nyuuzyou/wikis huggingface dataset:
Integral of Bounded Measurable Function with respect to Finite Signed Measure is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f : X \to \R$ be a bounded $\Sigma$-measurable function.
Let $\mu$ be a finite signed measure on $\struct {X, \Sigma}$. 
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$. 
Then the $\mu$-integral of $f$ defined by: 
:$\ds \int f \rd \mu = \int f \rd \mu^+ - \int f \rd \mu^-$
is well-defined.
== Proof ==
We show that $f$ is $\mu^+$-integrable and $\mu^-$-integrable.
We will then have: 
:$\ds -\infty < \int f \rd \mu^+ < \infty$
and:
:$\ds -\infty < \int f \rd \mu^- < \infty$ 
So that: 
:$\ds \int f \rd \mu^+ - \int f \rd \mu^-$ 
is well-defined.
Since $f$ is bounded, there exists $M > 0$ such that: 
:$\size {\map f x} \le M$
for each $x \in X$.
From Jordan Decomposition of Finite Signed Measure, we have: 
:$\mu^+$ and $\mu^-$ are finite.
That is: 
:$\map {\mu^+} X < \infty$
and:
:$\map {\mu^-} X < \infty$
We therefore have: 
so:
So:
:$\size f$ is $\mu^+$-integrable
and:
:$\size f$ is $\mu^-$-integrable.
From Characterization of Integrable Functions, we have: 
:$f$ is $\mu^+$-integrable and $\mu^-$-integrable
as required.
Category:Integral of Bounded Measurable Function with respect to Finite Signed Measure
Integral of Characteristic Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $E \in \Sigma$ be a measurable set, and let $\chi_E: X \to \R$ be its characteristic function.
Then $\map {I_\mu} {\chi_E} = \map \mu E$, where $\map {I_\mu} {\chi_E}$ is the $\mu$-integral of $\chi_E$.
=== Corollary ===
== Proof ==
Let $a_1 = 1$ and $E_1 = E$.
As in the definition of standard representation, denote $a_0 = 0$ and $E_0 = X \setminus E_1$.
Then for $x \in X$, we have:
:$\map {\chi_E} x = 0 \cdot \map {\chi_{E_0} } x + 1 \cdot \map {\chi_{E_1} } x$
since $E_1 = E$.
Hence $\chi_E = a_0 \chi_{E_0} + a_1 \chi_{E_1}$ is a standard representation for $\chi_E$.
Thus, by definition of $\mu$-integral:
:$\ds \int \chi_E \rd \mu = a_0 \map \mu {E_0} + a_1 \map \mu {E_1} = \map \mu E$
as desired.
== Sources ==
* : $9.3 \ \text{(i)}$
Category:Integral of Positive Simple Function
Category:Characteristic Functions
Category:Integral of Characteristic Function
Integral of Characteristic Function/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $E \in \Sigma$ be a measurable set, and let $\chi_E: X \to \R$ be its characteristic function.
Then:
:$\ds \int \chi_E \rd \mu = \map \mu E$
where the integral sign denotes the $\mu$-integral of $\chi_E$.
== Proof ==
By Integral of Characteristic Function, have:
:$\map {I_\mu} {\chi_E} = \map \mu E$
where $\map {I_\mu} {\chi_E}$ is the $\mu$-integral of $\chi_E$.
From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, it also holds that:
:$\ds \int \chi_E \rd \mu = \map {I_\mu} {\chi_E}$
Combining these equalities gives the result.
== Sources ==
* : $9.8 \ \text{(i)}$, $\S 9$: Problem $2$
Category:Integral of Characteristic Function
Category:Integral of Positive Measurable Function
Integral of Compactly Supported Function

