avewright/proofwiki-math huggingface dataset:
Theorem: Decomposition of Complex Measure into Finite Signed Measures

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Then there exists unique finite signed measures $\mu_R$ and $\mu_I$ such that: 
:$\mu = \mu_R + i \mu_I$

Proof:
=== Existence ===
For each $A \in \Sigma$ define the function $\mu_R : X \to \R$ by: 
:$\map {\mu_R} A = \map \Re {\map \mu A}$
Similarly, for each $A \in \Sigma$ define the function $\mu_I : X \to \R$ by: 
:$\map {\mu_I} A = \map \Im {\map \mu A}$
Clearly we have: 
{{begin-eqn}}
{{eqn	| l = \map \mu A
		| r = \map \Re {\map \mu A} + i \map \Im {\map \mu A}
		| c = {{Defof|Real Part}}, {{Defof|Imaginary Part}}
}}
{{eqn	| r = \map {\mu_R} A + i \map {\mu_I} A
}}
{{end-eqn}}
for each $A \in \Sigma$.
So:
:$\mu = \mu_R + \mu_I$
It remains to show that $\mu_R$ and $\mu_I$ are finite signed measures.
We verify both of the conditions for a signed measure in turn. 
We have: 
:$\map \mu \O = 0$
so:
:$\map {\mu_R} A = \map \Re 0 = 0$
and:
:$\map {\mu_I} A = \map \Im 0 = 0$
verifying $(1)$ for $\mu_R$ and $\mu_I$. 
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets. 
Then: 
{{begin-eqn}}
{{eqn	| l = \map {\mu_R} {\bigcup_{n \mathop = 1}^\infty D_n}
		| r = \map \Re {\map \mu {\bigcup_{n \mathop = 1}^\infty D_n} } 
}}
{{eqn	| r = \map \Re {\sum_{n \mathop = 1}^\infty \map \mu {D_n} }
		| c = since $\mu$ is countably additive
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map \Re {\map \mu {D_n} } 
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map {\mu_R} {D_n}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \map {\mu_I} {\bigcup_{n \mathop = 1}^\infty D_n}
		| r = \map \Im {\map \mu {\bigcup_{n \mathop = 1}^\infty D_n} }
}}
{{eqn	| r = \map \Im {\sum_{n \mathop = 1}^\infty \map \mu {D_n} }
		| c = since $\mu$ is countably additive
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map \Im {\map \mu {D_n} }
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map {\mu_I} {D_n}
}}
{{end-eqn}}
verifying $(2)$ for $\mu_R$ and $\mu_I$.
So $\mu_R$ and $\mu_I$ are signed measures. 
Since $\mu$ is a complex-valued function, we have: 
:$\cmod {\map \mu X} < \infty$
We then have, from Modulus Larger than Real Part: 
:$\size {\map \Re {\map \mu X} } < \infty$ 
so:
:$\size {\map {\mu_R} X} < \infty$
Similarly, from Modulus Larger than Imaginary Part:
:$\size {\map \Im {\map \mu X} } < \infty$
so:
:$\size {\map {\mu_I} X} < \infty$
So:
:$\mu_R$ and $\mu_I$ are finite.
{{qed|lemma}}
Theorem: Measurable Function Zero A.E. iff Absolute Value has Zero Integral/Corollary

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a non-negative integrable function.
Let $A, B \in \Sigma$ have $A \subseteq B$. 
Then: 
:$\ds \int_A f \rd \mu = \int_B f \rd \mu$
{{iff}}: 
:$f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.

Proof:
We can write: 
:$B = A \cup \paren {B \setminus A}$
From Integral of Positive Measurable Function over Disjoint Union, we have: 
:$\ds \int_B f \rd \mu = \int_A f \rd \mu + \int_{B \setminus A} f \rd \mu$
Since: 
:$\ds \int_B f \rd \mu = \int_A f \rd \mu$
we get: 
:$\ds \int_{B \setminus A} f \rd \mu = 0$
From the definition of the $\mu$-integral over $A$, we have: 
:$\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu$
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have: 
:$\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu = 0$ {{iff}} $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere. 
So:
:$\ds \int_A f \rd \mu = \int_B f \rd \mu$
{{iff}}:
:$f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere. 
{{qed}}
Category:Measurable Function Zero A.E. iff Absolute Value has Zero Integral
3nq35tgu7yh5s0le5325k2q67567owl
Theorem: Jordan Decomposition of Finite Signed Measure

Let $\struct {X, \Sigma}$ be measurable.
Let $\mu$ be a finite signed measure on $\struct {X, \Sigma}$. 
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then $\mu^+$ and $\mu^-$ are finite measures.

