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Mathematics LibreTexts

8.6: Integrable Functions. Convergence Theorems

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    32373
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    I. Some important theorems apply to integrable functions.

    Theorem \(\PageIndex{1}\) (linearity of the integral)

    If \(f, g : S \rightarrow E^{*}\) are integrable on a set \(A \in \mathcal{M}\) in \((S, \mathcal{M}, m),\) so is

    \[p f+q g\]

    for any \(p, q \in E^{1},\) and

    \[\int_{A}(p f+q g)=p \int_{A} f+q \int_{A} g;\]

    in particular,

    \[\int_{A}(f \pm g)=\int_{A} f \pm \int_{A} g.\]

    Proof

    By Problem 5 in §5,

    \[\overline{\int}_{A} f+\overline{\int}_{A} g \geq \overline{\int}_{A}(f+g) \geq \underline{\int}_{A}(f+g) \geq \underline{\int}_{A} f + \underline{\int}_{A} g.\]

    (Here

    \[\overline{\int}_{A} f, \underline{\int}_{A} f, \overline{\int}_{A} g, \text { and } \underline{\int}_{A} g\]

    are finite by integrability; so all is orthodox.)

    As

    \[\overline{\int}_{A} f=\underline{\int}_{A} f \text { and } \overline{\int}_{A} g=\underline{\int}_{A} g,\]

    the inequalities turn into equalities, so that

    \[\int_{A} f+\int_{A} g=\overline{\int}_{A}(f+g)=\underline{\int}_{A}(f+g).\]

    Using also Theorem 1(e)(e') from §5, we obtain the desired result for any \(p, q \in E^{1}. \quad \square\)

    Theorem \(\PageIndex{2}\)

    A function \(f : S \rightarrow E^{*}\) is integrable on \(A\) in \((S, \mathcal{M}, m)\) iff

    (i) it is \(m\)-measurable on \(A,\) and

    (ii) \(\overline{\int}_{A} f\) (equivalently \(\overline{\int}_{A}|f|\)) is finite.

    Proof

    If these conditions hold, \(f\) is integrable on \(A\) by Theorem 3 of §5.

    Conversely, let

    \[\overline{\int}_{A} f=\underline{\int}_{A} f \neq \pm \infty.\]

    Using Lemma 2 in §5, fix measurable maps \(g\) and \(h\) (\(g \leq f \leq h\)) on \(A,\) with

    \[\int_{A} g=\int_{A} f=\int_{A} h \neq \pm \infty.\]

    By Theorem 3 in §5, \(g\) and \(h\) are integrable on \(A;\) so by Theorem 1,

    \[\int_{A}(h-g)=\int_{A} h-\int_{A} g=0.\]

    As

    \[h-g \geq h-f \geq 0,\]

    we get

    \[\int_{A}(h-f)=0,\]

    and so by Theorem 1(h) of §5, \(h-f=0\) a.e. on \(A\).

    Hence \(f\) is almost measurable on \(A,\) and

    \[\int_{A} f \neq \pm \infty\]

    by assumption. From formula (1), we then get

    \[\int_{A} f^{+} \text { and } \int_{A} f^{-}<\infty,\]

    and hence

    \[\int_{A}|f|=\int_{A}\left(f^{+}+f^{-}\right)=\int_{A} f^{+}+\int_{A} f^{-}<\infty\]

    by Theorem 1 and by Theorem 2 of §2. Thus all is proved.\(\quad \square\)

    Simultaneously, we also obtain the following corollary.

    Corollary \(\PageIndex{1}\)

    A function \(f : S \rightarrow E^{*}\) is integrable on \(A\) iff \(f^{+}\) and \(f^{-}\) are.

    Corollary \(\PageIndex{2}\)

    If \(f, g : S \rightarrow E^{*}\) are integrable on \(A,\) so also are

    \[f \vee g, f \wedge g,|f|, \text { and } k f \text { for } k \in E^{1},\]

    with

    \[\int_{A} k f=k \int_{A} f.\]

    Exercise!