== Theorem ==
Let $f : \R \to \R$ be a continuous real function.
Let $K \subset \R$ be a compact subset, say, $\closedint a b$.
Let $K$ be the support of $f$:
:$\map \supp f = K$.
Then:
:$\ds \int_{- \infty}^\infty \map f x \rd x = \int_a^b \map f x \rd x$
== Proof ==
Category:Calculus
Integral of Constant/Definite

== Theorem ==
Let $c$ be a constant.
:$\ds \int_a^b c \rd x = c \paren {b - a}$
=== Corollary ===
== Proof ==
Let $f_c: \R \to \R$ be the constant function.
By definition:
:$\forall x \in \R: \map {f_c} x = c$
Thus:
:$\map \sup {f_c} = \map \inf {f_c} = c$
So from Darboux's Theorem, we have:
:$\ds c \paren {b - a} \le \int_a^b c \rd x \le c \paren {b - a}$
Hence the result.
== Sources ==
* : $\S 13.5$
Category:Definite Integrals
Integral of Dirac Delta Function by Continuous Function over Reals

== Theorem ==
Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Then:
:$\ds \int_{- \infty}^{+ \infty} \map \delta x \map g x \rd x = \map g 0$
== Proof ==
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$
We have that:
From Darboux's Theorem:
:$\ds m \paren {\epsilon - \paren {-\epsilon} } \le \int_{-\epsilon}^\epsilon \map g x \rd x \le M \paren {\epsilon - \paren {-\epsilon}}$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint {-\epsilon} \epsilon$.
Hence:
:$\ds 2 m \epsilon \le \int_{-\epsilon}^\epsilon \map g x \rd x \le 2 M \epsilon$
and so dividing by $2 \epsilon$:
:$\ds m \le \dfrac 1 {2 \epsilon} \int_{-\epsilon}^\epsilon \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g 0$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+\infty} \map {F_\epsilon} x \map g x \rd x = \map g 0$
But by :
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+\infty} \map {F_\epsilon} x \map g x \rd x = \ds \int_{-\infty}^{+\infty} \map \delta x \map g x \rd x$
Hence the result.
Category:Dirac Delta Function
Integral of Dirac Delta Function over Reals

== Theorem ==
Let $\map \delta x$ denote the Dirac delta function.
Then:
:$\ds \int_{-\infty}^{+\infty} \map \delta x \rd x = 1$
== Proof ==
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$
We have that:
Then:
Hence the result.
== Sources ==
* 
Category:Dirac Delta Function
Integral of Distribution Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space and $f$ be a $\mu$-measurable function. Let $p > 0, r \geq 0$.
For $\lambda > 0$, let $E_\lambda = \set {x \in X: \size {\map f x} > \lambda}$, so that $\map m \lambda = \map \mu {E_\lambda}$ is the distribution function of $f$.
Then:
:$\ds \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda = \int_X \size f^{p + r} \rd \mu$
and in particular:
:$\ds \int_0^\infty p \lambda^{p - 1} \map m \lambda \rd \lambda = \int_X \size f^p \rd \mu$
== Proof ==
We have that for any measurable $A \in \Sigma$:
:$\map \mu A = \ds \int_A 1 \rd \mu$
Therefore, for $\lambda > 0$:
:$\map \mu {E_\lambda} = \ds \int_{E_\lambda} 1 \rd \mu$
which can also be written:
:$\map \mu {E_\lambda} = \ds \int_{E_\lambda} \size f^0 \rd \mu$
Therefore, taking $r = 0$ in the above, we obtain:
Category:Measure Theory
Integral of Exponent of Half Square over Reals

