu_2$ and $\int f^-\,d\mu_2$ are finite $\mu_1$-a.e.).