Proof:
From the definition of Jordan decomposition, we have: 
:$\mu = \mu^+ - \mu^-$
with at least one of $\mu^+$ and $\mu^-$ finite.
From the definition of a finite signed measure, we have: 
:$\cmod {\map \mu X} < \infty$
We show that: 
:if exactly one of $\mu^+$ and $\mu^-$ is finite, then $\mu$ is not a finite signed measure
that is: 
:if exactly one of $\mu^+$ and $\mu^-$ is finite, then $\map \mu X = \infty$ or $\map \mu X = -\infty$.
At which point we will have a contradiction, giving:
:$\mu^+$ and $\mu^-$ are both finite.
{{AimForCont}} suppose exactly one of $\mu^+$ and $\mu^-$ is finite.
Consider first the case of $\mu^+$ non-finite.
Then $\mu^-$ is finite, so:
:$\map {\mu^-} X < \infty$
Since $\mu^+$ is not finite, we have: 
:$\map {\mu^+} X = \infty$
From the definition of extended real subtraction, we then have: 
:$\map \mu X = \map {\mu^+} X - \map {\mu^-} X = \infty$
which contradicts: 
:$\cmod {\map \mu X} < \infty$
So we have a contradiction in the case of $\mu^+$ non-finite.
Now suppose that $\mu^-$ is non-finite.
Then $\mu^+$ is finite, so:
:$\map {\mu^+} X < \infty$
Since $\mu^-$ is not finite, we have: 
:$\map {\mu^-} X = \infty$
From the definition of extended real subtraction, we then have: 
:$\map \mu X = \map {\mu^+} X - \map {\mu^-} X = -\infty$
which contradicts: 
:$\cmod {\map \mu X} < \infty$
So we have a contradiction in the case of $\mu^-$ non-finite.
So we must have that both $\mu^+$ and $\mu^-$ are finite, completing the proof.
{{qed}}
Category:Finite Signed Measures
8kbspkv5ccniolol3raslshlb8rrr8g
Theorem: Integral of Bounded Measurable Function with respect to Finite Signed Measure is Well-Defined

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: 
{{begin-eqn}}
{{eqn	| l = \int \size f \rd \mu^+
		| o = \le
		| r = \int M \rd \mu^+
		| c = Measure is Monotone
}}
{{eqn	| r = M \int \rd \mu^+
		| c = Integral of Positive Measurable Function is Positive Homogeneous
}}
{{eqn	| r = M \map {\mu^+} X
		| c = Integral of Characteristic Function: Corollary
}}
{{eqn	| o = <
		| r = \infty
}}
{{end-eqn}}
so:
{{begin-eqn}}
{{eqn	| l = \int \size f \rd \mu^-
		| o = \le
		| r = \int M \rd \mu^-
		| c = Measure is Monotone
}}
{{eqn	| r = M \int \rd \mu^-
		| c = Integral of Positive Measurable Function is Positive Homogeneous
}}
{{eqn	| r = M \map {\mu^-} X
		| c = Integral of Characteristic Function: Corollary
}}
{{eqn	| o = <
		| r = \infty
}}
{{end-eqn}}
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.
{{qed}}
Category:Integral of Bounded Measurable Function with respect to Finite Signed Measure
lzuo2hce2cdmums7lrgl5rvby3sc8jq
Theorem: Positive Part of Vertical Section of Function is Vertical Section of Positive Part

Let $X$ and $Y$ be sets. 
Let $f : X \times Y \to \overline \R$ be a function.
Let $x \in X$. 
Then: 
:$\paren {f_x}^+ = \paren {f^+}_x$
where:
:$f_x$ denotes the $x$-vertical function of $f$
:$f^+$ denotes the positive part of $f$.

