Definition:Integrable Function

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Definition

Darboux

Let $f$ be bounded on $\closedint a b$.


Suppose that:

$\ds \underline {\int_a^b} \map f x \rd x = \overline {\int_a^b} \map f x \rd x$

where $\ds \underline {\int_a^b}$ and $\ds \overline {\int_a^b}$ denote the lower Darboux integral and upper Darboux integral, respectively.


Then the definite (Darboux) integral of $f$ over $\closedint a b$ is defined as:

$\ds \int_a^b \map f x \rd x = \underline {\int_a^b} \map f x \rd x = \overline {\int_a^b} \map f x \rd x$


$f$ is formally defined as (properly) integrable over $\closedint a b$ in the sense of Darboux, or (properly) Darboux integrable over $\closedint a b$.


More usually (and informally), we say:

$f$ is (Darboux) integrable over $\closedint a b$.


Riemann

Let $\Delta$ be a finite subdivision of $\closedint a b$, $\Delta = \set {x_0, \ldots, x_n}$, $x_0 = a$ and $x_n = b$.

Let there for $\Delta$ be a corresponding sequence $C$ of sample points $c_i$, $C = \tuple {c_1, \ldots, c_n}$, where $c_i \in \closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.

Let $\map S {f; \Delta, C}$ denote the Riemann sum of $f$ for the subdivision $\Delta$ and the sample point sequence $C$.


Then $f$ is said to be (properly) Riemann integrable on $\closedint a b$ if and only if:

$\exists L \in \R: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall$ finite subdivisions $\Delta$ of $\closedint a b: \forall$ sample point sequences $C$ of $\Delta: \norm \Delta < \delta \implies \size {\map S {f; \Delta, C} - L} < \epsilon$

where $\norm \Delta$ denotes the norm of $\Delta$.


The real number $L$ is called the Riemann integral of $f$ over $\closedint a b$ and is denoted:

$\ds \int_a^b \map f x \rd x$


Unbounded Above Positive Real Function

Let $f: \R \to \R$ be a positive real function.

Let $f$ be unbounded above on the open interval $\openint a b$.

Let $f_m$ denote the function $f$ truncated by $m$ for $m \in \Z_{>0}$.


Suppose that $f_m$ is Darboux integrable on $\openint a b$ for all $m \in \Z_{>0}$.

Suppose also that the following limit exists:

$\ds \lim_{m \mathop \to \infty} \int_a^b \map {f_m} x \rd x$


Then $f$ is integrable on $\openint a b$ and can be expressed in the conventional notation of the Darboux integral:

$\ds \int_a^b \map f x \rd x$


Unbounded Real Function

Let $f: \R \to \R$ be a real function.

Let $f$ be unbounded on the open interval $\openint a b$.

Let:

$f^+$ denote the positive part of $f$
$f^-$ denote the negative part of $f$

that is:

\(\ds \map {f^+} x\) \(:=\) \(\ds \max \set {0, \map f x}\)
\(\ds \map {f^-} x\) \(:=\) \(\ds -\min \set {0, \map f x}\)


Let $f^+$ and $f^-$ both be integrable on $\openint a b$.


Then $f$ is integrable on $\openint a b$ and its (definite) integral is understood to be:

$\ds \int_a^b \map f x \rd x := \int_a^b \map {f^+} x \rd x - \int_a^b \map {f^-} x \rd x$


Complex Riemann Integrable Function

Let $\mathbb I := \closedint a b$ be a closed real interval.

Let $f: \mathbb I \to \C$ be a bounded complex-valued function.


Define the real function $x: \mathbb I \to \R$ by:

$\forall t \in \mathbb I: \map x t = \map \Re {\map f t}$

Define the real function $y: \mathbb I \to \R$ by:

$\forall t \in \mathbb I: \map y t = \map \Im {\map f t}$

where:

$\map \Re {\map f t}$ denotes the real part of the complex number $\map f t$
$\map \Im {\map f t}$ denotes the imaginary part of $\map f t$.


Suppose that both $x$ and $y$ are Riemann integrable over $\mathbb I$.


Then the complex (Riemann) integral of $f$ over $\mathbb I$ is defined as:

$\ds \int_a^b \map f t \rd t = \int_a^b \map \Re {\map f t} \rd t + i \int_a^b \map \Im {\map f t} \rd t$


$f$ is formally defined as (properly) complex integrable over $\mathbb I$ in the sense of Riemann, or (properly) complex Riemann integrable over $\mathbb I$.


More usually (and informally), we say:

$f$ is (Riemann) complex integrable over $\mathbb I$.


Locally Integrable Function

Let $f : \R^d \to \C$ be a function.

Let $K \subseteq \R^d$ be a compact subset.

Suppose:

$\ds \forall K \subseteq \R^d : \int_K \size {\map f {\mathbf x} } \rd {\mathbf x} < \infty$

where $\mathbf x \in \R^d$ and $\rd {\mathbf x}$ and $\rd {\mathbf x} = \rd x_1 \rd x_2 \ldots \rd x_d$ is the volume element in $\R^d$.


Then $f$ is called the locally integrable function.


Measure Space

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f \in \MM_{\overline \R}, f: X \to \overline \R$ be a measurable function.


Then $f$ is said to be $\mu$-integrable if and only if:

$\ds \int f^+ \rd \mu < +\infty$

and

$\ds \int f^- \rd \mu < +\infty$

where $f^+$, $f^-$ are the positive and negative parts of $f$, respectively.


The integral signs denote $\mu$-integration of positive measurable functions.


$p$-Integrable Function

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f \in \MM_{\overline \R}, f: X \to \overline \R$ be a measurable function.

Let $p \ge 1$ be a real number.


Then $f$ is said to be $p$-integrable in respect to $\mu$ if and only if:

$\ds \int \size f^p \rd \mu < +\infty$

is $\mu$-integrable.


Lebesgue

Let $\lambda^n$ be a Lebesgue measure on $\R^n$ for some $n > 0$.

Let $f: \R^n \to \overline \R$ be an extended real-valued function.


Then $f$ is said to be Lebesgue integrable if and only if it is $\lambda^n$-integrable.

Similarly, for all real numbers $p \ge 1$, $f$ is said to be Lebesgue $p$-integrable if and only if it is $p$-integrable under $\lambda^n$.


Also see

  • Results about integrable functions can be found here.


Sources

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