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3.24 Properties and examples of integrable functions

Theorem

Let \(f: [a, b] \to \mathbb{R}\) be a bounded function. Assume there is a sequence of partitions \(\{P_n\}_{n \geq 1}\) of \([a, b]\) such that \(\lim\limits_{n \to +\infty}L(f, P_{n}) = \lim\limits_{n \to +\infty}U(f, P_{n}) = L\)

Then \(f\) is integrable and \(\displaystyle\int_{a}^{b}f(x)dx = L\)

Main idea: It's important to recall the definition of \(L(f)\) and \(U(f)\) since we want to prove \(L(f)=U(f)\)

Also we know the limit thus we want to put limit on something.

Instead of proving \(L(f)=U(f)\), we can consider proving \(L(f)=L\), then \(U(f)=L\)

To prove \(L(f)=L\;\char"27FA \;L(f)-\lim_{n\to\infty}U(f,P_{n})=0\;\char"27FA \;\lim_{n\to\infty} (L(f)-U(f,P_{n}))=0\)

And we know \(\lim_{n\to\infty}(L(f,P_{n})-U(f,P_{n}))=0\) and \(U(f,P_{n})-L\left(f\right)\leq U\left(f,P_{n}\right)-L\left(f,P_{n}\right)\), then we can put limit and use squeeze theorem

Proof

Let \(\{P_n\}_{n\geq 1}\) be a sequence of partitions with \(\lim_{n\to\infty}U(f,P_{n})=L\) and \(\lim_{n\to\infty}L(f,P_{n})=L\)

We know \(\{U(f,P_{n})\}_{n\geq 1}\) and \(\{L(f,P_{n})\}_{n\geq 1}\) are sequences

We want to prove \(U(f)=L(f)\)

Since \(U(f) = \inf \{ U(f, P) \mid P \in \mathcal{P}\}\) and \(L(f) = \sup \{ L(f, P) \mid P \in \mathcal{P} \}\), then \(L(f)\leq U(f,P),\forall P\in \mathcal{P}\)

In particular, \(L(f)\leq U(f,P_{n})\) \(\forall n\geq 1\), then \(0\leq U(f,P_{n})-L(f)\) (*)

Since \(L(f)=\sup_{p\in\mathcal{P}}\{L(f,P)\}\), then \(L(f,P)\leq L(f)\), then \(L(f,P_{n})\leq L(f)\)

Thus from () we have \(0\leq U(f,P_{n})-L(f)\leq U(f,P_{n})-L(f,P_{n})\) (*)

Then since (**), we put limit on it \(0\leq\lim_{n\to\infty}\left(U\left(f,P_{n}\right)-L\left(f\right)\right)\leq\lim _{n\to\infty}\left(U\left(f,P_{n}\right)-L\left(f,P_{n}\right)\right)=\lim_{n\to\infty} U\left(f,P_{n}\right)-\lim_{n\to\infty}L\left(f,P_{n}\right)=0\)

Thus \(\lim_{n\to\infty}\left(U\left(f,P_{n}\right)-L\left(f\right)\right)=0\Rightarrow L\left(f\right)=\lim_{n\to\infty}U\left(f,P_{n}\right)=L\)

Then using that \(0\leq U(f)-L(f,P_{n})\leq U\left(f,P_{n}\right)-L\left(f,P_{n}\right)\)

Put the limit, \(0\leq \lim_{n\to\infty}(U(f)-L(f,P_{n}))\leq\lim_{n\to\infty}( U\left(f,P_{n}\right)-L\left(f,P_{n}\right))=0\)​

We see that \(U(f)=\lim_{n\to\infty}L(f,P_n)=L\)

Thus \(L(f)=P(f)=L\)

Remark

Note that the thesis of the theorem is stating that in case both limits exist and are equal, then they have to coincide with the supremum of the lower sums and the infimum of the upper sums, respectively.

