Customized local convergence

The convergence locally to measure , sometimes called convergence locally in measure called, is a convergence concept of measure theory for functional consequences . It is the weakest notion of convergence used in measure theory. Sometimes it is also used in probability theory , where it is referred to as stochastic convergence ; However, depending on the source situation, this type of convergence can also designate the equivalent convergence to measure of probability .

definition

A dimension space and for measurable functions are given . Then the sequence of functions is called convergent locally according to measure against if for every set with and all that ${\displaystyle (X,{\mathcal {A}},\mu )}$ ${\displaystyle f,f_{n}\colon X\to \mathbb {K} }$ ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ ${\displaystyle f}$ ${\displaystyle A\in {\mathcal {A}}}$ ${\displaystyle \mu (A)<\infty }$ ${\displaystyle \varepsilon >0}$

${\displaystyle \lim _{n\to \infty }\mu (\{\left|f_{n}-f\right|\geq \varepsilon \}\cap A)=0}$

is. Then you write ${\displaystyle f_{n}\to f{\text{ lokal n. M.}}}$

example

Designates the characteristic function and defines the sequence of functions as ${\displaystyle \chi _{A}(x)}$

${\displaystyle f_{n}:=\chi _{[n,n+1]}}$,

this function sequence converges on the measure space locally to measure up to 0. For each Borel set in finite Lebesgue measure the series converges , and it follows , that is . ${\displaystyle (\mathbb {R} ,{\mathcal {B}}(\mathbb {R} ),\lambda )}$ ${\displaystyle A}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle \sum _{n}\lambda (A\cap [n,n+1])=\lambda (A\cap [0,\infty ))\leq \lambda (A)<\infty }$ ${\displaystyle \lambda (A\cap [n,n+1])\rightarrow 0}$ ${\displaystyle \lambda (\{|f_{n}-0|\geq \varepsilon \}\cap A)\rightarrow 0}$

properties

• Converge locally to measure to or , then also to locally to measure and to locally to measure. ${\displaystyle f_{n},g_{n}}$ ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle f_{n}+g_{n}}$ ${\displaystyle f+g}$ ${\displaystyle \alpha f_{n}}$ ${\displaystyle \alpha f}$
• If the sequence of functions converges locally according to measure against and against , then and locally μ- agree almost everywhere . That means for everyone with is μ-almost everywhere. ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle A}$ ${\displaystyle \mu (A)<\infty }$ ${\displaystyle f\chi _{A}=g\chi _{A}}$

Relationship to other convergence terms

Customized convergence

The convergence to measure implies the convergence locally to measure. For if the measure of the set becomes arbitrarily small on the basic set , it also becomes arbitrarily small on the section with any set of finite size. ${\displaystyle \{\left|f_{n}-f\right|\geq \varepsilon \}}$ ${\displaystyle X}$

However, the reverse is generally not true. So the sequence of functions converges

${\displaystyle f_{n}=\chi _{[n,n+1)}}$

on the dimensional space locally to measure towards 0, but not to measure. Because for is ${\displaystyle (\mathbb {R} ,{\mathcal {B}}(\mathbb {R} ),\lambda )}$ ${\displaystyle \varepsilon \in (0,1]}$

${\displaystyle \mu (\{|f_{n}-f|\geq \varepsilon \})=\lambda (\{|\chi _{[n,n+1)}-0|\geq \varepsilon \})=1}$

for everyone . So the sequence of functions does not converge according to measure to 0. But if you now consider a with and , then they are disjoint and it applies ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle A\in {\mathcal {B}}(\mathbb {R} )}$ ${\displaystyle \lambda (A)<\infty }$ ${\displaystyle A_{n}=A\cap [n,n+1)}$ ${\displaystyle A_{n}}$

${\displaystyle A\supset \bigcup _{n\in \mathbb {N} }A_{n}{\text{ und somit }}\infty >\mu (A)\geq \sum _{n=1}^{\infty }\mu (A_{n})}$.

