5.26 Integration on sets

In this lecture we will see how to use iterated integrals to compute Riemann integrals on rectangles and more general sets.

Integration on sets

A bounded set \(S \subseteq \mathbb{R}^2\) has area if its characteristic function \(\chi_S\) is Riemann integrable in some rectangle \(R\) containing \(S\). In case \(S\) has area, we denote it by \(\displaystyle a(S) = \int_{R} \chi_{S} = \int_{S} 1\)

As we noted before, this definition does not depend on the rectangle one chooses.

We have also another way to determine when a set has area

Theorem

Let \(S \subset \mathbb{R}^2\) be bounded set. Then \(S\) has area if and only if its boundary \(\partial S\) has measure zero.

---

Now let us consider functions on sets.

Let \(S \subseteq \mathbb{R}^{2}\) be a bounded set and \(f: S \to \mathbb{R}\) a function. Let \(R = [a,b]\times[c,d] \subseteq \mathbb{R}^2\) be a rectangle that contains \(S\) and define the function \(f^{*}:\mathbb{R}\to\mathbb{R}\text{ by }f^{*}(x,y)= \begin{cases} f(x,y) & \text{if }(x,y)\in S \\ 0 & \text{if }R\setminus S \end{cases}\)

​

Using this notion, we define when a function \(f\) is integrable on a bounded set \(S\).

---

Definition

Let \(S \subseteq \mathbb{R}^2\) be a bounded set. A function \(f: S \to \mathbb{R}\) is integrable on \(S\) if for every rectangle \(R = [a,b] \times [c,d] \subseteq \mathbb{R}^2\) containing \(S\), the function \(f^*: R \to \mathbb{R}\) on \(R\) is integrable.

In such a case, we write \(\displaystyle\int_{S} f = \int_{R} f^{*}\)

Remark

The definition does not depend on the rectangle one takes.

Exercise: Prove it!

---

Let us list some properties (without proof) that ensure us that a function is Riemann integrable.

Theorem

Let \(S \subset \mathbb{R}^n\) be bounded set with area and let \(f,g : S \rightarrow \mathbb{R}\) be bounded functions. Then

1. If \(f\) is continuous, then \(f\) is integrable.

2. \(f\) is integrable if and only if the set of its points of discontinuity has measure zero

3. If \(f,g\) are integrable on \(S\) and \(f(x,y) \leq g(x,y) \forall (x,y) \in S\), then \(\displaystyle \int_{S} f \leq \int_{S} g\)

4. If \(T \subset \mathbb{R}^2\) is a set with area with \(T \cap S = \emptyset\) and \(f\) can be extended to \(T\) so that \(f\) is integrable on \(T\), then \(f\) is integrable on \(T \cup S\) and \(\displaystyle\int_{S \cup T}f = \int_{S} f + \int_{T} f\)

5. (The integral is a linear map on the vector space of integrable functions.)

   If \(f,g\) are integrable on \(S\), then the function \(af + bg\) is integrable on \(S\) for all \(a,b \in \mathbb{R}\).
   In addition, \(\displaystyle \int_{S} af + bg = a \int_{S} f + b \int_{S} g\)

6. If \(f,g\) are continuous and non-negative on \(S\), then there exists \((x_0,y_0) \in S\) such that \(\displaystyle\int_{S} fg = f(x_{0},y_{0}) \int_{S} g\)

Theorem

Let \(R = [a, b] \times [c, d] \subseteq \mathbb{R}^2\) be a rectangle and \(f: R \to \mathbb{R}\) be a Riemann integrable function. Assume that the integral \(\displaystyle F(x) = \int_{c}^{d} f(x,y)dy\) exists for all \(x \in [a, b]\) and the function \(F: [a, b] \to \mathbb{R}\) is integrable. Then

\[ \int_{R} f = \int_{a}^{b} F(x)dx = \int_{a}^{b} \left(\int_{c}^{d} f(x,y) dy\right ) dx \]

Proof

To prove \(\displaystyle \int_{R}f=\int^{b}_{a}F(x)dx\), we prove that \(\displaystyle\left|\int_{R}f-\int^{b}_{a}F(x)dx\right|<\varepsilon,\forall \varepsilon >0\)