Proof:
Fix $x \in X$.
Then, we have, for each $y \in Y$:
{{begin-eqn}}
{{eqn	| l = \map {\paren {f^+}_x} y
		| r = \map {f^+} {x, y}
}}
{{eqn	| r = \max \set {0, \map f {x, y} } 
		| c = {{Defof|Positive Part}}
}}
{{eqn	| r = \max \set {0, \map {f_x} y}
		| c = {{Defof|Vertical Section of Function}}
}}
{{eqn	| r = \map {\paren {f_x}^+} y
		| c = {{Defof|Positive Part}}
}}
{{end-eqn}}
{{qed}}
Category:Vertical Section of Functions
Category:Positive Parts
lwt5i9veib2cxj3328hpyth4fkfhf1o
Theorem: Negative Part of Vertical Section of Function is Vertical Section of Negative Part

Let $X$ and $Y$ be sets. 
Let $f : X \times Y \to \overline \R$ be a function.
Let $x \in X$. 
Then: 
:$\paren {f_x}^- = \paren {f^-}_x$
where:
:$f_x$ denotes the $x$-vertical function of $f$
:$f^-$ denotes the negative part of $f$.

Proof:
Fix $x \in X$.
Then, we have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {f^-}_x} y
		| r = \map {f^-} {x, y}
}}
{{eqn	| r = -\min \set {0, \map f {x, y} } 
		| c = {{Defof|Negative Part}}
}}
{{eqn	| r = -\min \set {0, \map {f_x} y}
		| c = {{Defof|Vertical Section of Function}}
}}
{{eqn	| r = \map {\paren {f_x}^-} y
		| c = {{Defof|Negative Part}}
}}
{{end-eqn}}
{{qed}}
Category:Vertical Section of Functions
Category:Negative Parts
qofzj2aditq2wwqmngb23f86urtvign
Theorem: Positive Part of Horizontal Section of Function is Horizontal Section of Positive Part

Let $X$ and $Y$ be sets. 
Let $f : X \times Y \to \overline \R$ be a function.
Let $y \in Y$. 
Then: 
:$\paren {f^y}^+ = \paren {f^+}^y$
where:
:$f^y$ denotes the $y$-horizontal function of $f$
:$f^+$ denotes the positive part of $f$.

Proof:
Fix $y \in Y$.
Then, we have, for each $x \in X$:
{{begin-eqn}}
{{eqn	| l = \map {\paren {f^+}^y} x
		| r = \map {f^+} {x, y}
}}
{{eqn	| r = \max \set {0, \map f {x, y} } 
		| c = {{Defof|Positive Part}}
}}
{{eqn	| r = \max \set {0, \map {f^y} x}
		| c = {{Defof|Horizontal Section of Function}}
}}
{{eqn	| r = \map {\paren {f^y}^+} x
		| c = {{Defof|Positive Part}}
}}
{{end-eqn}}
{{qed}}
Category:Horizontal Section of Functions
Category:Positive Parts
9bpirq5i896ix412jujc68olprk1ugh
Theorem: Negative Part of Horizontal Section of Function is Horizontal Section of Negative Part

Let $X$ and $Y$ be sets. 
Let $f : X \times Y \to \overline \R$ be a function.
Let $y \in Y$. 
Then: 
:$\paren {f^y}^- = \paren {f^-}^y$
where:
:$f^y$ denotes the $y$-horizontal function of $f$
:$f^-$ denotes the negative part of $f$.

Proof:
Fix $y \in Y$.
Then, we have, for each $x \in X$:
{{begin-eqn}}
{{eqn	| l = \map {\paren {f^-}^y} x
		| r = \map {f^-} {x, y}
}}
{{eqn	| r = -\min \set {0, \map f {x, y} } 
		| c = {{Defof|Negative Part}}
}}
{{eqn	| r = -\min \set {0, \map {f^y} x}
		| c = {{Defof|Horizontal Section of Function}}
}}
{{eqn	| r = \map {\paren {f^y}^-} x
		| c = {{Defof|Negative Part}}
}}
{{end-eqn}}
{{qed}}
Category:Horizontal Section of Functions
Category:Negative Parts
i074z6no34l8ybh8h3t5j59eqdmu6w8
Theorem: Characteristic Function of Null Set is A.E. Equal to Zero

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N$ be a $\mu$-null set. 
Then: 
:$\chi_N = 0$ $\mu$-almost everywhere.
where $\chi_N$ is the characteristic function of $N$.

