nyuuzyou/wikis huggingface dataset:
Lebesgue 1-Space is Subset of Tempered Distribution Space

== Theorem ==
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 both $f$ and $\phi$.
Furthermore:
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:
Therefore:
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:
:$T_f \in \map {\SS'} \R$
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Tempered Distributions
Lebesgue Decomposition Theorem

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$. 
=== Finite Measure ===
=== Complex Measure ===
=== Finite Signed Measure ===
=== $\sigma$-Finite Measure ===
== Uniqueness ==
Category:Lebesgue Decomposition Theorem
Category:Mutually Singular Measures
Category:Absolutely Continuous Measures
Lebesgue Decomposition Theorem/Complex Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$. 
Let $\nu$ be a complex measure on $\struct {X, \Sigma}$. 
Then there exists complex measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that: 
:$(1): \quad \nu_a$ is absolutely continuous with respect to $\mu$
:$(2): \quad \nu_s$ and $\mu$ are mutually singular
:$(3): \quad \nu = \nu_a + \nu_s$
== Proof ==
Let $\cmod \nu$ be the variation of $\nu$.
From Variation of Complex Measure is Finite Measure, $\cmod \nu$ is a finite measure.
Then from Lebesgue Decomposition Theorem for Finite Measures, there exists finite measures $\cmod \nu_a$ and $\cmod \nu_s$ on $\struct {X, \Sigma}$ such that: 
:$(1): \quad \cmod \nu_a$ is absolutely continuous with respect to $\mu$
:$(2): \quad \cmod \nu_s$ and $\mu$ are mutually singular
:$(3): \quad \cmod \nu = \cmod \nu_a + \cmod \nu_s$
More precisely, the proof of this theorem grants that there exists a $\mu$-null set $N \subseteq X$ such that: 
:$\map {\cmod \nu_a} A = \map {\cmod \nu} {N^c \cap A}$
and:
:$\map {\cmod \nu_s} A = \map {\cmod \nu} {N \cap A}$
for each $A \in \Sigma$, with $\cmod \nu_a$ and $\cmod \nu_s$ having the desired properties. 
Let $\nu_a$ be the intersection measure of $\nu$ by $N^c$.
Let $\nu_s$ be the intersection measure of $\nu$ by $N$.
We verify that $\nu_a$ and $\nu_s$ are our desired complex measures. 
From Variation of Intersection Measure, we have:
:$\cmod {\nu_a} = \cmod \nu_a$
and:
:$\cmod {\nu_s} = \cmod \nu_s$ 
Note that $(1)$ is precisely the definition of the complex measure $\nu_s$ being absolutely continuous with respect to $\mu$.
Hence we have $(1)$. 
Note that $(2)$ is precisely the definition of the complex measure $\nu_s$ being mutually singular to $\mu$.
Hence we have $(2)$.
We finally prove $(3)$. 
For each $A \in \Sigma$, $A \cap N$ and $A \cap N^c$ are disjoint with $\paren {A \cap N} \cup \paren {A \cap N^c} = A$, so we have:
:$\map \nu A = \map \nu {A \cap N} + \map \nu {A \cap N^c} = \map {\nu_a} A + \map {\nu_s} A$ 
Hence we have $(3)$. 
== Sources ==
* : $4.3$: Singularity
Category:Lebesgue Decomposition Theorem
Category:Complex Measures
Lebesgue Decomposition Theorem/Finite Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$. 
Let $\nu$ be a finite measure on $\struct {X, \Sigma}$. 
Then there exist finite measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that: 
:$(1): \quad \nu_a$ is absolutely continuous with respect to $\mu$
:$(2): \quad \nu_s$ and $\mu$ are mutually singular