Since \(f\) is integrable, then \(\forall \varepsilon>0\), there exists a partition \(P=\{R_{ij}\}\) of \(R=[a,b]\times [c,d]\) s.t. \(M_{ij}=\sup\{f(x,y):(x,y)\in R_{ij}\}\) and \(m_{ij}=\inf\{f(x,y):(x,y)\in R_{ij}\}\)

Then \(L(f,P)=\sum^{n}_{i=1}\sum^{k}_{j=1}m_{ij}a(R_{ij})\) and \(U(f,P)=\sum^{n}_{i=1}\sum^{k}_{j=1}M_{ij}a(R_{ij})\)

Assume \(P=\{x_{0}=a,x_{1},...,x_{n}=b\}\times \{y_{0}=c,...,y_{k}=d\}\) and \(R_{ij}=[x_{i-1},x_{i}]\times [y_{j-1},y_{j}]\)

Thus \(a(R_{ij})=(x_i-x_{i-1})(y_j-y_{j-1})\) and \(U(f,P)-L(f,P)<\frac{\varepsilon}{2}\) and \(\displaystyle L(f,P)\leq\int_{R}f\leq U(f,P)\)

---

Now in every \(R_{ij}\), we have \(m_{ij}\leq R_{ij}\leq M_{ij}\). Now for any \(t_{i}\in [x_{i-1},x_{i}]\) we have that \(\displaystyle F(t_{i})=\int^{d}_{c}f(t_{i},y)dy\)​

Since \(F(x) = \displaystyle\int_c^d f(x,y) dy\) is integrable and \([c,d] = [y_0, y_1] \cup [y_1, y_2] \cup \dots \cup [y_{k-1}, y_k]\)
\(\Rightarrow F(x) = \displaystyle\int_{c}^{d} f(x,y) dy =\) \(\displaystyle\int_{y_0}^{y_1}f(x,y) dy + \displaystyle\int_{y_1}^{y_2}f(x,y) d y + \dots + \displaystyle\int_{y_{k-1}}^{y_k}f(x,y) dy\)

Now, for \(t_i \in [x_{i-1}, x_i]\), we know that \(m_{ij} \le f(t_i, y) \le M_{ij}\) in \([x_{i-1}, x_i] \times [y_{j-1}, y_j]\)​\((y \in [y_{j-1}, y_j])\)
\(\Rightarrow \displaystyle\int_{y_{j-1}}^{y_j} m_{ij} dy \le \displaystyle\int_{y_{j-1}}^{y_j} f(t_i, y) dy \le \displaystyle\int_{y_{j-1}}^{y_j} M_{ij} dy\)
\(\Rightarrow m_{ij}(y_j - y_{j-1}) \le \displaystyle\int_{y_{j-1}}^{y_j} f(t_i, y) dy \le M_{ij}(y_j - y_{j-1})\)
\(\Rightarrow \sum_{j=1}^k m_{ij}(y_j - y_{j-1}) \le \sum_{j=1}^k \displaystyle\int_{y_{j-1}}^{y_j} f(t_i, y) dy \le \sum_{j=1}^k M_{ij}(y_j - y_{j-1})\)

\(\Rightarrow \sum_{j=1}^{k} m_{ij}(y_{j} - y_{j-1}) \le F(t_{i}) \le \sum_{j=1}^{k} M_{ij}(y_{j} - y_{j-1})\)
\(\Rightarrow \underbrace{\sum_{i=1}^n \sum_{j=1}^k m_{ij}(y_j - y_{j-1})(x_i - x_{i-1})}_{L(f,P)} \le \sum_{i=1}^n F(t_i)(x_i - x_{i-1})\) \(\le \underbrace{\sum_{i=1}^n \sum_{j=1}^k M_{ij}(y_j - y_{j-1})(x_i - x_{i-1})}_{U(f,P)}\)

\(\Rightarrow L(f,P) \le \sum_{i=1}^n F(t_i)(x_i - x_{i-1}) \le U(f,P)\)
But \(\{x_0, x_1, \dots, x_n\}\) is a partition of \([a,b]\) and \(t_i \in [x_{i-1}, x_i]\)
\(\Rightarrow \sum_{i=1}^n F(t_i)(x_i - x_{i-1})\) is a Riemann sum for \(F\)
We may assume that the partition is fine enough (if not we take a refinement)

