Customized convergence

The convergence to measure , even convergence to the extent or convergence in the level called, is a convergence concept of measure theory for functional consequences . The probabilistic counterpart of this type of convergence is also called convergence in probability or stochastic convergence, but sometimes the definition of convergence locally made to measure is also used there.

definition

Let a dimension space and measurable functions be given . Then the sequence of functions is convergent to measure or convergent to the extent to if for all rule, ${\displaystyle (X,{\mathcal {A}},\mu )}$ ${\displaystyle f,(f_{n})_{n\in \mathbb {N} }\colon X\to \mathbb {K} }$ ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ ${\displaystyle f}$ ${\displaystyle \varepsilon >0}$

${\displaystyle \lim _{n\to \infty }\mu (\{x\in X:\left|f_{n}(x)-f(x)\right|\geq \varepsilon \})=0}$

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

Relationship to other types of convergence

Convergence in the pth mean

From the convergence in the p-th mean follows the convergence to measure, because it is

${\displaystyle \mu (\{|f_{n}-f|\geq \varepsilon \})\leq {\tfrac {1}{\varepsilon ^{p}}}\int _{X}|f_{n}-f|^{p}\mathrm {d} \mu ={\tfrac {1}{\varepsilon ^{p}}}\Vert f_{n}-f\Vert _{p}^{p}}$.

According to Vitali's theorem of convergence , the convergence in the p-th mean is equivalent to the convergence according to measure and the equal integrability 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 to 0 in measure, 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

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

Just as little can be dispensed with the convergence to measure, because one chooses and the measure space , then the sequence of functions is that 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 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. ${\displaystyle \varepsilon <{\tfrac {1}{2}}}$ ${\displaystyle f}$ ${\displaystyle \lambda (\{f_{n}-f\leq \varepsilon \})}$

Almost uniform convergence

The almost uniform convergence is automatically followed by the custom-made convergence. Because by definition, the almost uniform convergence corresponds to the uniform convergence on the complement of a set with for anything . Consequently there is an index so that for all . So is for anything and thus the sequence converges according to measure. ${\displaystyle A}$ ${\displaystyle \mu (A)<\delta }$ ${\displaystyle \delta }$ ${\displaystyle N(\varepsilon )}$ ${\displaystyle A\supset \{|f_{n}-f|\geq \varepsilon \}}$ ${\displaystyle n\geq N(\varepsilon )}$ ${\displaystyle \mu (\{|f_{n}-f|\geq \varepsilon \})\leq \delta }$ ${\displaystyle \delta }$

Conversely, however, the convergence made to measure does not generally result in an almost uniform convergence. For example, consider the sequence of intervals

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

and defines the sequence of functions ${\displaystyle ([0,1],{\mathcal {B}}([0,1]),\lambda )}$

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

so this sequence converges to 0 according to measure, since a deviation of the sequence of functions from 0 is only possible on the intervals. The width of the intervals and thus the size of the sets on which the sequence of functions deviates from 0 converge to 0. However, the sequence does not converge almost uniformly, since it applies to with that ${\displaystyle A\in {\mathcal {A}}}$ ${\displaystyle \mu (A)<\delta <1}$

${\displaystyle \sup _{x\in [0,1]\setminus A}f_{n}(x)={\begin{cases}1&{\text{falls }}I_{n}\setminus A\neq \emptyset \\0&{\text{falls }}I_{n}\setminus A={\emptyset }\end{cases}}}$.

But since this is firmly chosen and the constantly "wander", the sequence oscillates and can therefore not converge almost uniformly. ${\displaystyle A}$ ${\displaystyle I_{n}}$

Pointwise convergence μ-almost everywhere

From the point-wise convergence μ-almost everywhere follows the convergence to measure for finite dimensional spaces. The conclusion follows Yegorov's theorem , that from the convergence μ-almost everywhere (in the finite case) the almost uniform convergence follows, from this in turn the convergence to measure follows.

On the finiteness of the measure space it can not be waived, as the (more detail below under investigation) function sequence on shows. It converges point by point to 0, but not to measure. ${\displaystyle f_{n}=\chi _{[n,n+1]}}$ ${\displaystyle (\mathbb {R} ,{\mathcal {B}}(\mathbb {R} ),\lambda )}$

However, the convergence does not apply, so convergence to measure does not result in convergence almost everywhere. An example can be constructed as follows: Consider the intervals

${\displaystyle [0,1],[0,{\tfrac {1}{2}}],[{\tfrac {1}{2}},1],[0,{\tfrac {1}{3}}],[{\tfrac {1}{3}}{\tfrac {2}{3}}],[{\tfrac {2}{3}},1],[0,{\tfrac {1}{4}}],\dots }$,

numbers them with the natural numbers and names this sequence . Then the sequence of functions converges ${\displaystyle (I_{n})_{n\in \mathbb {N} }}$ ${\displaystyle f_{n}(x)=\chi _{I_{n}}(x)}$

on the measurement space made 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 it takes the values ​​0 and 1 infinitely often, so it cannot converge point by point. ${\displaystyle ([0,1],{\mathcal {B}}([0,1]),\lambda |_{[0,1]})}$ ${\displaystyle \varepsilon \in (0,1]}$ ${\displaystyle \lim _{n\to \infty }\lambda (\{f_{n}\geq \varepsilon \})=\lim _{n\to \infty }\lambda (I_{n})=0}$ ${\displaystyle x}$ ${\displaystyle I_{n}}$ ${\displaystyle I_{n}}$ ${\displaystyle \chi _{I_{n}}}$

Customized local convergence

Custom-made convergence implies custom-made convergence locally . 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.

Weak convergence of dimensions

From the convergence according to the measure of a sequence of functions, under certain circumstances it is possible to infer the weak convergence of the series of image measures .

If there is a measure space , if there is a finite measure and if the sequence of functions converges towards measure , then the sequence of image measures converges weakly towards . ${\displaystyle (X,{\mathcal {A}},\mu )}$ ${\displaystyle \mu }$ ${\displaystyle f_{n}}$ ${\displaystyle f}$ ${\displaystyle \nu _{n}:=\mu _{f_{n}}}$ ${\displaystyle \nu :=\mu _{f}}$

The image dimensions are then dimensions on . More generally, this statement can also be shown for sequences of functions with values ​​in separable metric spaces . ${\displaystyle \mathbb {R} }$

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. ${\displaystyle (X,d)}$ ${\displaystyle \left|f_{n}-f\right|\geq \varepsilon }$ ${\displaystyle d(f_{n},f)\geq \varepsilon }$ ${\displaystyle \{d(f_{n},f)\geq \varepsilon \}}$ ${\displaystyle (X,d)}$ ${\displaystyle {\mathcal {B}}(X)}$ ${\displaystyle (X,{\mathcal {B}}(X))}$

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 .