# 关于Fubini's Theorem证明的一些讨论

Posted by Derek on March 5, 2019

# 1. Fubini's Theorem是啥

Theorem. Assume that $R=[a, b] \times [c,d]$ is a rectangle in the $xy$ plane. Let $f$ be a function on $R$ and assume that

(i) $f$ is integrable on $R$.

(ii) $\forall y \in [c, d],$ the function $f_y: [a, b] \to \mathbb{R}$ defined by $f_y(x)=f(x, y)$ integrable on $[a, b].$

(iii) The function $g: [c, d] \to \mathbb{R}$ defined by $g(y)=\int_a^bf(x, y)\ \mathrm{d}x$ is integrable on $[c, d].$

Then $$\iint_Rf\ \mathrm{d}A=\int_c^d\left(\int_a^bf(x, y)\ \mathrm{d}x\right)\ \mathrm{d}y.$$

# 2. 进一步强化定理与证明

Theorem. Assume that $R=[a, b] \times [c,d]$ is a rectangle in the $xy$ plane. Let $f$ be a function on $R$ and assume that

(i) $f$ is integrable on $R$.

(ii) $\forall y \in [c, d],$ the function $f_y: [a, b] \to \mathbb{R}$ defined by $f_y(x)=f(x, y)$ integrable on $[a, b].$

Then $$\iint_Rf\ \mathrm{d}A=\int_c^d\left(\int_a^bf(x, y)\ \mathrm{d}x\right)\ \mathrm{d}y.$$

Proof. Take a partition $P=\lbrace x_0, ..., x_J; y_0, ..., y_K\rbrace,$ where $x_0=a, x_J=b, y_0=c, y_K=d.$ Define $m_{jk}, M_{jk}, \Delta x_j, \Delta y_k$ as usual, where $j\in\lbrace0, ..., J\rbrace, k\in\lbrace0, ..., K\rbrace.$

Since $\forall y\in[c, d],\ f(x, y)$ is integrable on $[a, b],$ then let $g(\xi_k)=\int_a^bf(x, \xi_k)\ \mathrm{d}x,$ where $\xi_k\in[y_{k-1}, y_k].$ Then we have $$\sum\limits_{j=1}^Jm_{jk}\Delta x_j \leq g(\xi_k)=\int_a^bf(x, \xi_k)\ \mathrm{d}x \leq \sum\limits_{j=1}^JM_{jk}\Delta x_j.\tag{1}$$ Therefore, $$\sum\limits_{j=1}^J\sum\limits_{k=1}^Km_{jk}\Delta x_j \Delta y_k \leq \sum\limits_{k=1}^Kg(\xi_k) \Delta y_k \leq \sum\limits_{j=1}^J\sum\limits_{k=1}^KM_{jk}\Delta x_j \Delta y_k.$$ Let $\lambda=\max\lbrace\Delta x_i, \Delta y_k\rbrace.$ Since $f$ is integral on $R$, then we have $$\lim\limits_{\lambda\to0}\sum\limits_{j=1}^J\sum\limits_{k=1}^Km_{jk}\Delta x_j \Delta y_k=\lim\limits_{\lambda\to0}\sum\limits_{j=1}^J\sum\limits_{k=1}^KM_{jk}\Delta x_j \Delta y_k=\iint_Rf\ \mathrm{d}A.$$ By Squeeze Theorem, we have $$\lim\limits_{\lambda\to0}\sum\limits_{k=1}^Kg(\xi_k) \Delta y_k=\iint_Rf\ \mathrm{d}A.$$ By the definition of integral, we know since $\lim\limits_{\lambda\to0}\sum\limits_{k=1}^Kg(\xi_k) \Delta y_k$ exits, then $g(y)$ is integrable on $[c, d].$

Since $g(y):=\int_a^bf(x, y)dx$ is integrable on $[c, d],$ then by Fubini's Theorem, $$\iint_Rf\ \mathrm{d}A=\int_c^dg(y)\ \mathrm{d}y=\int_c^d\left(\int_a^bf(x, y)\ \mathrm{d}x\right)\ \mathrm{d}y,$$

$\square$

## 2.1 一些说明

Proof. Take the same notations.

Since $f(x, y)$ is integrable on $[a, b], \forall y\in[c, d],$ then $f(x, y)$ is also integrable on $[x_{j-1}, x_j], \forall y\in[y_{k-1}, y_k].$ Therefore, $$m_{jk}(x_j-x_{j-1})\leq\int_{x_{j-1}}^{x_j}f(x, y)\ \mathrm{d}x.$$ Since $[a, b]=\bigcup\limits^J_{j=1}[x_{j-1}, x_j],$ and each $x_j$ is distinct, then any intersection of two adjacent intervals has zero content. Thus, $$\int_a^bf(x, y)\ \mathrm{d}x=\sum\limits_{j=1}^J\int_{x_{j-1}}^{x_j}f(x, y)\ \mathrm{d}x.$$ Hence, we have $$\sum\limits_{j=1}^Jm_{jk}(x_j-x_{j-1})\leq\sum\limits_{j=1}^J\int_{x_{j-1}}^{x_j}f(x, y)dx=\int_a^bf(x, y)\ \mathrm{d}x, \forall y\in[y_{k-1}, y_k].$$

$\square$