    For products \(f g,\) this holds if \(f\) or \(g\) is bounded. In fact, we have the following theorem.

    Theorem \(\PageIndex{3}\) (weighted law of the mean)

    Let \(f\) be \(m\)-measurable and bounded on A. Set

    \[p=\inf f[A] \text { and } q=\sup f[A].\]

    Then if \(g\) is \(m\)-integrable on \(A,\) so is \(f g,\) and

    \[\int_{A} f|g|=c \int_{A}|g|\]

    for some \(c \in[p, q]\).

    If, further, \(f\) also has the Darboux property on A (Chapter 4, §9), then \(c=f(x_{0})\) for some \(x_{0} \in A.\)

    Proof

    By assumption,

    \[\left(\exists k \in E^{1}\right) \quad|f| \leq k\]

    on \(A.\) Hence if \(\int_{A}|g|=0\),

    \[
    \left|\int_{A} f|g|\right| \leq \int_{A}|f g| \leq k \int_{A}|g|=0;\]

    so any \(c \in[p, q]\) yields

    \[\int_{A} f|g|=c \int_{A}|g|=0.\]

    If, however, \(\int_{A}|g| \neq 0,\) the number

    \[c=\left(\int_{A} f|g|\right) / \int_{A}|g|\]

    is the required constant.

    Moreover, as \(f\) and \(g\) are \(m\)-measurable on \(A,\) so is \(f g;\) and as

    \[\left|\int_{A} f g\right| \leq|c| \int_{A}|g|<\infty,\]

    \(f g\) is integrable on \(A\) by Theorem 2.

    Finally, if \(f\) has the Darboux property and if \(p<c<q\) (with \(p, q\) as above), then

    \[f(x)<c<f(y)\]

    for some \(x, y \in A\) (why?); hence by the Darboux property, \(f\left(x_{0}\right)=c\) for some \(x_{0} \in A.\)

    If, however,

    \[c \leq \inf f[A]=p,\]

    then

    \[(f-c)|g| \geq 0\]

    and

    \[\int_{A}(f-c)|g|=\int_{A} f|g|-c \int_{A}|g|=0 \text { (why?);}\]

    so by Theorem 1(h) in §5, \(f-c=0\) a.e. on \(A.\) Then surely \(f\left(x_{0}\right)=c\) for some \(x_{0} \in A\) (except the trivial case \(m A=0\)). This also implies \(c \in f[A] \in[p, q]\).

    Proceed similarly in the case \(c \geq q. \quad \square\)

    Corollary \(\PageIndex{3}\)

    If \(f\) is integrable on \(A \in \mathcal{M},\) it is so on any \(B \subseteq A(B \in \mathcal{M})\).

    Proof

    Apply Theorem 1(f) in §5, and Theorem 3 of §5, to \(f^{+}\) and \(f^{-}. \quad \square\)

    II. Convergence Theorems. If \(f_{n} \rightarrow f\) on \(A\) (pointwise, a.e., or uniformly), does it follow that

    \[\int_{A} f_{n} \rightarrow \int_{A} f ?\]

    To give some answers, we need a lemma.

    Lemma \(\PageIndex{1}\)

    If \(f \geq 0\) on \(A \in \mathcal{M}\) and if

    \[\underline{\int}_{A} f>p \in E^{*},\]

    there is an elementary and nonnegative map \(g\) on \(A\) such that

    \[\int_{A} g>p,\]

    and \(g<f\) on \(A\) except only at those \(x \in A\) (if any) at which

    \[f(x)=g(x)=0.\]

    (We then briefly write \(g \subset f\) on \(A.\))

    Proof

    By Lemma 1 in §5, there is an elementary and nonnegative map \(G \leq f\) on \(A,\) with

    \[\underline{\int}_{A} f \geq \int_{A} G>p.\]

    For the rest, proceed as in Problem 7 of §4, replacing \(f\) by \(G\) there.\(\quad \square\)

    Theorem \(\PageIndex{4}\) (monotone convergence)

    If \(0 \leq f_{n} \nearrow f\) (a.e.) on \(A \in \mathcal{M},\) i.e.,

    \[0 \leq f_{n} \leq f_{n+1} \quad (\forall n),\]

    and \(f_{n} \rightarrow f\) (a.e.) on \(A,\) then

    \[\overline{\int}_{A} f_{n} \nearrow \overline{\int}_{A} f.\]

    Proof for \(\mathcal{M}\)-measurable \(f_{n}\) and \(f\) on \(A.\)

    By Corollary 2 in §5, we may assume that \(f_{n} \nearrow f\) (pointwise) on \(A\) (otherwise, drop a null set).