Thus it is because otherwise the series would diverge. It then follows ${\displaystyle \lim _{n\to \infty }\lambda (A_{n})=0}$

${\displaystyle \lim _{n\to \infty }\mu (\{|f_{n}-f|\geq \varepsilon \}\cap A)=\lim _{n\to \infty }\mu (\{|\chi _{[n,n+1)}-0|\geq \varepsilon \}\cap A)=\lim _{n\to \infty }\lambda (A_{n})=0.}$

Thus the function sequence converges locally according to measure to the 0.

On finite measure spaces , convergence locally according to measure also follows from convergence according to measure, so both concepts of convergence are equivalent. This follows directly from the fact that the basic set already has finite measure. Since the function sequence converges locally according to measure, it accordingly also converges on the basic set and thus also according to measure.

Pointwise convergence μ-almost everywhere

From the point-wise convergence μ-almost everywhere , the convergence follows locally to measure. Because limits to the measure space on a lot with a so considers the measure space . This restricted measure space is a finite measure space, so Jegorow's theorem applies there . This provides the almost uniform convergence on the restricted dimensional space, which in turn implies the convergence to measure. Since this conclusion holds for every restriction to sets of finite measure, the function sequence converges to locally according to measure. ${\displaystyle A}$ ${\displaystyle \mu (A)<\infty }$ ${\displaystyle (A,{\mathcal {A}}|_{A},\mu |_{A})}$ ${\displaystyle (X,{\mathcal {A}},\mu )}$

The convergence does not apply, however, so convergence does not follow from convergence locally according to measure, convergence almost everywhere. An example can be constructed as follows: Consider the intervals

${\displaystyle (I_{n})_{n\in \mathbb {N} }=[0,1],[0,{\tfrac {1}{2}}],[{\tfrac {1}{2}},1],[0,{\tfrac {1}{3}}],[{\tfrac {1}{3}},{\tfrac {2}{3}}],[{\tfrac {2}{3}},1],[0,{\tfrac {1}{4}}],[{\tfrac {1}{4}},{\tfrac {2}{4}}],\dots }$

Then the sequence of functions converges

${\displaystyle f_{n}(x)=\chi _{I_{n}}(x)}$

on the measure space locally according to measure towards 0, because for is . But the sequence of functions does not converge point-by-point almost everywhere to 0, because an arbitrary one is contained in an infinite number and is also not contained in an infinite number . Thus, at every point, the values ​​0 and 1 take on infinitely often, so it cannot converge. ${\displaystyle ([0,1],{\mathcal {B}}([0,1]),\lambda |_{[0,1]})}$ ${\displaystyle \varepsilon \in (0,1]}$ ${\displaystyle \lim _{n\to \infty }\lambda (\{f_{n}\geq \varepsilon \})=\lim _{n\to \infty }\lambda (I_{n})=0}$ ${\displaystyle x}$ ${\displaystyle I_{n}}$ ${\displaystyle I_{n}}$ ${\displaystyle \chi _{I_{n}}}$

Convergence in the pth mean

According to Vitali's convergence theorem , a sequence is convergent in the p-th mean if and only if it is locally convergent by measure and is equally integrable in the p-th mean .

The possibility of equal integration cannot be dispensed with, as the following example illustrates. One sets and defines the sequence of functions ${\displaystyle p=1}$

${\displaystyle f_{n}=n^{2}\chi _{[0,1/n]}}$.

on the measure space , this converges locally by measure to 0, because for is ${\displaystyle ([0,1],{\mathcal {B}}([0,1]),\lambda |_{[0,1]})}$ ${\displaystyle \varepsilon \in (0,1]}$

${\displaystyle \lim _{n\to \infty }\lambda (\{n^{2}\chi _{[0,1/n]}\geq \varepsilon \})=\lim _{n\to \infty }{\frac {1}{n}}=0}$.

But it cannot be integrated equally (in the first mean) because it is

${\displaystyle \inf _{a\in [0,\infty )}\sup _{f\in (f_{n})_{n\in \mathbb {N} }}\int _{\{a<|f|\}}|f|\mathrm {d} \lambda =\infty }$

Following Vitali's convergence theorem, it is also not (in the first mean) convergent to 0, because it is

${\displaystyle \lim _{n\to \infty }\int _{[0,1]}|f_{n}|\mathrm {d} \lambda =\lim _{n\to \infty }n^{2}\cdot {\frac {1}{n}}=\infty }$.