Example

\(f: [0, b] \to \mathbb{R}\), \(f(x) = x\)image​

\(m_i = \inf \{ f(x) \mid x \in [x_{i-1}, x_i] \} = x_{i-1} \quad \forall i\) and \(L_i = m_i (x_i - x_{i-1}) = x_{i-1} \frac{b}{n} \quad \forall i\)

\(M_i = \sup \{ f(x) \mid x \in [x_{i-1}, x_i] \} = x_i\) and \(U_i = M_i (x_i - x_{i-1}) = x_i \frac{b}{n}\)

Let \(P_n=(x_0,...,x_n)\) where \(x_i-x_{i-1}=\frac{b}{n}\), then \(P_1 = (0, b)\), \(P_2 = (0, \frac{b}{2}, b)\), \(P_3 = (0, \frac{b}{3}, \frac{2b}{3}, b)\), \(P_n = (0, \frac{b}{n}, \frac{2b}{n}, \dots, b)\)

\(L(f, P_n) = \sum_{i=1}^{n} m_i (x_i - x_{i-1})\) \(= \sum_{i=1}^{n} x_{i-1} \frac{b}{n}\) \(= \frac{b}{n} \sum_{i=1}^{n} x_{i-1}\) \(= \frac{b}{n} \sum_{i=1}^{n} (i-1) \frac{b}{n}\) \(= \frac{b^2}{n^2} \sum_{i=1}^{n} (i-1)\) \(= \frac{b^2}{n^2} \sum_{j=0}^{n-1} j\) \(= \frac{b^2}{n^2} \frac{n(n-1)}{2}\) \(\Rightarrow L(f, P_n) = \frac{b^2}{2} \frac{n(n-1)}{n^2}\)

\(\Rightarrow \lim_{n \to \infty} L(f, P_n) = \lim_{n \to \infty} \frac{b^2}{2} \frac{n(n-1)}{n^2} = \frac{b^2}{2}\)

\(U(f, P_n) = \sum_{i=1}^{n} U_i = \sum_{i=1}^{n} M_i (x_i - x_{i-1}) = \sum_{i=1}^{n} x_i \frac{b}{n}\) \(= \frac{b}{n} \sum_{i=1}^{n} x_i = \frac{b}{n} \sum_{i=1}^{n} i \frac{b}{n}\)

\(= \frac{b^2}{n^2} \sum_{i=1}^{n} i = \frac{b^2}{n^2} \frac{n(n+1)}{2}\) \(\Rightarrow U(f, P_n) = \frac{b^2}{2} \frac{n(n+1)}{n^2}\)

\(\Rightarrow \lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{b^2}{2} \frac{n(n+1)}{n^2} = \frac{b^2}{2}\) \(= \lim_{n \to \infty} L(f, P_n)\)

\(\Rightarrow U(f) = L(f) = \frac{b^2}{2}\)

\(\Rightarrow f(x) = x \text{ is integrable}\), \(\displaystyle\int_{0}^{b} x dx = \frac{b^{2}}{2}\)

Example 2

\(f(x) =\begin{cases} 1, & x \in [0, \frac{1}{2}) \cup (\frac{1}{2},1] \\ 2, & x = \frac{1}{2}\end{cases}\), let \(P_n = \left\{ 0, \frac{1}{2} - \frac{1}{n}, \frac{1}{2}, \frac{1}{2} + \frac{1}{n}, 1 \right\}\)

image​

\(x_1 - x_0 = \frac{1}{2} - \frac{1}{n} - 0 = \frac{1}{2} - \frac{1}{n}\)
\(x_2 - x_1 = \frac{1}{2} - \left( \frac{1}{2} - \frac{1}{n} \right) = \frac{1}{n}\)
\(x_3 - x_2 = \frac{1}{2} + \frac{1}{n} - \frac{1}{2} = \frac{1}{n}\)
\(x_4 - x_3 = 1 - \left( \frac{1}{2} + \frac{1}{n} \right) = \frac{1}{2} - \frac{1}{n}\)

\(m_1 = \inf f(x) \quad \text{for } x \in [x_0, x_1] = 1\)
\(m_2 = \inf f(x) \quad \text{for } x \in [x_1, x_2] = 1\)
\(m_3 = 1\)
\(m_4 = 1\)

\(L(f, P_n) = \sum_{i=1}^{n} m_i (x_i - x_{i-1}) = m_1 \left(\frac{1}{2} - \frac{1}{n}\right) + m_2 \left(\frac{1}{4}\right) + m_3 \left(\frac{1}{n}\right) + m_4 \left(\frac{1}{2} - \frac{1}{n}\right)\)