    By Theorem 1(c) of §5, \(0 \leq f_{n} \nearrow f\) implies

    \[0 \leq \int_{A} f_{n} \leq \int_{A} f,\]

    and so

    \[\lim _{n \rightarrow \infty} \int_{A} f_{n} \leq \int_{A} f.\]

    The limit, call it \(p,\) exists in \(E^{*},\) as \(\left\{\int_{A} f_{n}\right\} \uparrow.\) It remains to show that

    \[p \geq \overline{\int}_{A} f=\underline{\int}_{A} f.\]

    (We know that

    \[\overline{\int}_{A} f=\underline{\int}_{A} f,\]

    by the assumed measurability of \(f;\) see Theorem 3 in §5.)

    Suppose

    \[\underline{\int}_{A} f>p.\]

    Then Lemma 1 yields an elementary and nonnegative map \(g \subset f\) on \(A,\) with

    \[p<\int_{A} g.\]

    Let

    \[A_{n}=A\left(f_{n} \geq g\right), \quad n=1,2, \ldots.\]

    Then \(A_{n} \in \mathcal{M}\) and

    \[A_{n} \nearrow A=\bigcup_{n=1}^{\infty} A_{n}.\]

    For if \(f(x)=0,\) then \(x \in A_{1},\) and if \(f(x)>0,\) then \(f(x)>g(x),\) so that \(f_{n}(x)>g(x)\) for large \(n;\) hence \(x \in A_{n}.\)

    By Note 4 in §5, the set function \(s=\int g\) is a measure, hence continuous by Theorem 2 in Chapter 7, §4. Thus

    \[\int_{A} g=s A=\lim _{n \rightarrow \infty} s A_{n}=\lim _{n \rightarrow \infty} \int_{A_{n}} g.\]

    But as \(g \leq f_{n}\) on \(A_{n},\) we have

    \[\int_{A_{n}} g \leq \int_{A_{n}} f_{n} \leq \int_{A} f_{n}.\]

    Hence

    \[\int_{A} g=\lim \int_{A_{n}} g \leq \lim \int_{A} f_{n}=p,\]

    contrary to \(p<\int_{A} g.\) This contradiction completes the proof.\(\quad \square\)

    Lemma \(\PageIndex{2}\) (Fatou)

    If \(f_{n} \geq 0\) on \(A \in \mathcal{M}(n=1,2, \ldots),\) then

    \[\overline{\int}_{A} \underline{\lim} f_{n} \leq \underline{\lim } \overline{\int}_{A} f_{n}.\]

    Proof

    Let

    \[g_{n}=\inf _{k \geq n} f_{k}, \quad n=1,2, \ldots;\]

    so \(f_{n} \geq g_{n} \geq 0\) and \(\left\{g_{n}\right\} \uparrow\) on \(A.\) Thus by Theorem 4,

    \[\overline{\int}_{A} \lim g_{n} = \lim \overline{\int}_{A} g_{n} = \underline{\lim} \overline{\int}_{A} g_{n} \leq \underline{\lim} \overline{\int}_{A} f_{n}.\]

    But

    \[\lim _{n \rightarrow \infty} g_{n}=\sup _{n} g_{n}=\sup _{n} \inf _{k \geq n} f_{k}=\lim _{n} f_{n}.\]

    Hence

    \[\overline{\int}_{A} \underline{\lim} f_{n} = \overline{\int}_{A} \lim g_{n} \leq \underline{\lim} \overline{\int}_{A} f_{n},\]
    as claimed.\(\quad \square\)