== Theorem ==
:$\ds \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = \sqrt {2 \pi}$
== Proof ==
Let $t = \dfrac {x^2} 2$.
Then:
We have that $e^{- x^2 / 2}$ is an even function.
From Definite Integral of Even Function: Corollary:
:$\ds \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = 2 \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x$
Hence the result.
== Sources ==
* : $\S 17.7 \ (6)$
Category:Gaussian Integral
Integral of Increasing Function Composed with Measurable Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a $\sigma$-finite measure space.
Let $f: X \to \R_{\ge 0}$ be a positive measurable function.
Let $\phi: \R_{\ge 0} \to \R_{\ge 0}$ be a continuously differentiable, increasing function such that $\map \phi 0 = 0$.
Then:
:$\ds \int \phi \circ f \rd \mu = \int_0^\infty \map {\phi'} t \map F t \rd t$
where:
:$F$ is the survival function of $f$
:$\ds \int_0^\infty$ denotes an improper integral.
=== Corollary ===
== Proof ==
== Sources ==
* : $13.13$
* : $1.1.1$: The Distribution Function
Category:Survival Functions
Category:Integral of Positive Measurable Function
Category:Integral of Increasing Function Composed with Measurable Function
Integral of Increasing Function Composed with Measurable Function/Corollary

== Corollary to Integral of Increasing Function Composed with Measurable Function ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$ such that $p \ge 1$.
Let $f: X \to \R$ be a $p$-integrable function.
Then:
:$\ds \norm f_p^p = \int_0^\infty p t^{p - 1} \map F t \rd t$
where $F$ is the survival function of $\size f$.
== Proof ==
== Sources ==
* : $1.1.1$: The Distribution Function
Category:Survival Functions
Category:Integral of Positive Measurable Function
Category:Integral of Increasing Function Composed with Measurable Function
Integral of Integrable Function is Additive

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$ be $\mu$-integrable functions.
Then $f + g$ is $\mu$-integrable, with:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$
=== Corollary 1 ===
=== Corollary 2 ===
=== Complex Function ===
== Proof ==
From Pointwise Sum of Measurable Functions is Measurable:
:$f + g$ is $\Sigma$-measurable.
From Function Measurable iff Positive and Negative Parts Measurable we have that: 
:$\paren {f + g}^+$ and $\paren {f + g}^-$ are $\Sigma$-measurable.
Since $f$ and $g$ are $\mu$-integrable, we have that $f$ and $g$ are also $\Sigma$-measurable.
Then from Function Measurable iff Positive and Negative Parts Measurable we have that:
:$f^+$, $f^-$, $g^+$ and $g^-$ are $\Sigma$-measurable.
We now want to show that: 
:$\ds \int \paren {f + g}^+ \rd \mu < \infty$
and:
:$\ds \int \paren {f + g}^- \rd \mu < \infty$
From Bound for Positive Part of Pointwise Sum of Functions, we have: 
:$\paren {f + g}^+ \le f^+ + g^+$
Then: 
From Bound for Negative Part of Pointwise Sum of Functions, we have: 
:$\paren {f + g}^- \le f^- + g^-$
Then: 
Now, we will show that: 
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$ 
From Difference of Positive and Negative Parts, we have: 
:$f + g = \paren {f + g}^+ - \paren {f + g}^-$
On the other hand, we have: 
That is: 
:$\paren {f^+ + g^+} - \paren {f^- + g^-} = \paren {f + g}^+ - \paren {f + g}^-$
From Integral of Positive Measurable Function is Additive, we have: 
:$f^+ + g^+$ and $f^- + g^-$ are $\mu$-integrable.
From Integral of Integrable Function is Additive: Lemma, we then have: 
:$\ds \int \paren {f^+ + g^+} \rd \mu - \int \paren {f^- + g^-} \rd \mu = \int \paren {f + g}^+ \rd \mu - \int \paren {f + g}^- \rd \mu$ 
From Integral of Positive Measurable Function is Additive, we have: 
:$\ds \int \paren {f^+ + g^+} \rd \mu = \int f^+ \rd \mu + \int g^+ \rd \mu$
and:
:$\ds \int \paren {f^- + g^-} \rd \mu = \int f^- \rd \mu + \int g^- \rd \mu$ 
Hence: 
== Sources ==
* : $10.4 \ \text{(ii)}$
Category:Integral of Integrable Function
Category:Integral of Integrable Function is Additive
Integral of Integrable Function is Additive/Complex Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.
Let $f, g : X \to \C$ be a $\mu$-integrable function.
Then $f + g$ is $\mu$-integrable and:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$ 
== Proof ==
From Addition of Real and Imaginary Parts, we have:
:$\map \Re {f + g} = \map \Re f + \map \Re g$
and:
:$\map \Im {f + g} = \map \Im f + \map \Im g$
Since $f$ is $\mu$-integrable:
:$\map \Re f$ and $\map \Im f$ are $\mu$-integrable.
Similarly since $g$ is $\mu$-integrable:
:$\map \Re g$ and $\map \Im g$ are $\mu$-integrable.
From Integral of Integrable Function is Additive, we have:
:$\map \Re f + \map \Re g = \map \Re {f + g}$ is $\mu$-integrable
with:
:$\ds \int \map \Re {f + g} \rd \mu = \int \map \Re f \rd \mu + \int \map \Re g \rd \mu$ 
Also from Integral of Integrable Function is Additive, we have:
:$\map \Im f + \map \Im g = \map \Im {f + g}$ is $\mu$-integrable
with:
:$\ds \int \map \Im {f + g} \rd \mu = \int \map \Im f \rd \mu + \int \map \Im g \rd \mu$ 
Hence $f + g$ is $\mu$-integrable.
Further, we have:
Category:Integral of Integrable Function is Additive
Integral of Integrable Function is Additive/Corollary 1