\(= \frac{1}{2} - \frac{1}{n} + \frac{1}{4} + \frac{1}{n} + \frac{1}{2} - \frac{1}{n} + \frac{1}{4}\) \(= 1 \Rightarrow \lim L(f, P_n) = 1\)

\(M_1 = \sup \{ f(x) \mid x \in [x_0, x_1] \} = 1\)
\(M_2 = 2\)
\(M_3 = 2\)
\(M_4 = 1\)

\(U(f, P_n) = 1 \left(\frac{1}{2} - \frac{1}{4} \right) + 2 \left(\frac{1}{4}\right) + 2 \left(\frac{1}{4}\right) + 1 \left(\frac{1}{2} - \frac{1}{n} \right)\)

\(= \frac{1}{2} - \frac{1}{n} + \frac{2}{4} + \frac{2}{4} + \frac{1}{2} - \frac{1}{n}\)

\(= \lim_{n \to \infty} 1 + \frac{2}{n} = 1\)

\(\Rightarrow L(f) = U(f) = 1\)

Remark

In the theorem above, the converse also holds. That is, if \(f\) is an integrable function, then there exists a sequence of partitions \(\{P_n\}_{n \geq 1}\) of \([a, b]\) such that \(\lim_{n \to \infty} L(f, P_n) = \lim_{n \to \infty} U(f, P_n)\)

Proof

Main idea: Use this theorem and set \(\varepsilon=\frac1n\) which is convergent

Since \(f\) is integrable, then there exists a partition \(P_{\epsilon}\) of \([a,b]\) such that \(U(f, P_{\epsilon}) - L(f, P_{\epsilon}) < \epsilon\)

Then take \(\varepsilon=\frac1n\), then we have \(U(f,P_{n})-L(f,P_{n})<\frac{1}{n}\)

Then \(\lim_{n\to\infty}[ U(f,P_{n})-L(f,P_{n})]<\lim_{n\to\infty}\frac{1}{n}\)​

Then \(\lim_{n\to\infty}U(f,P_{n})-\lim_{n\to\infty}L(f,P_{n})<0\)

Thus \(\lim_{n\to\infty}U(f,P_{n})<\lim_{n\to\infty}L(f,P_{n})\)

Also since we know \(U(f,P_n)\geq L(f,P_n)\), then \(\lim_{n\to\infty}U(f,P_{n})>\lim_{n\to\infty}L(f,P_{n})\)

Thus \(\lim_{n \to \infty} L(f, P_n) = \lim_{n \to \infty} U(f, P_n)\)

Remark.

The integral does not see the isolated point.

Our next goal is to prove that any continuous function is integrable.

For this, we need to recall some notions first.

Definition.

A function \(f: A \to \mathbb{R}\) defined on a subset \(A\) of \(\mathbb{R}\) is uniformly continuous on \(A\) if

\(\forall\varepsilon>0,\exists\delta>0,\forall x,y\in A,\left|x-y\right|<\delta:|f (x)-f(y)|<\varepsilon\)​

Remark.

Note that this definition does not depend on the point on \(A\). In some sense, it is about global continuity of the function \(f\). That is, the images of two points \(f(x)\) and \(f(y)\) are close if \(x\) and \(y\) are close enough.

The relation with the continuity of the function at one point \(x = a\) is that the \(\delta\) may depend on the point \(x = a\).

For example, considering the function \(f(x) = x^2\), taking \(\varepsilon = \frac{1}{2}\), for \(x = 3\), we need a smaller \(\delta\) than for \(x = 1\).

image​

If we want to compare the \(\delta\)'s corresponding to two (or finitely many points), we may take the smaller \(\delta\). But if we have infinitely many points, we are in trouble since this smaller \(\delta\) might be \(0\).

A good notion of finiteness for infinitely many points is satisfied when we consider functions on closed intervals, or more generally on compact sets.

Recall that a compact set of \(\mathbb{R}\) is a closed and bounded subset. In particular, a closed interval is a compact set.