Proof:
Let $x \in X$ be such that: 
:$\map {\chi_N} x \ne 0$
Then. since $\map {\chi_N} x \in \set {0, 1}$: 
:$\map {\chi_N} x = 1$
which is equivalent to: 
:$x \in N$ 
from the definition of a characteristic function.
So:
:if $x \in X$ is such that $\map {\chi_N} x \ne 0$, then $x \in N$.
Since $N$ is a $\mu$-null set, we have: 
:$\chi_N = 0$ $\mu$-almost everywhere.
{{qed}}
Category:Characteristic Functions
Category:Measure Theory
Category:Characteristic Function of Null Set is A.E. Equal to Zero
a04ilba9fx2eb8wpr547c9px5vngoso
Theorem: Pointwise Addition preserves A.E. Equality

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g, F, G : X \to \overline \R$ be functions with: 
:$f = F$ $\mu$-almost everywhere
and:
:$g = G$ $\mu$-almost everywhere
and the pointwise sums $f + g$ and $F + G$ well-defined.
Then: 
:$f + g = F + G$ $\mu$-almost everywhere.

Proof:
Since:
:$f = F$ $\mu$-almost everywhere
there exists a $\mu$-null set $N_1 \subseteq X$ such that: 
:if $x \in X$ has $\map f x \ne \map F x$ then $x \in N_1$.
Since: 
:$g = G$ $\mu$-almost everywhere
there exists a $\mu$-null set $N_2 \subseteq X$ such that: 
:if $x \in X$ has $\map G x \ne \map G x$ then $x \in N_2$ 
Note that if $x \in X$ is such that: 
:$\map f x = \map F x$
and:
:$\map g x = \map G x$ 
then: 
:$\map f x + \map g x = \map F x + \map G x$
whenever this sum is well-defined.
By the Rule of Transposition, we therefore have: 
:if $x \in X$ has $\map f x + \map g x \ne \map F x + \map G x$ then $\map f x \ne \map F x$ or $\map g x \ne \map G x$.
That is, whenever $x \in X$ has $\map f x + \map g x \ne \map F x + \map G x$, we have: 
:$x \in N_1 \cup N_2$
From Null Sets Closed under Countable Union, we have: 
:$N_1 \cup N_2$ is $\mu$-null.
So:
:$f + g = F + G$ $\mu$-almost everywhere.
{{qed}}
Category:Almost-Everywhere Equality Relation
sz3sowr5smr31kaa7mlwr3gjpfjf16k
Theorem: Lebesgue 1-Space is Subset of Tempered Distribution Space

Let $\map {L^1} \R$ be the Lebesgue $1$-space.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then in the distributional sense:
:$\map {L^1} \R \subseteq \map {\SS'} \R$
That is:
:$T_f \subseteq \map {\SS'} \R$
where $f \in \map {L^1} \R$.

Proof:
Let $f \in \map {L^1} \R$.
By definition of the Lebesgue space:
:$\ds \norm f_1 = \int_\R \size {\map f x} \rd x < \infty$
where $\norm {\, \cdot \,}_1$ denotes the 1-seminorm.
Let $\phi \in \map \SS \R$ be a Schwartz test function, where $\map \SS \R$ is the Schwartz space.
Let $T_f : \map \SS \R \to \R$ be a functional such that:
:$\ds \map {T_f} \phi = \int_\R \map f x \map \phi x \rd x$
By Integral Operator is Linear, $\map {T_f} \phi$ is linear {{WRT}} both $f$ and $\phi$.
Furthermore:
{{begin-eqn}}
{{eqn | l = \size {\map {T_f} \phi}
      | r = \size {\int_\R \map f x \map \phi x \rd x}
}}
{{eqn | o = \le
      | r = \int_\R \size {\map f x} \size {\map \phi x} \rd x
}}
{{eqn | o = \le
      | r = \sup_{x \mathop \in \R} \size {\map \phi x} \int_\R \size {\map f x} \rd x
      | c = {{Defof|Supremum of Real-Valued Function}}
}}
{{eqn | r = \norm f_1 \norm \phi_\infty
      | c = {{Defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions }}
}}
{{end-eqn}}
Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence in the Schwartz space.
Suppose $\sequence {\phi_n}_{n \mathop \in \N}$ converges in $\map \SS \R$:
:$\phi_n \stackrel \SS {\longrightarrow} \mathbf 0$
where $\mathbf 0$ denotes the zero Schwartz test function.
That is:
:$\ds \forall l, m \in \N : \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^l \map {\phi_n^{\paren m} } x} = 0$
Set $m = 0$ and $l = 0$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \norm {\phi_n}_\infty  
      | r = \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {\map {\phi_n} x}
}}
{{eqn | r = 0
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \size {\map {T_f} {\phi_n} }
      | o = \le
      | r = \lim_{n \mathop \to \infty} \norm f_1 \norm {\phi_n}_\infty
}}
{{eqn | r = 0
}}
{{end-eqn}}
Thus:
:$\ds \lim_{n \mathop \to \infty} \map {T_f} {\phi_n} = 0$
On the other hand:
:$\map {T_f} {\mathbf 0} = 0$
Hence, from the definition of the tempered distribution it follows that:
:$T_f \in \map {\SS'} \R$
{{qed}}
Theorem: Characteristic Function of Null Set is A.E. Equal to Zero/Corollary