So we have \(\displaystyle \left|\int^b_aF(x)dx-\sum^n_{i=1}F(t_i)(x_i-x_{i-1})\right|<\frac{\varepsilon}{2}\) because \(F\) is integrable

Then \(\displaystyle\int^{b}_{a}F(x)dx\leq \sum^{n}_{i=1}F(t_{i})(x_{i}-x_{i-1})+\frac{\varepsilon}{2}\leq U(f,P)+\frac{\varepsilon}{2}\)

Similarly, \(\displaystyle L(f,P)-\frac{\varepsilon}{2}\leq\int^{b}_{a}F(x)dx\). Since above \(\displaystyle L(f,P)\leq\int_{R}f\leq U(f,P)\), then \(\displaystyle\int_{R}f-\int^{b}_{a}F(x)dx\leq U(f,P)-\int^b_aF(x)dx\leq U(f,P)-\left(L(f,P)-\frac{\varepsilon}{2}\right)\leq \varepsilon\)

---

Remark

1. The theorem above also holds if we interchange the values of \(x\) and \(y\). Write the statement as exercise.

2. Now we have a concrete way to compute Riemann integrals on rectangles when the function satisfies certain properties.

   For example, these properties are satisfied when \(f\) is continuous on \(R\). Note that in such a case,

   \(\text{Riemann integral on } R = \text{iterated integral on } R\) (for \(f\) continuous).

3. The theorem above also holds for more general sets than rectangles. These sets have to have area.

In what follows, we study some sets on which the theorem holds; these are called elementary sets.

Elementary sets

We introduce now simple or normal sets. These are sets which can be described as the points between the graph of two functions on the variable \(y\), or the points between the graph of two functions on the variable \(x\), or both.

Definitions

Let \(D \subseteq \mathbb{R}^2\) be a bounded set.

1. We say that \(D\) is \(y\)-simple if there are continuous functions \(\varphi_1, \varphi_2: [a, b] \rightarrow \mathbb{R}\) such that \(D=\{(x,y)\in\mathbb{R}^{2}\ |\ x\in[a,b],\varphi_{1}(x)\leq y\leq\varphi_{2}(x)\}\)

   ​

2. We say that \(D\) is \(x\)-simple if there are continuous functions \(\psi_1, \psi_2: [c, d] \to \mathbb{R}\) such that \(D = \{(x, y) \in \mathbb{R}^{2} \mid y \in [c, d], \psi_{1}(y) \le x \le \psi _{2}(y)\}\)

   ​

3. We say that \(D\) is simple if it is \(x\)-simple and \(y\)-simple.

   ​

   \(B(0,0,2) = \{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 \leq 4 \}\)
   ​\(= \{ (x,y) \in \mathbb{R}^2 \mid y^2 \leq 4 - x^2 \}\)
   ​\(= \{ (x,y) \in \mathbb{R}^2 \mid -\sqrt{4-x^2} \leq y \leq \sqrt{4-x^2} \}\)
   ​\(= \{ (x,y) \in \mathbb{R}^{2}\mid -2 \leq x \leq 2, \varphi_{1}(x) \leq y \leq \varphi _{2}(x) \}\)

   ---

   ​

   \(R= [a,b] \times [c,d]\) \(= \left\{ (x,y) \in \mathbb{R}^{2} \middle| \overset{\psi_1(y)}{a}\le x \le \overset {\psi_2(y)}{b}, \overset{\varphi_1(x)}{c}\le y \le \overset{\varphi_2(x)}{d}\right \}\)

4. We refer to any set as above as an elementary region.

Theorem

Let \(D \subseteq \mathbb{R}^2\) be an elementary region. Then \(D\) has area. Moreover,

1. If \(D\) is y-simple, then \(\displaystyle a(D) = \int_{a}^{b} \varphi_{2}(x) - \varphi_{1}(x) \, dx\)

2. If \(D\) is x-simple, then \(\displaystyle a(D) = \int_{c}^{d} \psi_{2}(x) - \psi_{1}(x) \, dx\)

Proof. Exercise.