Theorem (Abbott, Theorem 4.4.8)

A function that is continuous on a compact set \(K\) is uniformly continuous on \(K\)​ .

Recall also that a continuous function on a compact set \(K\) is bounded and has maximum and minimum values.

\(\exists x_m,x_M\in K\) such that \(f(x_m)\leq f(x)\leq f(x_M),\forall x\in k\)

Theorem.

If \(f: [a, b] \to \mathbb{R}\) is a continuous function on \([a, b]\), then \(f\) is integrable on \([a, b]\).

Proof Idea: Use the properties of continuous functions on closed intervals to better describe the lower and upper sums.

\([x_i, x_{i+1}] \text{ is a compact set } \Rightarrow f \text{ has a minimum and a maximum.}\)

\(P = (x_0, x_1, \dots, x_n) \text{ any partition}\)

\(y_{i}\in[x_{i-1},x_{i}],f(y_{i})=\min\left\lbrace f(x)\mid x\in[x_{i-1},x_{i}]\right \rbrace=\inf\{f(x)\mid x\in[x_{i-1},x_{i}]\}=m_{i}\)

\(z_{i}\in[x_{i-1},x_{i}],f(z_{i})=\max\left\lbrace f(x)\mid x\in[x_{i-1},x_{i}]\right \rbrace=\sup\{f(x)\mid x\in[x_{i-1},x_{i}]\}=M_{i}\)

Moreover, \(f\) is uniformly continuous on \([a, b]\).

For \(\epsilon > 0\), there exists \(\delta > 0\) such that \(|f(x)-f(y)|<\frac{\epsilon}{b-a},\text{if }|x-y|<\delta,\forall x,y\in[a,b]\)

Let \(P_{\epsilon}= (x_{0}, x_{1}, x_{2}, \dots, x_{n}) \text{ be a partition of }[a, b] \text{ such that }|x_{i}- x_{i-1}| < \delta\)

In particular \(|f(z_{i})-f(y_{i})|=f(z_{i})-f(y_{i})<\frac{\epsilon}{b-a}\)

Now, \(0 \leq U(f, P_\epsilon) - L(f, P_\epsilon) = \sum_{i=1}^{n} M_i (x_i - x_{i-1}) - \sum_{i=1}^{n} m_i (x_i - x_{i-1})\)

\(= \sum_{i=1}^{n} f(z_i) (x_i - x_{i-1}) - \sum_{i=1}^{n} f(y_i) (x_i - x_{i-1})\)

\(= \sum_{i=1}^{n} (f(z_i) - f(y_i)) (x_i - x_{i-1})\)

\(< \sum_{i=1}^{n} \frac{\epsilon}{b - a} (x_i - x_{i-1})\) \(=\frac{\epsilon}{b-a}\sum_{i=1}^{n}(x_{i}-x_{i-1})\)

\(= \frac{\epsilon}{b-a} ( (x_1 - a) + (x_2 - x_1) + \dots + (x_{n-1} - x_{n-2}) + (b - x_{n-1}) )\)

\(= \frac{\epsilon}{b-a} (b-a) = \epsilon\)

\(\Rightarrow 0 \leq U(f, P_\epsilon) - L(f, P_\epsilon) < \epsilon\)

\(\Rightarrow f \text{ is integrable}\)

Examples:

  1. Every polynomial \(p(X) \in \mathbb{R}[X]\) is integrable on any closed interval \([a, b] \subset \mathbb{R}\).

  2. The functions \(\sin(x), \cos(x)\) and \(\exp(x)\) are integrable on any closed interval \([a, b] \subset \mathbb{R}\). These are given by uniformly convergent series, hence they are continuous functions.

  3. More generally, any power series is integrable for any closed interval \([a, b] \subset (-R, R)\), where \(R\) is the radius of convergence.
    For example,
    \(f(x) = \sum_{n=0}^{\infty} x^n \quad \forall x \in \left[ -\frac{1}{2}, \frac{1}{2} \right],\)​\(g(x) = \frac{1}{1 - x} = \sum_{n=1}^{\infty} \frac{x^n}{n} \quad \forall x \in \left[ -\frac{1}{2}, \frac{1}{2} \right].\)

Now, the crucial point is to find techniques to actually compute the values of these integrals.