== Corollary ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Suppose that the pointwise sum $f + g$ is well-defined.
Then $f + g$ is $\mu$-integrable, with:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$
== Proof ==
Define: 
:$A = \set {x \in X : \map f x \in \R}$
and:
:$B = \set {x \in X : \map g x \in \R}$ 
From Set of Points for which Measurable Function is Real-Valued is Measurable, we have that: 
:$A$ and $B$ are $\Sigma$-measurable. 
From $\sigma$-Algebra Closed under Countable Intersection, we have that: 
:$A \cap B$ is $\Sigma$-measurable.
From Integral of Integrable Function over Measurable Set is Well-Defined, we have that: 
:$f \cdot \chi_{A \cap B}$ and $g \cdot \chi_{A \cap B}$ are $\mu$-integrable.
Since whenever $x \in X$ is such that $\size {\map f x} = +\infty$ or $\size {\map g x} = +\infty$, we have $\map {\chi_{A \cap B} } x = 0$, these are both functions $X \to \R$. 
So, we can apply Integral of Integrable Function is Additive to obtain: 
:$f \cdot \chi_{A \cap B} + g \cdot \chi_{A \cap B}$ is $\mu$-integrable.
and
:$\ds \int \paren {f \cdot \chi_{A \cap B} + g \cdot \chi_{A \cap B} } \rd \mu = \int f \cdot \chi_{A \cap B} \rd \mu + \int g \cdot \chi_{A \cap B} \rd \mu$
Rewrite: 
:$f \cdot \chi_{A \cap B} + g \cdot \chi_{A \cap B} = \paren {f + g} \cdot \chi_{A \cap B}$
From Integrable Function is A.E. Real-Valued, we have that: 
:$X \setminus A$ and $X \setminus B$ are $\mu$-null sets.
Then by Null Sets Closed under Countable Union, we have that: 
:$\paren {X \setminus A} \cup \paren {X \setminus B}$ is $\mu$-null.
So, from Complement of Intersection, we have: 
:$X \setminus \paren {A \cap B}$ is $\mu$-null.
So if $x \in X$ has: 
:$\map {\paren {f + g} } x \map {\chi_{A \cap B} } x \ne \map {\paren {f + g} } x$
we have: 
:$\map {\chi_{A \cap B} } x = 0$
and:
:$x \in X \setminus \paren {A \cap B}$, which is $\mu$-null.
So: 
:$\paren {f + g} \cdot \chi_{A \cap B} = f + g$ $\mu$-almost everywhere.
So from A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have that: 
:$f + g$ is $\mu$-integrable
and:
:$\ds \int \paren {f + g} \cdot \chi_{A \cap B} \rd \mu = \int \paren {f + g} \rd \mu$
Similarly we have: 
:$f \cdot \chi_{A \cap B} = f$ $\mu$-almost everywhere.
So from A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have: 
:$\ds \int f \cdot \chi_{A \cap B} \rd \mu = \int f \rd \mu$
Since we also have:
:$g \cdot \chi_{A \cap B} = g$ $\mu$-almost everywhere
we also get: 
:$\ds \int g \cdot \chi_{A \cap B} \rd \mu = \int g \rd \mu$
Putting it all together we get: 
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$
Category:Integral of Integrable Function is Additive
Integral of Integrable Function is Additive/Corollary 2