:$(3): \quad \nu = \nu_a + \nu_s$
== Proof ==
Define the set $\NN_\mu$ by: 
:$\NN_\mu = \set {B \in \Sigma : \map \mu B = 0}$
Since $\nu$ is a finite measure, there exists $M \ge 0$ such that: 
:$\map \nu A \le M$
for all $A \in \Sigma$.
So by the Continuum Property:
:the supremum of $\set {\map \nu B : B \in \NN_\mu}$ exists as a real number $L$.
By the definition of the supremum, for each $n \in \N$ there exists $B_n \in \NN_\mu$ such that: 
:$L - \dfrac 1 n \le \map \nu {B_n} \le L$
Then, from the Squeeze Theorem:
:$\ds \lim_{n \mathop \to \infty} \map \nu {B_n} = L$
Now set: 
:$\ds N = \bigcup_{n \mathop = 1}^\infty B_n$
From Null Sets Closed under Countable Union, we have that: 
:$N$ is $\mu$-null.
So:
:$N \in \NN_\mu$ 
Further, from Set is Subset of Union, we have:
:$B_n \subseteq N$ for each $N \in \N$
giving: 
:$\map \nu {B_n} \le \map \nu N \le L$ for each $n \in \N$
from the definition of the supremum and Measure is Monotone. 
Taking $n \to \infty$, we obtain: 
:$\map \nu N = L$ 
by Limits Preserve Inequalities. 
Explicitly, we have found that: 
:$\map \nu N = \sup \set {\map \nu B : B \in \NN_\mu}$
Let $\nu_a$ be the intersection measure of $\nu$ by $N^c$.
Let $\nu_s$ be the intersection measure of $\nu$ by $N$.
Since $\nu$ is finite, so are $\nu_a$ and $\nu_s$ from Intersection Measure of Finite Measure is Finite Measure.
Then, for each $A \in \Sigma$ we have:
so:
:$\nu = \nu_a + \nu_s$ 
verifying $(3)$.
We have: 
and $\map \mu N = 0$.
So $\nu_s$ is concentrated on $N$ and $\mu$ is concentrated on $N^c$.
So $\mu$ and $\nu_s$ are mutually singular, verifying $(2)$.
We finish by showing that $\nu_a$ is absolutely continuous with respect to $\mu$.
To simplify notation, first consider a $\Sigma$-measurable $B \subseteq N^c$ such that $\map \mu B = 0$.
That is, $B \in \NN_\mu$. 
 that $\map \nu B \ne 0$.
In particular, $\map \nu B > 0$. 
Then, we have:
However, from Null Sets Closed under Countable Union, we have: 
:$N \cup B$ is $\mu$-null.
So: 
:$\map \nu {N \cup B} \in \set {\map \nu B : B \in \NN_\mu}$, contradicting that $\map \nu N$ is the supremum of this set.
So we must have $\map \nu B = 0$.
Now suppose that general $B \in \Sigma$ has $\map \mu B = 0$.
Then, from Intersection is Subset, we have: 
:$B \cap N^c \subseteq B$
From Null Sets Closed under Subset, we have: 
:$\map \mu {B \cap N^c} = 0$
while $B \cap N^c$ is a $\Sigma$-measurable subset of $N^c$, so: 
:$\map \nu {B \cap N^c} = 0$
so:
:$\map {\nu_a} B = 0$
So $\nu_a$ is absolutely continuous with respect to $\mu$.
Hence the result.
Category:Lebesgue Decomposition Theorem
Category:Measures
Lebesgue Decomposition Theorem/Finite Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$. 
Let $\nu$ be a finite signed measure on $\struct {X, \Sigma}$. 
Then there exist finite signed measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that: 
:$(1): \quad \nu_a$ is absolutely continuous with respect to $\mu$
:$(2): \quad \nu_s$ and $\mu$ are mutually singular
:$(3): \quad \nu = \nu_a + \nu_s$
== Proof ==
This follows immediately from Finite Signed Measure is Complex Measure.
== Sources ==
* : $4.3$: Singularity
Category:Lebesgue Decomposition Theorem
Category:Signed Measures
Lebesgue Decomposition of Finite Borel Measure

