We outline the proof of Fubini's Theorem for integrable functions. The complete proof requires several steps building from simple to general functions.

Given: $f \in L^1(X \times Y, \mu \otimes \nu)$

Step 1 (Indicator functions): For $E = A \times B$ where $A \in \mathcal{F}$ and $B \in \mathcal{G}$: $\int_{X \times Y} \chi_E \, d(\mu \otimes \nu) = (\mu \otimes \nu)(A \times B) = \mu(A) \nu(B)$

The iterated integrals give: $\int_X \left(\int_Y \chi_E(x, y) \, d\nu(y)\right) d\mu(x) = \int_X \chi_A(x) \nu(B) \, d\mu(x) = \mu(A) \nu(B)$

Thus Fubini holds for indicators of rectangles.

Step 2 (Simple functions): By linearity, Fubini holds for simple functions $s = \sum_{i=1}^{n} a_i \chi_{E_i}$ where $E_i$ are finite unions of rectangles.

Step 3 (Non-negative functions): For non-negative measurable $f$, approximate by increasing simple functions $s_n \uparrow f$. By the Monotone Convergence Theorem applied three times (once for each integral), Fubini holds for $f$. This is essentially Tonelli's Theorem.

Step 4 (Integrable functions): For general $f \in L^1$, write $f = f^+ - f^-$ where $f^+ = \max(f, 0)$ and $f^- = \max(-f, 0)$. Both $f^+$ and $f^-$ are non-negative and integrable.

By Step 3, Fubini holds for $f^+$ and $f^-$. By linearity: $\int_{X \times Y} f \, d(\mu \otimes \nu) = \int_{X \times Y} f^+ \, d(\mu \otimes \nu) - \int_{X \times Y} f^- \, d(\mu \otimes \nu)$

Applying Fubini to each term: $= \int_X \left(\int_Y f^+(x, y) \, d\nu(y)\right) d\mu(x) - \int_X \left(\int_Y f^-(x, y) \, d\nu(y)\right) d\mu(x)$

By linearity of integration: $= \int_X \left(\int_Y f(x, y) \, d\nu(y)\right) d\mu(x)$