Since \(D\) is y-simple, then \(D=\{(x,y)\in \R^2:a\leq x\leq b,\psi_1(x)\leq y\leq \psi_2(x)\}\)

Consider \(R=[a,b]\times [c,d]\) and \(c\) is the lower bound of \(\psi_{1}(x)\) and \(d\) is the upper bounded of \(\psi_{2}(x)\)​. \(f^{*}:R\to R\text{ by }f^{*}(x,y)= \begin{cases} f(x,y) & \text{if }(x,y)\in S \\ 0 & \text{if }R\setminus S \end{cases}\)

Then \(\displaystyle\int_{D}f=\int_{R}f^{*}=\int_{a}^{b}\left(\int_{\psi_1(x)}^{\psi_2(x)}1\mathrm{d}x\right) =\int^b_a\left(\psi_2(x)-\psi_1(x)\right)dx\)

With the definitions and results above, we have the following theorem which is a generalization of the previous one over rectangles.

Theorem (Integration over elementary regions)

Let \(D \subseteq \mathbb{R}^2\) be an elementary region and \(f: D \to \mathbb{R}\) be an integrable function.

1. If \(D\) is y-simple with \(D = \{(x,y) \in \mathbb{R}^2 \mid x \in [a,b], \varphi_1(x) \leq y \leq \varphi_2(x)\}\) and the integral \(\displaystyle F(x) = \int_{\varphi_1(x)}^{\varphi_2(x)}f(x,y) \, dy\) exists for all \(x \in [a,b]\) then \(\displaystyle \int_{S} f = \int_{a}^{b} \left(\int_{\varphi_1(x)}^{\varphi_2(x)}f(x,y) \, dy \right ) dx\)

2. If \(D\) is \(x\)-simple with \(D = \{(x,y) \in \mathbb{R}^2 \ | \ y \in [c, d], \psi_1(y) \leq x \leq \psi_2(y)\}\) and the integral \(\displaystyle G(y) = \int_{\psi_1(y)}^{\psi_2(y)}f(x,y) \ dx\) exists for all \(y \in [c,d]\) then \(\displaystyle \int_{S} f = \int_{c}^{d} \left(\int_{\psi_1(y)}^{\psi_2(y)}f(x,y) \ dx \right) dy\)

3. If \(D\) is simple and both integrals \(F(x)\) and \(G(y)\) exist (for example, when \(f\) is continuous), one can compute the integral using either 1. or 2. above.

Proof

1

By assumption, \(D\) is y-simple, \(D=\{(x,y)\in\mathbb{R}^{2}\ |\ x\in[a,b],\varphi_{1}(x)\leq y\leq\varphi_{2}(x)\}\) where \(\varphi_1,\varphi_2:[a,b]\to \R\) continuous.

Then \(\displaystyle \int_{D}f=\int_{R}f^{*}\) where \(D\subseteq R\) rectangle where \(R=[a,b]\times [c,d]\)

And \(f^{*}(x,y)= \begin{cases} f(x,y) & \text{if }(x,y)\in D \\ 0 & \text{if }R\setminus D \end{cases}\), idea: compute \(\displaystyle\int_{R}f^{*}=\int^{b}_{a}F(x)dx\)

Using the theorem above: \(\displaystyle F(x)=\int^{d}_{c}f^{*}(x,y)dy\) exists and being integrable.

Now \(\displaystyle F(x)=\int^{b}_{a}f^{*}(x,y)dy= \int_{c}^{\varphi_1(x)}f^{*}(x,y) dy + \int_{\varphi_1(x)} ^{\varphi_2(x)}f^{*}(x,y) dy + \int_{\varphi_2(x)}^{d}f^{*}(x,y) dy = \int_{\varphi_1(x)} ^{\varphi_2(x)}f(x,y) dy\)

Then \(\displaystyle \int_{D} f=\int^{b}_{a}F(x)dx=\int^{b}_{a}\left(\int^{\varphi_2(x)}_{\varphi_1(x)}f(x,y)dy\right)dx\)

Example

\(T = \{(x,y) \in \mathbb{R}^2 \mid 0 \le x \le \frac{\pi}{2}, 0 \le y \le x\}\)

​

This is \(\text{y-simple}\), Compute \(\displaystyle \int_{T}f \text{ with }f(x,y) = x^{3}y + \cos(x)\)

\(f(x,y) = x^3y + \cos(x) \text{ is continuous on every rectangle}\)

\(\displaystyle F(x)=\int^x_0(x^3y+\cos(x))dy\) exists and it is continuous, then calculate \({\displaystyle\int_{T}f=\int_0^{\frac{\pi}{2}}F\left(x\right)dx}\)

‍