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3.19 Riemann Sum

Recall, the lower and upper sums of \(f\) with respect to the partition \(P\) are given by

\(L(f, P) = \displaystyle\sum_{i=1}^{n}m_{i}(x_{i}- x_{i-1})\), \(U(f, P) = \displaystyle\sum_{i=1}^{n}M_{i} (x_{i} - x_{i-1})\)

By definition, we have that: \(L(f, P) \leq U(f, P)\) for any partition \(P = (x_0, x_1, \dots, x_n)\) of \([a, b]\).

Riemann sum

A Riemann sum of \(f\) with respect to the partition \(P = (x_0, x_1, \dots, x_n)\) of \([a, b]\) such that taking an arbitrary point in the interval \([x_{i-1},x_i]\) instead of the infimum or the supremum: \(S(f,P)=\displaystyle\sum_{i=1}^{n}f\left(z_{i}\right)(x_{i}-x_{i-1})\) where \(z_{i}\in[x_{i-1},x_{i}]\)

Since \(m_{i}\leq f(z_{i})\leq M_{i}\) for all \(1\leq i\leq n\), we obtain \(L(f, P) \leq S(f, P) \leq U(f, P)\)

Question

How do we compare sums associated with different partitions?

If \(P_1\) and \(P_2\) are partitions of a closed interval \([a, b]\), then the partition \(P_1 \cup P_2\) is a refinement of both partitions \(P_1\) and \(P_2\): \(P_1 \subseteq P_1 \cup P_2 \quad \text{and} \quad P_2 \subseteq P_1 \cup P_2\)

In this sense, the set of partitions of the interval is a Partially Ordered Set (POSET) when ordered by inclusion.

This means maybe we can't directly compare two point in set, but we can compare to \(P_1\cup P_2\)

We say that the partition \(P = P_1 \cup P_2\) is a common refinement of \(P_1\) and \(P_2\).

Lemma

If \(P_1\) and \(P_{2}\) are two partitions of \([a, b]\), then \(L(f, P_1) \leq U(f, P_2)\)

Proof

We use this lemma. In particular, \(L(f,P_{1})\leq L(f,P_{1}\cup P_{2})\) and \(U(f,P_{1}\cup P_{2})\leq U(f,P_{2})\)

And we know \(L(f,P_{1}\cup P_{2})\leq U(f,P_{1}\cup P_{2})\), then we get \(L(f,P_{1})\leq\ldots\leq U(f,P_{2})\)​

Definition

Let \(\mathcal{P}\) be the collection of all possible partitions of the interval \([a, b]\). The upper integral of \(f\) is defined as: \(U(f) = \inf \{ U(f, P) \mid P \in \mathcal{P}\}\)

This is because by lemma it's bounded and by completeness axiom in \(\R\)​\, then we have a infimum

Similarly, the lower integral of \(f\) is defined as: \(L(f) = \sup \{ L(f, P) \mid P \in \mathcal{P} \}\)

Lemma

For any bounded function \(f\) defined on a closed interval \([a, b]\), we have \(L(f) \leq U(f)\)

Proof

We know \(U(f) = \inf \{ U(f, P) \mid P \in \mathcal{P}\}\), then \(U(f)\leq U(f,P),\forall P\in\mathcal{P}\)

And \(L(f)=\sup\{L(f,P)\mid P\in\mathcal{P}\}\), then \(L(f,P)\leq L(f),\forall P\in\mathcal{P}\)

But we know that \(L(f,P)\leq U(f,Q)\), \(\forall P,Q\in \mathcal{P}\) by this

Then \(L(f,P)\leq\inf\{U(f,Q)\mid Q\in\mathcal{P}\}=U\left(f\right)\)

Then \(L(f)=\sup\{L(f,P)\mid P\in\mathcal{P}\}\leq U\left(f\right)\)​

Riemann Integrability

A bounded function \(f\) defined on the interval \([a,b]\) is Riemann-integrable if \(U(f) = L(f)\).

In this case, we define \(\displaystyle \int_{a}^{b}f\) or \(\displaystyle\int_{a}^{b}f(x) \, dx\) to be this common value; namely, \(\displaystyle\int_{a}^{b}f = U(f) = L(f).\)

This symbol is because \(\inf\left\lbrace\underbrace{\sum_{i=1}^{n}}_{\int}\underbrace{M_{i}}_{f(x)}\underbrace{(x_{i}-x_{i-1})} _{dx}\right\rbrace=\displaystyle\int_{a}^{b}f(x)\,dx\)

Remark

If the function \(f\) also satisfies that \(f(x) \geq 0\) for all \(x \in [a, b]\), then\(\displaystyle\int_{a}^{b}f(x)dx\) represents the area between the graph of \(f\) and the \(x\)-axis.

