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Mathematics LibreTexts

7.1: Upper and Lower Integrals

  • Page ID
    22678
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    Definition

    Given a closed interval \([a, b] \subset \mathbb{R}\) with \(a<b,\) we call any finite subset of \([a, b]\) which includes both \(a\) and \(b\) a partition of \([a, b]\).

    For convenience, whenever we consider a partition \(P\) of an interval \([a, b]\) we will index the elements in increasing order, starting with \(0 .\) That is, if \(|P|=n+1\) and \(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\},\) then

    \[a=x_{0}<x_{1}<x_{2}<\cdots<x_{n}=b.\]

    Definition

    Suppose \(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}\) is a partition of \([a, b]\) and \(f:[a, b] \rightarrow \mathbb{R}\) is bounded. For \(i=1,2, \ldots, n,\) let

    \[m_{i}=\inf \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\}\]

    and

    \[M_{i}=\sup \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\}.\]

    We call

    \[L(f, P)=\sum_{i=1}^{n} m_{i}\left(x_{i}-x_{i-1}\right)\]

    the lower sum of \(f\) determined by \(P\) and

    \[U(f, P)=\sum_{i=1}^{n} M_{i}\left(x_{i}-x_{i-1}\right)\]

    the upper sum of \(f\) determined by \(P .\)

    Definition

    If \(P_{1}\) and \(P_{2}\) are both partitions of \([a, b]\) and \(P_{1} \subset P_{2},\) then we call \(P_{2}\) a refinement of \(P_{1}\).

    Definition

    If \(P_{1}\) and \(P_{2}\) are both partitions of \([a, b],\) then we call the partition \(P=P_{1} \cup P_{2}\) the common refinement of \(P_{1}\) and \(P_{2}\).

    lemma \(\PageIndex{1}\)

    Suppose \(P_{1}=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}\) is a partition of \([a, b], s \in(a, b)\), \(s \notin P_{1},\) and \(f:[a, b] \rightarrow \mathbb{R}\) is bounded. If \(P_{2}=P_{1} \cup\{s\},\) then \(L\left(f, P_{1}\right) \leq L\left(f, P_{2}\right)\) and \(U\left(f, P_{2}\right) \leq U\left(f, P_{1}\right)\).

    Proof

    Suppose \(x_{i-1}<s<x_{i}\) and let

    \[\begin{aligned} w_{1} &=\inf \left\{f(x): x_{i-1} \leq x \leq s\right\}, \\ W_{1} &=\sup \left\{f(x): x_{i-1} \leq x \leq s\right\}, \\ w_{2} &=\inf \left\{f(x): s \leq x \leq x_{i}\right\}, \\ W_{2} &=\sup \left\{f(x): s \leq x \leq x_{i}\right\}, \\ m_{i} &=\inf \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\}, \end{aligned}\]

    and

    \[M_{i}=\sup \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\}.\]

    Then \(w_{1} \geq m_{i}, w_{2} \geq m_{i}, W_{1} \leq M_{i},\) and \(W_{2} \leq M_{i} .\) Hence

    \[\begin{aligned} L\left(f, P_{2}\right)-L\left(f, P_{1}\right) &=w_{1}\left(s-x_{i-1}\right)+w_{2}\left(x_{i}-s\right)-m_{i}\left(x_{i}-x_{i-1}\right) \\ &=w_{1}\left(s-x_{i-1}\right)+w_{2}\left(x_{i}-s\right)-m_{i}\left(s-x_{i-1}\right) \\ & \quad-m_{i}\left(x_{i}-s\right) \\ &=\left(w_{1}-m_{i}\right)\left(s-x_{i-1}\right)+\left(w_{2}-m_{i}\right)\left(x_{i}-s\right) \\ & \geq 0 \end{aligned}\]

    and

    \[\begin{aligned} U\left(f, P_{1}\right)-U\left(f, P_{2}\right)=& M_{i}\left(x_{i}-x_{i-1}\right)-W_{1}\left(s-x_{i-1}\right)-W_{2}\left(x_{i}-s\right) \\=& M_{i}\left(s-x_{i-1}\right)+M_{i}\left(x_{i}-s\right)-W_{1}\left(s-x_{i-1}\right) \\ &-W_{2}\left(x_{i}-s\right) \\=&\left(M_{i}-W_{1}\right)\left(s-x_{i-1}\right)+\left(M_{i}-W_{2}\right)\left(x_{i}-s\right) \\ \geq & 0. \end{aligned}\]

    Thus \(L\left(f, P_{1}\right) \leq L\left(f, P_{2}\right)\) and \(U\left(f, P_{2}\right) \leq U\left(f, P_{1}\right)\). \(\quad\) Q.E.D.

    Proposition \(\PageIndex{2}\)

    Suppose \(P_{1}\) and \(P_{2}\) are partitions of \([a, b],\) with \(P_{2}\) a refinement of \(P_{1} .\) If \(f:[a, b] \rightarrow \mathbb{R}\) is bounded, then \(L\left(f, P_{1}\right) \leq L\left(f, P_{2}\right)\) and \(U\left(f, P_{2}\right) \leq U\left(f, P_{1}\right)\).

    Proof

    The proposition follows immediately from repeated use of the previous lemma. \(\quad\) Q.E.D.

    Proposition \(\PageIndex{3}\)

    Suppose \(P_{1}\) and \(P_{2}\) are partitions of \([a, b] .\) If \(f:[a, b] \rightarrow \mathbb{R}\) is bounded, then \(L\left(f, P_{1}\right) \leq U\left(f, P_{2}\right)\).

    Proof

    The result follows immediately from the definitions if \(P_{1}=P_{2} .\) Otherwise, let \(P\) be the common refinement of \(P_{1}\) and \(P_{2} .\) Then

    \[L\left(f, P_{1}\right) \leq L(f, P) \leq U(f, P) \leq U\left(f, P_{2}\right).\]

    Q.E.D.

    Definition

    Suppose \(a<b\) and \(f:[a, b] \rightarrow \mathbb{R}\) is bounded. We call

    \[\underline{\int_{a}^{b}} f=\sup \{L(f, P): P \text { is a partition of }[a, b]\}\]

    the lower integral of \(f\) over \([a, b]\) and

    \[\overline{\int_{a}^{b}} f=\inf \{U(f, P): P \text { is a partition of }[a, b]\}\]

    the upper integral of \(f\) over \([a, b]\).

    Note that both the lower integral and the upper integral are finite real numbers since the lower sums are all bounded above by any upper sum and the upper sums are all bounded below by any lower sum.

    Proposition \(\PageIndex{4}\)

    Suppose \(a<b\) and \(f:[a, b] \rightarrow \mathbb{R}\) is bounded. Then

    \[\underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f.\]

    Proof

    Let \(P\) be a partition of \([a, b] .\) Then for any partition \(Q\) of \([a, b],\) we have \(L(f, Q) \leq U(f, P) .\) Hence \(U(f, P)\) is an upper bound for any lower sum, and so

    \[\underline{\int_{a}^{b}} f \leq U(f, P).\]

    But this shows that the lower integral is a lower bound for any upper sum. Hence

    \[\int_{a}^{b} f \leq \int_{a}^{b} f.\]

    Q.E.D.

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