== Theorem ==
Let $\map \BB \R$ be the Borel $\sigma$-algebra of the real number line $\R$.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$. 
Let $\mu$ be a finite Borel measure on $\struct {\R, \map \BB \R}$.
Then there exists:
:a discrete measure $\mu_{pp}$
:a measure $\mu_{sc}$ that is continuous and mutually singular with respect to $\lambda$
:a measure $\mu_{ac}$ that is absolutely continuous with respect to $\lambda$
such that $\mu = \mu_{pp} + \mu_{sc} + \mu_{ac}$. 
== Proof ==
Let:
:$C = \set {x \in \R : \map \mu {\set x} > 0}$ 
We argue that $C$ is countable.
Note that:
:$\ds C = \bigcup_{n \mathop = 1}^\infty \set {x \in \R : \map \mu {\set x} > \frac 1 n}$
We show that each set:
:$\ds C_n = \set {x \in \R : \map \mu {\set x} > \frac 1 n}$
is finite.
 that $C_n$ is infinite for some $n$.
Let $D_n = \sequence {x_j}_{j \mathop \in \N}$ be a countably infinite subset of $C_n$.
Then we have:
:$\ds \map \mu {D_n} = \sum_{j \mathop = 1}^\infty \map \mu {\set {x_j} } \ge \sum_{j \mathop = 1}^\infty \frac 1 n = \infty$
This contradicts that $\mu$ is finite.
Hence $C_n$ is finite for each $n \in \N$. 
Hence from Countable Union of Finite Sets is Countable, $C$ is countable.
Define:
:$\map {\mu_{pp} } A = \map \mu {A \cap C}$
for each $A \in \Sigma$.
We have that:
:$\map {\mu_{pp} } {C^c} = \map \mu \O = 0$
By Measure is Discrete iff Concentrated on Countable Set, it follows that $\mu_{pp}$ is discrete. 
Now consider the intersection measure $\mu_{C^c}$.
From Lebesgue Decomposition Theorem, there exists finite measures $\mu_{ac}$ and $\mu_{sc}$ such that:
:$(1) \quad$ $\mu_{ac}$ is absolutely continuous with respect to $\lambda$
:$(2) \quad$ $\mu_{sc}$ and $\lambda$ are mutually singular
:$(3) \quad$ $\mu_{C^c} = \mu_{sc} + \mu_{ac}$.
We show that $\mu_{sc}$ is continuous.
Let $x \in \R$.
If $x \in C$ then we have $\set x \cap C^c = \O$ and $\map {\mu_{ac} } {\set x} = 0$ and hence $\map {\mu_{sc} } {\set x} = \map \mu \O = 0$. 
If $x \not \in C$, then we have that $\set x \cap C^c = \set x$ and $\map \mu {\set x} = \map {\mu_{ac} } {\set x} = 0$. 
Hence $\map {\mu_{sc} } {\set x} = \map \mu \O = 0$. 
Hence $\map {\mu_{sc} } {\set x} = 0$ for each $x \in \R$. 
So $\mu_{sc}$ is continuous.
We have therefore obtained the desired decomposition.
== Sources ==
* : $4.3$: Singularity
Category:Lebesgue Decomposition Theorem
Category:Discrete Measures
Category:Mutually Singular Measures
Category:Absolutely Continuous Measures
Lebesgue Infinity-Space is Subset of Tempered Distribution Space

