[proofplan] We decompose $f$ into its positive and negative parts $f = f^+ - f^-$ and apply [Tonelli's Theorem](/theorems/3017) to each. Since $f \in L^1(X \times Y, \mu \otimes \nu)$, both $f^+$ and $f^-$ are non-negative integrable functions, so Tonelli yields finite iterated integrals for each. The finiteness of these iterated integrals implies that the sections $f_x^+$ and $f_x^-$ are $\nu$-integrable for $\mu$-a.e. $x$, giving $f_x \in L^1(Y, \nu)$ $\mu$-a.e. Subtracting the two Tonelli equalities produces the Fubini identity. [/proofplan]

[step:Decompose $f$ into positive and negative parts and apply Tonelli to each] Write $f = f^+ - f^-$ where $f^+ = \max\{f, 0\} \geq 0$ and $f^- = \max\{-f, 0\} \geq 0$. Since $f$ is $\mathcal{A} \otimes \mathcal{B}$-measurable, both $f^+$ and $f^-$ are $\mathcal{A} \otimes \mathcal{B}$-measurable and non-negative. Moreover, $|f| = f^+ + f^-$, so \begin{align*} \int_{X \times Y} f^+\, d(\mu \otimes \nu) \leq \int_{X \times Y} |f|\, d(\mu \otimes \nu) = \|f\|_{L^1(X \times Y)} < \infty, \end{align*} and the same bound holds for $f^-$. Both $f^+$ and $f^-$ are non-negative and in $L^1(X \times Y, \mu \otimes \nu)$. By [Tonelli's Theorem](/theorems/3017), applied to the non-negative measurable function $f^+$ on the $\sigma$-finite product space $(X \times Y, \mathcal{A} \otimes \mathcal{B}, \mu \otimes \nu)$: \begin{align*} \int_{X \times Y} f^+\, d(\mu \otimes \nu) = \int_X \left(\int_Y f^+(x, y)\, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f^+(x, y)\, d\mu(x)\right) d\nu(y). \end{align*} The same equality holds with $f^-$ in place of $f^+$: \begin{align*} \int_{X \times Y} f^-\, d(\mu \otimes \nu) = \int_X \left(\int_Y f^-(x, y)\, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f^-(x, y)\, d\mu(x)\right) d\nu(y). \end{align*} All six integrals above are finite (bounded by $\|f\|_{L^1} < \infty$). [/step]

[step:Establish almost-everywhere integrability of sections] Define $\varphi^+ : X \to [0, \infty]$ by $\varphi^+(x) = \int_Y f^+(x, y)\, d\nu(y)$ and $\varphi^- : X \to [0, \infty]$ by $\varphi^-(x) = \int_Y f^-(x, y)\, d\nu(y)$. By Tonelli's theorem, both $\varphi^+$ and $\varphi^-$ are $\mathcal{A}$-measurable. Moreover: \begin{align*} \int_X \varphi^+(x)\, d\mu(x) = \int_{X \times Y} f^+\, d(\mu \otimes \nu) < \infty. \end{align*} Since $\varphi^+ \geq 0$ and $\int_X \varphi^+\, d\mu < \infty$, the function $\varphi^+$ is finite $\mu$-a.e., meaning $\varphi^+(x) < \infty$ for $\mu$-a.e. $x \in X$. The same holds for $\varphi^-$. Let $N \subset X$ be the $\mu$-null set where either $\varphi^+(x) = \infty$ or $\varphi^-(x) = \infty$. For $x \in X \setminus N$ (which holds for $\mu$-a.e. $x$), both $\int_Y f^+(x, y)\, d\nu(y) < \infty$ and $\int_Y f^-(x, y)\, d\nu(y) < \infty$. Since $f_x = f_x^+ - f_x^-$ and both parts have finite integrals, we conclude: \begin{align*} \int_Y |f(x, y)|\, d\nu(y) = \int_Y f^+(x, y)\, d\nu(y) + \int_Y f^-(x, y)\, d\nu(y) = \varphi^+(x) + \varphi^-(x) < \infty. \end{align*} Hence $f_x \in L^1(Y, \nu)$ for $\mu$-a.e. $x \in X$. This establishes conclusion (1). By symmetry (interchanging the roles of $X$ and $Y$), $f^y \in L^1(X, \mu)$ for $\nu$-a.e. $y \in Y$, establishing conclusion (2). [/step]

[step:Subtract the Tonelli equalities and conclude] For $\mu$-a.e. $x \in X$ (specifically, for $x \in X \setminus N$), the section $f_x \in L^1(Y, \nu)$, and \begin{align*} \int_Y f(x, y)\, d\nu(y) = \int_Y f^+(x, y)\, d\nu(y) - \int_Y f^-(x, y)\, d\nu(y) = \varphi^+(x) - \varphi^-(x). \end{align*} Define $\varphi : X \to \mathbb{R}$ by $\varphi(x) = \varphi^+(x) - \varphi^-(x)$ for $x \in X \setminus N$ (and set $\varphi(x) = 0$ on $N$, say). Since $\varphi^+, \varphi^- \in L^1(X, \mu)$ (both have finite integrals, as shown above), the difference $\varphi = \varphi^+ - \varphi^- \in L^1(X, \mu)$. This establishes conclusion (3): the function $x \mapsto \int_Y f(x,y)\, d\nu(y)$ belongs to $L^1(X, \mu)$. Integrating $\varphi$ over $X$ and using the Tonelli equalities: \begin{align*} \int_X \left(\int_Y f(x, y)\, d\nu(y)\right) d\mu(x) &= \int_X \varphi^+(x)\, d\mu(x) - \int_X \varphi^-(x)\, d\mu(x) \\ &= \int_{X \times Y} f^+\, d(\mu \otimes \nu) - \int_{X \times Y} f^-\, d(\mu \otimes \nu) \\ &= \int_{X \times Y} (f^+ - f^-)\, d(\mu \otimes \nu) = \int_{X \times Y} f\, d(\mu \otimes \nu). \end{align*} The subtraction of the two integrals is justified because both are finite. By symmetry, the same argument with $X$ and $Y$ interchanged gives \begin{align*} \int_Y \left(\int_X f(x, y)\, d\mu(x)\right) d\nu(y) = \int_{X \times Y} f\, d(\mu \otimes \nu), \end{align*} establishing conclusion (4) and (5). [/step]