Example

  1. \(f: [0,3] \to \mathbb{R}\), \(f(x) = x\), \(A = \frac{9}{2}\)image​

    \(P_1 = (0,3)\), \(P_2 = (0,2,3)\), \(P_3 = (0,1,3)\), \(P_4 = (0,1,2,3)\)

    Then \(L(f, P_1) = 0\), \(U(f, P_1) = 3(3-0) = 9\)

    \(L(f, P_2) = 0(2-0) + 2(3-2) = 2\), \(U(f, P_2) = 2(2-0) + 3(3-2) = 7\)

    \(L(f, P_3) = 0(1-0) + 1(3-1) = 2\), \(U(f, P_3) = 1(1-0) + 3(3-1) = 7\)

    \(L(f, P_4) = 0(1-0) + 1(2-1) + 2(3-2) = 3\), \(U(f, P_4) = 1(1-0) + 2(2-1) + 3(3-2) = 6\)

    Let \(P = (x_0, x_1, \dots, x_n)\). For \(f(x) = x\), \(\inf \{ f(x) \mid x \in [x_{i-1}, x_i] \} = \inf \{ x \mid x \in [x_{i-1}, x_i] \} = x_{i-1} = m_i \quad \forall i\)

    \(L(f, P) = \displaystyle\sum_{i=1}^{n}m_{i}(x_{i}- x_{i-1}) =\sum_{i=1}^{n}x_{i-1} (x_{i}- x_{i-1})\) \(=\displaystyle\sum_{i=1}^{n}x_{i-1}x_{i}-x_{i-1}^{2}\)

    \(\displaystyle U(f, P) = \sum_{i=1}^{n}M_{i} (x_{i} - x_{i-1}) = \sum_{i=1}^{n}x_{i} (x_{i} - x _{i-1})\) \(\displaystyle= \sum_{i=1}^{n}x_{i}^{2}- x_{i}x_{i-1}\)

    We want to see \(L(f,P)=U(f,P)=\frac92\), for partition \(P\), we should consider the one we can handle instead of arbitrary one

    Consider \(P_{n}=(x_{0},...,x_{n}),\left|x_{i}-x_{i-1}\right|=\frac{b-a}{n}=\frac{3}{n}\)

    Then \({L(f,P)=\frac{3}{n}{{\displaystyle\sum_{i=1}^{n}}}x_{i-1}=\frac{3}{n}{{\displaystyle\sum_{i=1}^{n}}}\left(i-1\right)\frac{3}{n}=\frac{9}{n^{2}}{}{{\displaystyle\sum_{i=1}^{n}}}\left(i-1\right)=\frac{9}{n^{2}}\frac{n\left(n-1\right)}{2}=\frac{9n^{2}-9n}{2n^{2}}} =\frac{9}{2}-\frac{9}{2n}\)

    Similarly \(U(f,P)=\frac{9}{2}-\frac{9}{2n}\), when \(n\to\infty\), we have \(L(f,P)=U(f,P)=\frac92=A\)​

  2. True for any polynomial

    Any polynomial is Riemann integrable. Example: \(f(x) = x^2\).

  3. Let \(f: [0,1] \to \mathbb{R}\) be defined as \(f(x) =\begin{cases} 1, & x \in \mathbb{Q} \\0, & x \in \mathbb{R} - \mathbb{Q}\end{cases}\)

    Function is not integrable. \(L(f, P) = 0 \quad \forall P \in \mathcal{P}\), \(U(f, P) = 1 \quad \forall P \in \mathcal{P}\)

Theorem

A bounded function \(f\) is integrable on \([a,b]\) if and only if, for every \(\epsilon > 0\), there exists a partition \(P_{\epsilon}\) of \([a,b]\) such that \(U(f, P_{\epsilon}) - L(f, P_{\epsilon}) < \epsilon.\)

Proof

\(\Rightarrow\)) Assume that \(f\) is integrable. Then \(U(f) = L(f)\) where \(U(f) = \inf \{ U(f, P) \mid P \in \mathcal{P} \}\) and \(L(f) = \sup \{ L(f, P) \mid P \in \mathcal{P} \}\)

Given \(\epsilon > 0\), there exist partitions: \(P_1 \in \mathcal{P}\) such that \(U(f, P_1) \leq U(f) + \frac{\epsilon}{2}\)

\(P_2 \in \mathcal{P}\) such that \(L(f) - \frac{\epsilon}{2} \leq L(f, P_2)\) by definition of inf and supimage​

Let \(P_{\varepsilon} = P_1 \cup P_2 \Rightarrow\) \(L(f, P_2) \leq L(f, P_{\varepsilon})\), \(U(f, P_{\varepsilon}) \leq U(f, P_1)\)

\(\Rightarrow U(f,P_{\varepsilon})-L(f,P_{\varepsilon})\leq U(f,P_{1})-L(f,P_{\varepsilon} )\) \(\leq U(f, P_1) - L(f, P_2) \leq U(f) + \frac{\varepsilon}{2} - L(f, P_2)\) \(\leq U(f) + \frac{\varepsilon}{2} - (L(f) - \frac{\varepsilon}{2}) = U(f) + \frac{\varepsilon}{2} - L(f) + \frac{\varepsilon}{2} = \varepsilon\)

\(\Leftarrow\)) We want to show that \(U(f)=L(f)\)

We know \(U(f) \leq U(f, P_{\varepsilon})\), \(L(f) \geq L(f, P_{\varepsilon})\)

Hence \(0 \leq U(f) - L(f)\) \(\leq U(f, P_{\varepsilon}) - L(f)\) \(\leq U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) < \varepsilon\)

\(\Rightarrow U(f) - L(f) = 0\)