LEC CAL2 6

Exercise: if $f:[0,1]\to \mathbb{R}$ is continuous, show that ${\Gamma }_f$ has measure zero (use uniform continuity). Then, assume $f$ is Riemann integrable. Show again that ${\Gamma }_f$ has measure zero. Also true if the domain is a subset of ${\mathbb{R}}^n$.

Fubini’s theorem

Theorem 197.1(Fubini's Theorem).

Let $A\subseteq {\mathbb{R}}^n$ and $B\subseteq {\mathbb{R}}^m$ be closed rectangles. Let $f:A\times B\to \mathbb{R}$ be integrable. For $x\in A$ let $g_x:B\to \mathbb{R}$ be defined by $g_x(y)=f(x,y)$. Let

$$\mathcal{L}(x):=\underline{{\int }_B}{g}_x,\mathcal{U}(x):=\overline{{\int }_B}{g}_x.$$

Then, $\mathcal{L}$ and $\mathcal{U}$ are integrable on $A$ and

$${\int }_{A\times B}f={\int }_A\mathcal{L}={\int }_A\mathcal{U}.$$

Proof.

Let$P_A$ be a partition of $A$ and $P_B$ be a partition of $B$. Together they give a partition $P$ of $A\times B$.
Note that for $x\in S_A$, $m_{S_A\times S_B}\le m_{S_B}(g_x)$.

Then,

$$\begin{array}{rl}L(P,f)& =\sum _S{m}_S(f)\cdot \text{Vol}(S)\\ & =\sum _{S_A\times S_B}{m}_{S_A\times S_B}\cdot \text{Vol}(S_A)\text{Vol}(S_B)\\ & =\sum _{S_A}\left[\sum _{S_B}{m}_{S_A\times S_B}\text{Vol}(S_B)\right]\text{Vol}(S_A)\end{array}$$

For each $S_A$ in the sum, for each $x\in S_A$, we have $m_{S_A\times S_B}\le m_{S_B}(g_x)$, so

$$\sum _{S_B}{m}_{S_A\times S_B}\text{Vol}(S_B)\le \sum _{S_B}{m}_{S_B}(g_x)\text{Vol}(S_B)=L(P_B,g_x)\le \mathcal{L}(x).$$

It follows that

$$\sum _{S_B}{m}_{S_A\times S_B}\text{Vol}(S_B)\le \underset{x\in S_A}{\inf }\mathcal{L}(x)=m_{S_A}(\mathcal{L}).$$

Thus,

$$L(P,f)\le \sum _{S_A}{m}_{S_A}(\mathcal{L})\text{Vol}(S_A)=L(P_A,\mathcal{L}).$$

Now,

$$\begin{array}{r}L(P,f)\stackrel{1}{\le }L(P_A,\mathcal{L})\le U(P_A,\mathcal{L})\le U(P_A,\mathcal{U})\stackrel{2}{\le }U(P,f).\end{array}$$

We have proved $(1)$, and $(2)$ is analogous. Since $f$ is integrable, we have $\sup \{L(P,f)\}=\inf \{U(P,f)\}={\int }_{A\times B}f$. Hence,

$$\sup \{L(P_A,\mathcal{L})\}=\inf \{U(P_A,\mathcal{L})\}={\int }_{A\times B}f,$$

that is, $\mathcal{L}$ is integrable on $A$ and ${\int }_{A\times B}f={\int }_A\mathcal{L}$. The assertion for $\mathcal{U}$ follows from a symmetric argument.□

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