    Theorem \(\PageIndex{5}\) (dominated convergence)

    Let \(f_{n} : S \rightarrow E\) be \(m\)-measurable on \(A \in \mathcal{M} (n=1,2,\ldots).\) Let

    \[f_{n} \rightarrow f \text{ (a.e.) on } A.\]

    Then

    \[\lim _{n \rightarrow \infty} \int_{A}\left|f_{n}-f\right|=0,\]

    provided that there is a map \(g : S \rightarrow E^{1}\) such that

    \[\int_{A} g<\infty\]

    and

    \[(\forall n) \quad\left|f_{n}\right| \leq g \text { a.e. on } A.\]

    Proof

    Neglecting null sets, we may assume that

    \[\left|f_{n}\right| \leq g<\infty\]

    on \(A\) and \(f_{n} \rightarrow f\) (pointwise) on \(A;\) so \(|f| \leq g\) and

    \[\left|f_{n}-f\right| \leq\left|f_{n}\right|+|f| \leq 2 g\]

    on \(A.\) As \(|f|<\infty,\) we have

    \[\left|f_{n}-f\right| \rightarrow 0\]

    on \(A.\) Hence, setting

    \[h_{n}=2 g-\left|f_{n}-f\right| \geq 0,\]

    we get

    \[2 g=\lim _{n \rightarrow \infty} h_{n}=\underline{\lim} h_{n}.\]

    We may also assume that \(g\) is measurable on \(A.\) (If not, replace it by a measurable \(G \geq g,\) with

    \[\int_{A} G=\int_{A} g<\infty,\]

    by Lemma 2 in §5.) Then all

    \[h_{n}=2 g-\left|f_{n}-f\right|\]

    are measurable (even integrable) on \(A\).

    Thus by Lemma 2,

    \begin{aligned} \int_{A} 2 g = \int_{A} \underline{\lim} h_{n} & \leq \underline{\lim} \int_{A}\left(2 g-\left|f_{n}-f\right|\right) \\ &=\underline{\lim}\left(\int_{A} 2 g+\int_{A}\left(-\left|f_{n}-f\right|\right)\right) \\ &= \int_{A} 2 g+\underline{\lim}\left(-\int_{A}\left|f_{n}-f\right|\right) \\ &= \int_{A} 2 g-\overline{\lim} \int_{A}\left|f_{n}-f\right|. \end{aligned}
    (See Problems 5 and 8 in Chapter 2, §13.)

    Canceling \(\int_{A} 2 g\) (finite!), we have

    \[0 \leq-\overline{\lim } \int_{A}\left|f_{n}-f\right|.\]

    Hence

    \[0 \geq \overline{\lim } \int_{A}\left|f_{n}-f\right| \geq \underline{\lim }\int_{A}\left|f_{n}-f\right| \geq 0,\]

    as \(\left|f_{n}-f\right| \geq 0.\) This yields

    \[0=\overline{\lim } \int_{A}\left|f_{n}-f\right|=\underline{\lim} \int_{A}\left|f_{n}-f\right|=\lim \int_{A}\left|f_{n}-f\right|,\]

    as required.\(\quad \square\)

    Note 1. Theorem 5 holds also for complex and vector-valued functions (for \(\left|f_{n}-f\right|\) is real).

    In the extended-real case, Theorems 1(g) in §5 and Theorems 1 and 2 in §6 yield

    \[\left|\int_{A} f_{n}-\int_{A} f\right|=\left|\int_{A}\left(f_{n}-f\right)\right| \leq \int_{A}\left|f_{n}-f\right| \rightarrow 0,\]

    i.e.,

    \[\int_{A} f_{n} \rightarrow \int_{A} f.\]

    Moreover, \(f\) is integrable on \(A,\) being measurable (why?), with

    \[\int_{A}|f| \leq \int_{A} g<\infty.\]

    For complex and vector-valued functions, this will follow from §7. Observe that Theorem 5, unlike Theorem 4, requires the \(m\)-measurability of the \(f_{n}\).