== Corollary ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Suppose that the pointwise difference $f - g$ is well-defined.
Then $f - g$ is $\mu$-integrable, with:
:$\ds \int \paren {f - g} \rd \mu = \int f \rd \mu - \int g \rd \mu$
== Proof ==
From Integral of Integrable Function is Homogeneous, we have that: 
:$-g$ is $\mu$-integrable.
From , we therefore have:
:$f - g$ is $\mu$-integrable
with:
:$\ds \int \paren {f - g} \rd \mu = \int f \rd \mu + \int \paren {-g} \rd \mu$
Then, from Integral of Integrable Function is Homogeneous, we have: 
:$\ds \int \paren {-g} \rd \mu = -\int g \rd \mu$
so:
:$\ds \int \paren {f - g} \rd \mu = \int f \rd \mu - \int g \rd \mu$
Category:Integral of Integrable Function is Additive
Integral of Integrable Function is Additive/Lemma

== Lemma ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $f_1, f_2, g_1, g_2 : X \to \R$ be positive $\mu$-integrable functions with: 
:$f_1 - f_2 = g_1 - g_2$
Then: 
:$\ds \int f_1 \rd \mu - \int f_2 \rd \mu = \int g_1 \rd \mu - \int g_2 \rd \mu$
== Proof ==
We have that: 
:$f_1 + g_2 = g_1 + f_2$
From Pointwise Sum of Measurable Functions is Measurable, we have that: 
:$f_1 + g_2$ and $g_1 + f_2$ are $\Sigma$-measurable.
So we have: 
:$\ds \int \paren {f_1 + g_2} \rd \mu = \int \paren {g_1 + f_2} \rd \mu$
So:
:$\ds \int f_1 \rd \mu + \int g_2 \rd \mu = \int g_1 \rd \mu + \int f_2 \rd \mu$
from Integral of Positive Measurable Function is Additive.
Since $f_1, f_2, g_1, g_2$ are all $\mu$-integrable, each term is finite, and so: 
:$\ds \int f_1 \rd \mu - \int f_2 \rd \mu = \int g_1 \rd \mu - \int g_2 \rd \mu$
Category:Integral of Integrable Function is Additive
Integral of Integrable Function is Homogeneous