== Theorem ==
Let $\map {L^\infty} \R$ be the Lebesgue infinity-space.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then in the distributional sense it holds that:
:$\map {L^\infty} \R \subseteq \map {\SS'} \R$
That is, every distribution defined by an element of $\map {L^\infty} \R$ is a tempered distribution.
== Proof ==
Let $f \in \map {L^\infty} \R$.
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Let $T_f : \map \SS \R \to \R$ be a mapping such that:
:$\ds \map {T_f} \phi = \int_\R \map f x \map \phi x \rd x$
Then:
Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence in $\map \SS \R$.
Suppose, $\sequence {\phi_n}_{n \mathop \in \N}$ converges in $\map \SS \R$ to $\mathbf 0$:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$
That is:
:$\ds \forall l, m \in \N : \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^l \map {\phi_n^{\paren m} } x} = 0$
Specifically:
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {\map {\phi_n} x} = 0$
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^2 \map {\phi_n} x} = 0$
Then:
On the other hand:
:$\map {T_f} {\mathbf 0} = 0$
where $\mathbf 0$ denotes the zero Schwartz test function.
Hence:
:$T_f \in \map {\SS'} \R$
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Tempered Distributions
Lebesgue Integral is Extension of Darboux Integral

== Theorem == 
Let $f: \closedint a b \to \R$ be a Darboux integrable function.
Then it is also Lebesgue integrable, and furthermore:
:$\ds R \int_a^b \map f x \rd x = \int_{\closedint a b} f \rd \lambda$
where $\ds R \int_a^b$ is the Darboux integral and $\ds \int_{\closedint a b}$ is the Lebesgue integral.
== Proof ==
Since every step function is also a simple function, we have
:$\ds \map L P \le \sup_{\phi \mathop \le f} \int_a^b \map \phi x \rd x \le \inf_{\psi \mathop \ge f} \int_a^b \map \psi x \rd x \le \map U P$
where $\map L P$ and $\map U P$ are the lower Darboux sum and upper Darboux sum as defined in the definition of definite integral.
Since $f$ is Darboux integrable, the inequalities are all equalities and $f$ is measurable by basic properties of measurable functions.
== Sources ==
* : $11.8 \ \text{(i)}$
Category:Analysis
Category:Integral Calculus
Category:Measure Theory
Lebesgue Measure Invariant under Orthogonal Group

== Theorem ==
Let $M \in \map {\mathrm O} {n, \R}$ be an orthogonal matrix.
Let $\lambda^n$ be $n$-dimensional Lebesgue measure.
Then the pushforward measure $M_* \lambda^n$ equals $\lambda^n$.
== Proof ==
By Orthogonal Group is Subgroup of General Linear Group, $M \in \GL {n, \R}$.
From Pushforward of Lebesgue Measure under General Linear Group, it follows that:
:$M_* \lambda^n = \size {\det M^{-1} } \lambda^n$
Since $M^{-1} \in \map {\mathrm O} {n, \R}$ by Orthogonal Group is Group, Determinant of Orthogonal Matrix applies to yield:
:$\size {\det M^{-1} } = 1$
Hence the result.
== Sources ==
* : $7.9$
Category:Lebesgue Measure
Category:Orthogonal Groups
Lebesgue Measure is Diffuse

== Theorem ==
Let $\lambda^n$ be Lebesgue measure on $\R^n$.
Then $\lambda^n$ is a diffuse measure.
== Proof ==
A singleton $\set {\mathbf x} \subseteq \R^n$ is seen to be closed by combining:
:Euclidean Space is Complete Metric Space
:Metric Space is Hausdorff
:Corollary to Compact Subspace of Hausdorff Space is Closed
Hence by Closed Set Measurable in Borel Sigma-Algebra:
:$\set {\mathbf x} \in \map \BB {\R^n}$
where $\map \BB {\R^n}$ is the Borel $\sigma$-algebra on $\R^n$.
Write $\mathbf x + \epsilon = \tuple {x_1 + \epsilon, \ldots, x_n + \epsilon}$ for $\epsilon > 0$.
Then:
:$\ds \set {\mathbf x} = \bigcap_{m \mathop \in \N} \horectr {\mathbf x} {\mathbf x + \frac 1 m}$
where $\\horectr {\mathbf x} {\mathbf x + \dfrac 1 m}$ is a half-open $n$-rectangle.
By definition of Lebesgue measure, we have (for all $m \in \N$):
:$\ds \map {\lambda^n} {\horectr {\mathbf x} {\mathbf x + \frac 1 m} } = \prod_{i \mathop = 1}^n \frac 1 m = m^{-n}$
From Characterization of Measures, it follows that:
:$\ds \map {\lambda^n} {\set {\mathbf x} } = \lim_{m \mathop \to \infty} m^{-n}$
which equals $0$ from Sequence of Powers of Reciprocals is Null Sequence.
Therefore, for each $\mathbf x \in \R^n$:
:$\map {\lambda^n} {\set {\mathbf x} } = 0$
that is, $\lambda^n$ is a diffuse measure.
== Sources ==
* : $\S 4$: Problem $11 \ \text{(i)}$
* : $\S 6$: Problem $5 \ \text{(i)}$
Category:Diffuse Measures
Category:Lebesgue Measure
Lebesgue Measure is Invariant under Translations