    Note 2. Theorem 5 fails if there is no "dominating"

    \[g \geq\left|f_{n}\right| \text { with } \int_{A} g<\infty,\]

    even if \(f\) and the \(f_{n}\) are integrable.

    Example

    Let \(m\) be Lebesgue measure in \(A=E^{1}, f=0,\) and

    \[f_{n}=\left\{\begin{array}{ll}{1} & {\text { on }[n, n+1],} \\ {0} & {\text { elsewhere. }}\end{array}\right.\]

    Then \(f_{n} \rightarrow f\) and \(\int_{A} f_{n}=1;\) so

    \[\lim _{n \rightarrow \infty} \int_{A} f_{n}=1 \neq 0=\int_{A} f.\]

    The trouble is that any

    \[g \geq f_{n} \quad(n=1,2, \ldots)\]

    would have to be \(\geq 1\) on \(B=[1, \infty);\) so

    \[\int_{A} g \geq \int_{B} g=1 \cdot m B=\infty,\]

    instead of \(\int_{A} g<\infty\).

    This example also shows that \(f_{n} \rightarrow f\) alone does not imply

    \[\int_{A} f_{n} \rightarrow \int_{A} f.\]

    Theorem \(\PageIndex{6}\) (absolute continuity of the integral)

    Given \(f : S \rightarrow E\) with

    \[\overline{\int}_{A}|f|<\infty\]

    and \(\varepsilon>0,\) there is \(\delta>0\) such that

    \[\overline{\int}_{X}|f|<\varepsilon\]

    whenever

    \[m X<\delta \quad(A \supseteq X, X \in \mathcal{M}).\]

    Proof

    By Lemma 2 in §5, fix \(h \geq|f|,\) measurable on \(A,\) with

    \[\int_{A} h=\overline{\int}_{A}|f|<\infty.\]

    Neglecting a null set, we assume that \(|h|<\infty\) on \(A\) (Corollary 1 of §5). Now, \((\forall n)\) set

    \[g_{n}(x)=\left\{\begin{array}{ll}{h(x),} & {x \in A_{n}=A(h<n),} \\ {0,} & {x \in-A_{n}.}\end{array}\right.\]

    Then \(g_{n} \leq n\) and \(g_{n}\) is measurable on \(A\). (Why?)

    Also, \(g_{n} \geq 0\) and \(g_{n} \rightarrow h\) (pointwise) on \(A\).

    For let \(\varepsilon>0,\) fix \(x \in A,\) and find \(k>h(x).\) Then

    \[(\forall n \geq k) \quad h(x) \leq n \text { and } g_{n}(x)=h(x).\]

    So

    \[(\forall n \geq k) \quad\left|g_{n}(x)-h(x)\right|=0<\varepsilon.\]

    Clearly, \(g_{n} \leq h.\) Hence by Theorem 5

    \[\lim _{n \rightarrow \infty} \int_{A}\left|h-g_{n}\right|=0.\]

    Thus we can fix \(n\) so large that

    \[\int_{A}\left(h-g_{n}\right)<\frac{1}{2} \varepsilon.\]

    For that \(n,\) let

    \[\delta=\frac{\varepsilon}{2 n}\]

    and take any \(X \subseteq A(X \in \mathcal{M}),\) with \(m X<\delta\).

    As \(g_{n} \leq n\) (see above), Theorem 1(c) in §5 yields

    \[\int_{X} g_{n} \leq \int_{X}(n)=n \cdot m X<n \delta=\frac{1}{2} \varepsilon.\]

    Hence as \(|f| \leq h\) and

    \[\int_{X}\left(h-g_{n}\right) \leq \int_{A}\left(h-g_{n}\right)<\frac{1}{2} \varepsilon\]

    (Theorem 1(f) of §5), we obtain

    \[\overline{\int}_{X}|f| \leq \int_{X} h=\int_{X}\left(h-g_{n}\right)+\int_{X} g_{n}<\frac{1}{2} \varepsilon+\frac{1}{2} \varepsilon=\varepsilon,\]

    as required.\(\quad \square\)