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Let $\lambda \in \R$.
Let $\lambda f$ be the pointwise $\lambda$-multiple of $f$.
Then $\lambda f$ is $\mu$-integrable, and:
:$\ds \int \lambda f \rd \mu = \lambda \int f \rd \mu$
=== Complex Function ===
== Proof ==
First suppose that $\lambda \ge 0$. 
From Positive Part of Multiple of Function, we have: 
:$\paren {\lambda f}^+ = \lambda f^+$
From Negative Part of Multiple of Function, we have:
:$\paren {\lambda f}^- = \lambda f^-$
From Function Measurable iff Positive and Negative Parts Measurable, we have: 
:$f^-$ and $f^+$ are $\Sigma$-measurable.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\lambda f^-$, $\lambda f^+$ and $\lambda f$ are $\Sigma$-measurable. 
Then, we have: 
Similarly: 
So: 
:$\lambda f$ is $\mu$-integrable. 
We then have: 
Now suppose that $\lambda < 0$. 
From Positive Part of Multiple of Function, we have: 
:$\paren {\lambda f}^+ = -\lambda f^-$
From Negative Part of Multiple of Function, we have: 
:$\paren {\lambda f}^- = -\lambda f^+$ 
From Function Measurable iff Positive and Negative Parts Measurable, we have: 
:$f^-$ and $f^+$ are $\Sigma$-measurable.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\lambda f^-$, $\lambda f^+$ and $\lambda f$ are $\Sigma$-measurable.
Then we have: 
Similarly:
So: 
:$\lambda f$ is $\mu$-integrable. 
We then have: 
== Sources ==
* : $10.4 \ \text{(i)}$
Category:Integral of Integrable Function
Integral of Integrable Function is Homogeneous/Complex Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.
Let $f : X \to \C$ be a $\mu$-integrable function.
Let $\lambda \in \C$. 
Then $\lambda f$ is $\mu$-integrable and:
:$\ds \int \lambda f \rd \mu = \lambda \int f \rd \mu$ 
== Proof ==
We have:
From Real Part of Complex Product, we have:
:$\map \Re {\lambda f} = \map \Re \lambda \map \Re f - \map \Im \lambda \map \Im f$
From Imaginary Part of Complex Product, we have:
:$\map \Im {\lambda f} = \map \Im \lambda \map \Re f + \map \Re \lambda \map \Im f$
Since $f$ is $\mu$-integrable, $\map \Re f$ and $\map \Im f$ are $\mu$-integrable by definition.
From Integral of Integrable Function is Additive and Integral of Integrable Function is Homogeneous, $\map \Re {\lambda f}$ and $\map \Im {\lambda f}$ are both $\mu$-integrable and we have:
:$\ds \int \map \Re {\lambda f} \rd \mu = \map \Re \lambda \int \map \Re f \rd \mu - \map \Im \lambda \int \map \Im f \rd \mu$
and:
:$\ds \int \map \Im {\lambda f} \rd \mu = \map \Im \lambda \int \map \Re f \rd \mu + \map \Re \lambda \int \map \Im f \rd \mu$
Then $\lambda f$ is $\mu$-integrable.
We then have:
Category:Integral of Integrable Function is Homogeneous
Integral of Integrable Function is Monotone

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Suppose that $f \le g$ pointwise.
Then:
:$\ds \int f \rd \mu \le \int g \rd \mu$
== Proof ==
Since: 
:$f \le g$
we have that $g - f$ is well-defined with: 
:$g - f \ge 0$
From Integral of Integrable Function is Additive: Corollary 2, we have: 
:$g - f$ is $\mu$-integrable
with:
:$\ds \int \paren {g - f} \rd \mu = \int g \rd \mu - \int f \rd \mu$
Since: 
:$\ds \int \paren {g - f} \rd \mu \ge 0$
we have:
:$\ds \int g \rd \mu - \int f \rd \mu \ge 0$
This gives:
:$\ds \int g \rd \mu \ge \int f \rd \mu$
== Sources ==
* : $10.4 \ \text{(iv)}$
Category:Integral of Integrable Function
Integral of Integrable Function over Measurable Set is Well-Defined

