Proof of Fubini's Theorem 📂Calculus

Proof of Fubini's Theorem

Theorem¹

Let’s define the function $f : R \to \mathbb{R}$ on the 2-dimensional domain $R : [a,b] \times [c,d]$. If $f(x,\cdot)$ is integrable over $[c,d]$, and $f(\cdot,y)$ is integrable over $[a,b]$, and $f$ is integrable over $R$, then

$$ \iint _{R} f dA = \int_{a}^{b} \int_{c}^{d} f(x,y) dy dx = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy $$

Explanation

The integration domain $R$ obviously comes from a Rectangle. As it is always the case in analysis, to summarize for those of you who dislike reading long explanations, if it is integrable in each of the two orthogonal directions, then it does not matter in which order we integrate $f$. This might seem trivial for those in other fields who mostly deal with functions that satisfy these conditions, but it is an extremely important theorem. The fact that it is not just a property of double integrals but a theorem named after a person indicates its significance.

Proof

$$ \begin{align*} (L) \iint _{R} f dA \le & (L) \int_{a}^{b} \left( \int_{c}^{d} f(x,y) dy \right) dx \\ \le & (U) \int_{a}^{b} \left( \int_{c}^{d} f(x,y) dy \right) dx \\ \le & (U) \iint _{R} f dA \end{align*} $$

Let’s define the function $g : [a,b] \to \mathbb{R}$ as $\displaystyle g(x) := \int_{c}^{d} f(x,y) dy$. Since $f$ is integrable over $R$, i.e., $\displaystyle (L) \iint _{R} f dA = (U) \iint _{R} f dA$,

$$ \iint _{R} f dA = (U) \int_{a}^{b} g(x) dx = (L) \int_{a}^{b} g(x) dx $$

Since $\displaystyle (U) \int_{a}^{b} g(x) dx = (L) \int_{a}^{b} g(x) dx$, $g$ is integrable over $[a,b]$. Expressing it again,

$$ \begin{align*} \iint _{R} f dA =& (U) \int_{a}^{b} g(x) dx \\ =& (L) \int_{a}^{b} g(x) dx \\ =& \int_{a}^{b} \int_{c}^{d} f(x,y) dy dx \end{align*} $$

By swapping $x$ and $y$ and repeating the same process,

$$ \iint _{R} f dA = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy $$

Therefore,