Fubini's Theorem

Fubini's theorem is the rigorous justification for computing multiple integrals as iterated single integrals. It guarantees that the order of integration can be interchanged under mild conditions.

The Theorem

Theorem9.4Fubini's Theorem

Let $f$ be continuous on a rectangle $R = [a,b] \times [c,d]$ . Then the double integral exists and equals either iterated integral: $\iint_R f(x,y)\,dA = \int_a^b \left(\int_c^d f(x,y)\,dy\right)dx = \int_c^d \left(\int_a^b f(x,y)\,dx\right)dy$

Theorem9.5Fubini's Theorem for General Regions

Let $f$ be continuous on a region $D$ of type I: $D = \{(x,y) : a \leq x \leq b,\; g_1(x) \leq y \leq g_2(x)\}$ where $g_1, g_2$ are continuous. Then $\iint_D f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$ Analogously for type II regions where $x$ varies between functions of $y$ .

Changing the Order of Integration

ExampleSimplification by reversing order

Evaluate $\int_0^1 \int_x^1 e^{y^2}\,dy\,dx$ . The inner integral $\int_x^1 e^{y^2}\,dy$ has no closed form. Reversing: the region is $\{(x,y) : 0 \leq y \leq 1,\; 0 \leq x \leq y\}$ , so $\int_0^1 \int_0^y e^{y^2}\,dx\,dy = \int_0^1 y e^{y^2}\,dy = \frac{1}{2}(e - 1)$

ExampleFubini fails without absolute convergence

Consider $f(x,y) = \frac{x^2 - y^2}{(x^2 + y^2)^2}$ on $(0,1] \times (0,1]$ . One can verify: $\int_0^1\int_0^1 f(x,y)\,dy\,dx = \frac{\pi}{4}, \quad \int_0^1\int_0^1 f(x,y)\,dx\,dy = -\frac{\pi}{4}$ The iterated integrals differ because $\iint |f|\,dA = \infty$ . This shows Fubini's theorem requires integrability (or continuity) of $f$ .

RemarkTonelli's theorem for non-negative functions

Tonelli's theorem states that for measurable $f \geq 0$ , the double integral and both iterated integrals always agree (even if they are all $+\infty$ ). This is useful because it allows one to check integrability by computing either iterated integral of $|f|$ first, then applying Fubini.