== Theorem ==
Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Let $\mathbf x \in \R^n$.
Then $\lambda^n$ is translation invariant; that is, for all $B \in \map \BB {\R^n}$, have:
:$\map {\lambda^n} {\mathbf x + B} = \map {\lambda^n} B$
where $\mathbf x + B$ is the set $\set {\mathbf x + \mathbf b: \mathbf b \in B}$.
== Proof ==
Denote with $\tau_{\mathbf x}: \R^n \to \R^n$ the translation by $\mathbf x$.
From Translation in Euclidean Space is Measurable Mapping, $\tau_{\mathbf x}$ is $\map \BB {\R^n} \, / \, \map \BB {\R^n}$-measurable.
Consider the pushforward measure $\lambda^n_{\mathbf x} := \paren {\tau_{\mathbf x} }_* \lambda^n$ on $\map \BB {\R^n}$.
By Characterization of Euclidean Borel Sigma-Algebra, it follows that:
:$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho} }$
where $\sigma$ denotes generated $\sigma$-algebra, and $\JJ^n_{ho}$ is the set of half-open $n$-rectangles.
Let us verify the four conditions for Uniqueness of Measures, applied to $\lambda^n$ and $\lambda^n_{\mathbf x}$.
Condition $(1)$ follows from Half-Open Rectangles Closed under Intersection.
Condition $(2)$ is achieved by the sequence of half-open $n$-rectangles given by:
:$J_k := \hointr {-k} k^n$
For condition $(3)$, let $\horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho}$ be a half-open $n$-rectangle.
Since:
:$\map {\tau_{\mathbf x}^{-1} } {\horectr {\mathbf a} {\mathbf b} } = \mathbf x + \horectr {\mathbf a} {\mathbf b} = \horectr {\mathbf {a + x} } {\mathbf {b + x} }$
we have:
Finally, since:
:$\ds \map {\lambda^n} {J_k} = \prod_{i \mathop = 1}^n \paren {k - \paren {-k} } = \paren {2 k}^n$
the last condition, $(4)$, is also satisfied.
Whence Uniqueness of Measures implies that:
:$\lambda^n_{\mathbf x} = \lambda^n$
and since for all $B \in \map \BB {\R^n}$ we have:
:$\mathbf x + B = \map {\tau_{\mathbf x}^{-1} } B$
this precisely boils down to:
:$\map {\lambda^n} {\mathbf x + B} = \map {\lambda^n} B$
== Motivation ==
This theorem formalizes the physical intuition that the size of an object does not depend on its position.
== Sources ==
* : $4.9 \ \text{(i)}$
* : $5.8 \ \text{(i)}$
Category:Lebesgue Measure
Category:Translation Mappings
Lebesgue Measure of Scalar Multiple

== Theorem ==
Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Let $B \in \BB$.
Let $t \in \R_{>0}$.
Then:
:$\map {\lambda^n} {t \cdot B} = t^n \map {\lambda^n} B$
where $t \cdot B$ is the set $\set {t \mathbf b: \mathbf b \in B}$.
== Proof ==
It follows from Rescaling is Linear Transformation that the mapping $\mathbf x \mapsto t \mathbf x$ is a linear transformation.
Denote $t \, \mathbf I_n$ for the matrix associated to this linear transformation by Linear Transformation as Matrix Product.
From Determinant of Rescaling Matrix:
:$\map \det {t \, \mathbf I_n} = t^n$
From Inverse of Rescaling Matrix, $t \, \mathbf I_n$ is the inverse of $t^{-1} \mathbf I_n$.
Thus, it follows that:
Now recall $\map \det {\paren {t^{-1} \, \mathbf I_n}^{-1} } = \map \det {t \, \mathbf I_n} = t^n$.
Since $t > 0$, $\size {t^n} = t^n$, and the result follows.
== Sources ==
* : $\S 5$: Problem $8$
* : $\S 7$: Problem $7$
Category:Lebesgue Measure
Lebesgue Pre-Measure is Pre-Measure