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N$ be a $\mu$-null set. 
Then: 
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.
where $\chi_{X \setminus N}$ is the characteristic function of $X \setminus N$.

Proof:
From Characteristic Function of Set Difference, we have: 
:$\chi_{X \setminus N} = \chi_X - \chi_{X \cap N}$
From Intersection with Subset is Subset, we therefore have: 
:$\map {\chi_{X \setminus N} } x = 1 - \map {\chi_N} x$
for each $x \in X$.
From Characteristic Function of Null Set is A.E. Equal to Zero, we have: 
:$\chi_N = 0$ $\mu$-almost everywhere. 
So we have, from Pointwise Addition preserves A.E. Equality:
:$1 - \chi_N = 1$ $\mu$-almost everywhere.
So:
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.
{{qed}}
Category:Characteristic Function of Null Set is A.E. Equal to Zero
12378wgqmtcf0y4nfz2ipxv37jm0spf
Theorem: Pointwise Multiplication preserves A.E. Equality

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g, h : X \to \overline \R$ be functions with: 
:$f = g$ $\mu$-almost everywhere.
Then: 
:$f \times h = g \times h$ $\mu$-almost everywhere
where $f \times h$ and $g \times h$ are the pointwise products of $f$ and $h$, and $g$ and $h$ respectively.

Proof:
Since:
:$f = g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N \subseteq X$ such that: 
:if $x \in X$ has $\map f x \ne \map g x$ then $x \in N$.
Note that if $x \in X$ is such that: 
:$\map f x = \map g x$
then: 
:$\map f x \map h x = \map g x \map h x$
By the Rule of Transposition, we therefore have: 
:if $x \in X$ has $\map f x \map h x \ne \map g x \map h x$ then $\map f x \ne \map g x$.
So: 
:if $x \in X$ has $\map f x \map h x \ne \map g x \map h x$ then $x \in N$. 
Since $N$ is $\mu$-null, we therefore have: 
:$f \times h = g \times h$ $\mu$-almost everywhere.
{{qed}}
Category:Almost-Everywhere Equality Relation
le68idc91h9ksksuvpzkcd43oqio8h2
Theorem: Restriction of Measurable Function is Measurable on Trace Sigma-Algebra

Let $\struct {X, \Sigma}$ be a measurable space. 
Let $f : X \to \overline \R$ be a $\Sigma$-measurable functions. 
Let $E \in \Sigma$. 
Let $\Sigma_E$ be the trace $\sigma$-algebra of $E$ in $\Sigma$. 
Then the restriction $f \restriction_E$ is $\Sigma_E$-measurable.

Proof:
From the definition of a $\Sigma_E$-measurable function, we aim to show that: 
:$\set {x \in E : \map f x \le \alpha} \in \Sigma_E$
for each $\alpha \in \R$.
Let $\alpha \in \R$. 
We have: 
:$\set {x \in E : \map f x \le \alpha} = \set {x \in X : \map f x \le \alpha} \cap E$
Since $f$ is $\Sigma$-measurable, we have: 
:$\set {x \in X : \map f x \le \alpha} \in E$
So, from the definition of trace $\sigma$-algebra, we have: 
:$\set {x \in X : \map f x \le \alpha} \cap E \in \Sigma_E$
So we have: 
:$\set {x \in E : \map f x \le \alpha} \in \Sigma_E$
for each $\alpha \in \R$.
So:
:$f \restriction_E$ is $\Sigma_E$-measurable.
{{qed}}
Category:Measurable Functions
Category:Trace Sigma-Algebras
3imsynwrjy8zlmrm0ny7as0yymwmsmy
Theorem: Variation of Complex Measure is Finite Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Let $\cmod \mu$ be the variation of $\mu$. 
Then $\cmod \mu$ is a finite measure on $\struct {X, \Sigma}$.