Nor can the convergence locally to measure be dispensed with, because if one chooses and the measure space , then the sequence of functions is the one through ${\displaystyle p=1}$ ${\displaystyle ([0,1],{\mathcal {B}}([0,1]),\lambda |_{[0,1]})}$

${\displaystyle f_{n}:={\begin{cases}\chi _{[0;1/2]}&{\text{ für }}n{\text{ gerade }}\\\chi _{(1/2;1]}&{\text{ für }}n{\text{ ungerade }}\end{cases}}}$.

defined is equally integrable in the first mean, since it is majorized by the integrable function, which is constant 1. Due to its oscillating behavior, however, the sequence cannot converge locally to measure, because there is no function for the basic set and , so that it becomes small. With an analogous argument it then also follows that the function sequence does not converge in the first mean. ${\displaystyle \varepsilon <{\tfrac {1}{2}}}$ ${\displaystyle f}$ ${\displaystyle \lambda (\{f_{n}-f\leq \varepsilon \})}$

Weak convergence in L ^p

From the convergence locally made-to-measure, the weak convergence in . If a sequence from converges to locally according to measure and if the sequence of real numbers is bounded, then the sequence also converges weakly to . ${\displaystyle p\in (1,\infty )}$ ${\displaystyle {\mathcal {L}}^{p}}$ ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ ${\displaystyle {\mathcal {L}}^{p}}$ ${\displaystyle f\in {\mathcal {L}}^{p}}$ ${\displaystyle (\|f_{n}\|_{p})_{n\in \mathbb {N} }}$ ${\displaystyle f}$

For this statement is generally not correct, as the following example shows: If one considers the measure space , then the sequence converges ${\displaystyle p=1}$ ${\displaystyle ([0,1],{\mathcal {B}}([0,1]),\lambda |_{[0,1]}}$

${\displaystyle f_{n}=n\chi _{[0,1/n]}}$

locally made to measure towards 0 and it's for everyone . But then for the constant function it is off ${\displaystyle \|f_{n}\|_{1}=1}$ ${\displaystyle n}$ ${\displaystyle g=1}$ ${\displaystyle {\mathcal {L}}^{\infty }}$

${\displaystyle \int _{X}f_{n}g\mathrm {d} \lambda =1}$.

Thus the sequence does not converge weakly to 0.

More convergence terms

The local convergence according to measure is the weakest convergence concept for function sequences of the measure theory, all other convergence concepts imply the convergence locally according to measure. For example, implies uniform convergence μ-almost everywhere the almost uniform convergence , in turn, the convergence to measure and hence the convergence locally to measure. The inversions are generally wrong.

More general wording

Customized convergence can also be defined more generally for functions with values ​​in metric spaces . To do this, replace the term with . However, it must be ensured here that the quantities are measurable, otherwise the expression in the definition is not well-defined. The measurability of these quantities is guaranteed, for example, if a separable metric space and the associated Borel σ-algebra are selected and the measurement space is selected. ${\displaystyle \left|f_{n}-f\right|\geq \varepsilon }$ ${\displaystyle d(f_{n},f)\geq \varepsilon }$ ${\displaystyle \{d(f_{n},f)\geq \varepsilon \}}$ ${\displaystyle (X,d)}$ ${\displaystyle {\mathcal {B}}(X)}$ ${\displaystyle (X,{\mathcal {B}}(X))}$

literature

• Jürgen Elstrodt : Measure and integration theory . 6th, corrected edition. Springer-Verlag, Berlin Heidelberg 2009, ISBN 978-3-540-89727-9 , doi : 10.1007 / 978-3-540-89728-6 .
• Achim Klenke : Probability Theory . 3. Edition. Springer-Verlag, Berlin Heidelberg 2013, ISBN 978-3-642-36017-6 , doi : 10.1007 / 978-3-642-36018-3 .