== Theorem ==
The Lebesgue pre-measure $\lambda^n$ on the half-open $n$-rectangles $\JJ_{ho}^n$ is a pre-measure.
== Proof ==
We employ Characterization of Pre-Measures.
It is known that $\map {\lambda^n} \O = 0$ by definition of Lebesgue pre-measure.
The only possibility for two disjoint half-open $n$-rectangles to constitute a single, large half-open $n$-rectangle is when they are of the form:
:$\horectr {\mathbf a} {\mathbf b} \quad \horectr {\mathbf a'} {\mathbf b'}$
such that we have for some $i$ with $1 \le i \le n$:
:$i \ne j \implies a_j = a'_j$
:$i \ne j \implies b_j = b'_j$
:$i = j \implies a'_j = b_j$
which intuitively can be visualised as two cubes that together form one large bar, namely $\horectr {\mathbf a} {\mathbf b'}$.
In this situation, we have:
Thus it is verified that $\lambda^n$ is finitely additive.
Finally, suppose that $\horectr {\mathbf a_m} {\mathbf b_m} \downarrow \O$ is a decreasing sequence of sets, with limit $\O$.
Then there exists at least one $j$ with $1 \le j \le n$ such that:
:$\ds \lim_{m \mathop \to \infty} a_{m, j} = \lim_{m \mathop \to \infty} b_{m, j}$
which by Combination Theorem for Sequences is equivalent to:
:$\ds \lim_{m \mathop \to \infty} b_{m, j} - a_{m, j} = 0$
The fact that the sequence is decreasing means that, from Cartesian Product of Subsets, for all $m \in \N$, for all $1 \le i \le n$:
:$\hointr {a_{m, i} } {b_{m, i} } \subseteq \hointr {a_{1, i} } {b_{1, i} }$
and whence $b_{m, i} - a_{m, i} \le b_{1, i} - a_{m, 1}$.
Hence we have:
This verifies the last condition for Characterization of Pre-Measures, since $\lambda^n$ only takes finite values.
Hence $\lambda^n$ is a pre-measure.
== Sources ==
* : $6.5$
Category:Pre-Measures
Lebesgue Space is Vector Space

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \closedint 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be Lebesgue $p$-space on $\struct {X, \Sigma, \mu}$.
Then $\map {\LL^p} {X, \Sigma, \mu}$ is a vector subspace of $\map \MM {X, \Sigma, \R}$, the space of real-valued $\Sigma$-measurable functions on $X$.
In particular, it is a vector space over $\R$.
== Proof ==
From Space of Real-Valued Measurable Functions is Vector Space: 
:$\map \MM {X, \Sigma, \R}$ forms a vector space with pointwise addition and pointwise scalar multiplication.
From construction, we have: 
:$\map {\LL^p} {X, \Sigma, \mu} \subseteq \map \MM {X, \Sigma, \R}$
Since $0 \in \map {\LL^p} {X, \Sigma, \mu}$, we have $\map {\LL^p} {X, \Sigma, \mu} \ne \O$.
From the One-Step Vector Subspace Test, it suffices to show that for each $f, g \in \map {\LL^p} {X, \Sigma, \mu}$ and $\lambda \in \R$ we have: 
:$f + \lambda g \in \map {\LL^p} {X, \Sigma, \mu}$ 
=== Case 1: $1 \le p < \infty$ ===
Since: 
:$\ds \paren {\int \size g^p \rd \mu}^{1/p} < \infty$ 
we have: 
From Minkowski's Inequality on Lebesgue Space, we have: 
:$f + \lambda g \in \map {\LL^p} {X, \Sigma, \mu}$
as required.
=== Case 2: $p = \infty$ ===
From $g \in \map {\LL^\infty} {X, \Sigma, \mu}$, we have: 
:$\map \mu {\set {x \in X : \size {\map g x} \ge c} } = 0$ for some $c > 0$. 
Then we have: 
:$\map \mu {\set {x \in X : \size {\map g x} \ge c} } = \map \mu {\set {x \in X : \size {\lambda \map g x} \ge \size \lambda c} } = 0$
So $\lambda g \in \map {\LL^\infty} {X, \Sigma, \mu}$.
Then, from Minkowski's Inequality on Lebesgue Space, we have: 
:$f + \lambda g \in \map {\LL^\infty} {X, \Sigma, \mu}$
as required.
== Sources ==
* : $12.5 \ \text{(i)}$
Category:Lebesgue Spaces