Proof:
We first show that $\map {\cmod \mu} A \ge 0$ for each $A \in \Sigma$. 
Let $A \in \Sigma$.
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
Then, for each $A \in \Sigma$, we have: 
:$\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
We clearly have: 
:$\set A \in \map P A$ 
so:
:$\ds \size {\map \mu A} \in \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
From the definition of supremum, we therefore have: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} \ge \size {\map \mu A} \ge 0$
giving:
:$\map {\cmod \mu} A \ge 0$
We will now verify the three conditions in Characterization of Measures.
Theorem: Measures in Jordan Decomposition of Complex Measure are Finite

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$.
Let $\tuple {\mu_1, \mu_2, \mu_3, \mu_4}$ be the Jordan decomposition of $\mu$. 
Then:
:$\mu_1$, $\mu_2$, $\mu_3$ and $\mu_4$ are finite.

Proof:
Let $\mu_R$ be the real part of $\mu$.
Let $\mu_I$ be the imaginary part of $\mu$.
Then:
:$\tuple {\mu_1, \mu_2}$ is the Jordan decomposition of $\mu_R$
and:
:$\tuple {\mu_3, \mu_4}$ is the Jordan decomposition of $\mu_I$.
From the definition of the real part and imaginary part, we have that both $\mu_R$ and $\mu_I$ are finite signed measures.
Now applying Jordan Decomposition of Finite Signed Measure to $\mu_R$, we have: 
:$\mu_1$ and $\mu_2$ are finite.
Applying Jordan Decomposition of Finite Signed Measure to $\mu_I$, we have: 
:$\mu_4$ and $\mu_4$ are finite
hence the result.
{{qed}}
Category:Complex Measures
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Theorem: Absolute Continuity of Signed Measure in terms of Jordan Decomposition

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {\nu^+, \nu^-}$ be the Jordan decomposition of $\nu$.
Then $\nu$ is absolutely continuous with respect to $\mu$ {{iff}}:
:$\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.

Proof:
We have that $\nu$ is absolutely continuous with respect to $\mu$ {{iff}}:
:$\size \nu$ is absolutely continuous with respect to $\mu$
where $\size \nu$ is the variation of $\nu$.
From the definition of variation, we have: 
:$\size \nu = \nu^+ + \nu^-$ 
Suppose that $\nu$ is absolutely continuous with respect to $\mu$. 
Then: 
:$\nu^+ + \nu^-$ absolutely continuous with respect to $\mu$.
That is:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A + \map {\nu^-} A = 0$.
Since $\nu^+ \ge 0$ and $\nu^-$, we have that: 
:$\map {\nu^+} A + \map {\nu^-} A = 0$ implies that $\map {\nu^+} A = 0$ and $\map {\nu^-} A = 0$.
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = 0$
so:
:$\nu^+$ is absolutely continuous with respect to $\mu$.
We also have: 
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^-} A = 0$
so:
:$\nu^-$ is absolutely continuous with respect to $\mu$.
So:
:if $\nu$ is absolutely continuous with respect to $\mu$ then $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
{{qed|lemma}}
Suppose that: 
:$\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
Then: 
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = 0$ and $\map {\nu^-} A = 0$.
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A + \map {\nu^-} A = 0$
giving: 
:$\nu^+ + \nu^-$ is absolutely continuous with respect to $\mu$.
That is: 
:$\size \nu$ is absolutely continuous with respect to $\mu$.
So:
:if $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$ then $\nu$ is absolutely continuous with respect to $\mu$.
{{qed}}
Theorem: Variation of Signed Measure is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\size \mu$ be the variation of $\mu$. 
Then $\size \mu$ is a measure.

Proof:
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$. 
Then: 
:$\size \mu = \mu^+ + \mu^-$
So $\size \mu$ is a measure from Linear Combination of Measures.
{{qed}}
Category:Measures
Category:Variation of Signed Measure
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