== Theorem ==
 
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $E \in \Sigma$. 
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Then the $\mu$-integral of $f$ over $E$:
:$\ds \int_E f \rd \mu = \int f \cdot \chi_E \rd \mu$
is well-defined.
== Proof ==
We need to show that $f \cdot \chi_E$ is $\mu$-integrable, so that its $\mu$-integral is well-understood.
Since $f$ is $\mu$-integrable, it is certainly $\Sigma$-measurable.
From Characteristic Function Measurable iff Set Measurable, we have that: 
:$\chi_E$ is $\Sigma$-measurable.
Then by Pointwise Product of Measurable Functions is Measurable, we have: 
:$f \cdot \chi_E$ is $\Sigma$-measurable.
We now want to verify that: 
:$\ds \int \paren {f \cdot \chi_E}^+ \rd \mu < \infty$
and:
:$\ds \int \paren {f \cdot \chi_E}^- \rd \mu < \infty$
Note that from the definition of a characteristic function:
:$\map {\chi_E} x \in \set {0, 1}$
so, for $x \in X$:
:$\map {f \cdot \chi_E} x \ge 0$ $\map {\chi_E} x = 0$ or $\map {\chi_E} x = 1$ and $\map f x \ge 0$
and:
:$\map {f \cdot \chi_E} x \le 0$ $\map {\chi_E} x = 0$ or $\map {\chi_E} x = 1$ and $\map f x \le 0$
So, we have: 
:$\paren {f \cdot \chi_E}^+ = f^+ \cdot \chi_E$
and:
:$\paren {f \cdot \chi_E}^- = f^- \cdot \chi_E$
We can see that: 
:$f^+ \cdot \chi_E \le f^+$
and:
:$f^- \cdot \chi_E \le f^-$
Since $f$ is $\mu$-integrable, we have: 
:$\ds \int f^+ \rd \mu < \infty$
and:
:$\ds \int f^- \rd \mu < \infty$
Then, from Integral of Positive Measurable Function is Monotone, we have: 
:$\ds \int f^+ \cdot \chi_E \rd \mu \le \int f^+ \rd \mu < \infty$
and:
:$\ds \int f^- \cdot \chi_E \rd \mu \le \int f^- \rd \mu < \infty$
This gives:
:$\ds \int \paren {f \cdot \chi_E}^+ \rd \mu < \infty$
and:
:$\ds \int \paren {f \cdot \chi_E}^- \rd \mu < \infty$
as required.
Category:Integral of Integrable Function over Measurable Set
Integral of Integrable Function over Null Set

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Let $N$ be a $\mu$-null set.
Then:
:$\ds \int_N f \rd \mu = 0$
where $\ds \int_N$ signifies an integral over $N$.
== Proof ==
We have, from definition: 
:$\ds \int_N f \rd \mu = \int f \cdot \chi_N \rd \mu$ 
Note that if $x \in X \setminus N$, we have: 
:$\map {\chi_N} x = 0$
So, if: 
:$\map f x \map {\chi_N} x \ne 0$
we must have $x \in N$. 
Since $N$ is a null set, this gives:
:$f \cdot \chi_N = 0$ $\mu$-almost everywhere. 
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we therefore have: 
:$\ds \int \size {f \cdot \chi_N} \rd \mu = 0$ 
Then, from Triangle Inequality for Integrals: Corollary, we have:
:$\ds \int f \cdot \chi_N \rd \mu = 0$
That is: 
:$\ds \int_N f \rd \mu = 0$
== Sources ==
* : $10.9 \, \text{(ii)}$
Category:Integral of Integrable Function over Measurable Set
Integral of Laplace Transform

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le A$.
Let $\laptrans f = F$ denote the Laplace transform of $f$.
Then:
:$\ds \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$
wherever $\ds \lim_{t \mathop \to 0} \dfrac {\map f t} t$ and $\laptrans f$ exist.
== Proof ==
Let $\map g t := \dfrac {\map f t} t$.
Then:
The result follows.
== Examples ==
== Also see ==
* Laplace Transform of Integral
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $8$. Division by $t$: Theorem $1 \text{-} 13$
* : Chapter $1$: The Laplace Transform: Solved Problems: Division by $t$: $21$
* : $\S 32$: Table of General Properties of Laplace Transforms: $32.16$
Category:Laplace Transforms