Almost uniform convergence

The almost uniform convergence is a convergence concept of the measure theory for function sequences , which is based on the uniform convergence . The almost uniform convergence was first introduced by Hermann Weyl , at that time still under the name of essentially uniform convergence . The almost uniform convergence should not be confused with the uniform convergence μ-almost everywhere , this corresponds to the convergence in and is a stronger concept of convergence. ${\displaystyle {\mathcal {L}}^{\infty }}$

definition

Given a measurement space and measurable functions , where or is. Then the sequence of functions is almost uniformly convergent if for every one there, so is the function and result in the complement of converges uniformly . So it applies ${\displaystyle (X,{\mathcal {A}},\mu )}$ ${\displaystyle f,(f_{n})_{n\in \mathbb {N} }\colon X\to \mathbb {K} }$ ${\displaystyle \mathbb {K} =\mathbb {C} }$ ${\displaystyle \mathbb {K} =\mathbb {R} }$ ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ ${\displaystyle \delta >0}$ ${\displaystyle A\in {\mathcal {A}}}$ ${\displaystyle \mu (A)<\delta }$ ${\displaystyle A}$

${\displaystyle \lim _{n\rightarrow \infty }\,\sup _{x\in X\setminus A}\left|f_{n}(x)-f(x)\right|=0.}$

Relationship to other convergence terms

Uniform convergence μ-almost everywhere

The almost uniform convergence follows directly from the uniform convergence μ-almost everywhere . Because if the sequence of functions converges uniformly on the complement of a null set , then for all , the assertion follows directly. However, the reverse is generally not true. If one considers, for example, the measurement space and the sequence of functions , then this sequence converges uniformly to 0 for any small size on the interval and thus also almost uniformly to 0 on the interval . However, the sequence of functions does not converge evenly μ-almost everywhere. ${\displaystyle A}$ ${\displaystyle \mu (A)=0\leq \delta }$ ${\displaystyle \delta >0}$ ${\displaystyle ([0,1],{\mathcal {B}}([0,1]),\lambda |_{[0,1]})}$ ${\displaystyle f_{n}(x)=x^{n}}$ ${\displaystyle \epsilon >0}$ ${\displaystyle [0,1-\epsilon ]}$ ${\displaystyle [0,1]}$

Pointwise convergence μ-almost everywhere

From the almost uniform convergence follows the point-wise convergence μ-almost everywhere . Because by definition there is a set for every null sequence such that and that converges on uniformly. But then there is a null set and the sequence of functions converges point by point to the function ${\displaystyle \delta _{k}}$ ${\displaystyle A_{k}}$ ${\displaystyle \mu (A_{k})<\delta _{k}}$ ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$ ${\displaystyle X\setminus A_{k}}$ ${\displaystyle A=\cap _{k=1}^{\infty }A_{k}}$ ${\displaystyle (f_{n})_{n\in \mathbb {N} }}$

${\displaystyle f(x)=\lim _{n\to \infty }f_{n}(x)\chi _{X\setminus A}(x)}$

and thus almost everywhere.

In the case of a finite measurement space , Jegorow's theorem also provides the inverse, i.e. that the almost uniform convergence follows from the point-wise convergence μ-almost everywhere. Thus for finite dimensional spaces the point-wise convergence almost everywhere and the almost uniform convergence coincide. The following example shows that the conclusion from point-wise convergence almost everywhere to almost uniform convergence for non-finite measure spaces is generally wrong. Looking at the sequence of functions

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

on the measure space , this sequence of functions converges pointwise almost everywhere to 0, because for anything is forever ${\displaystyle (\mathbb {R} ,{\mathcal {B}}(\mathbb {R} ),\lambda )}$ ${\displaystyle x}$ ${\displaystyle n\geq x+2}$

${\displaystyle f_{n}(x)-0=\chi _{[n,n+1]}(x)-0=0}$.

But the sequence does not converge almost uniformly to 0, because it is forever with and thus ${\displaystyle A\in {\mathcal {A}}}$ ${\displaystyle 0\leq \mu (A)<1}$ ${\displaystyle [n,n+1]\setminus A\neq \emptyset }$

${\displaystyle \sup _{x\in \mathbb {R} \setminus A}|f_{n}(x)-0|=1}$

for everyone with . So there cannot be an almost uniform convergence. ${\displaystyle A}$ ${\displaystyle \mu (A)<1}$

Custom-made convergence and custom-made locally

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. ${\displaystyle A}$ ${\displaystyle I_{n}}$

Since the convergence made to measure directly follows the convergence locally made to measure , the almost uniform convergence also implies the convergence locally made to measure, but the reverse is also generally invalid.

General formulation

The almost uniform convergence can be defined analogously for functions with values ​​in a metric space . A sequence of functions is called almost uniformly convergent if there is a set with each such that ${\displaystyle (M,d)}$ ${\displaystyle \delta >0}$ ${\displaystyle A\in {\mathcal {A}}}$ ${\displaystyle \mu (A)<\delta }$

${\displaystyle \lim _{n\to \infty }\sup _{x\in X\setminus A}\ d(f_{n}(x),f(x))=0}$

applies.

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 .
• Norbert Kusolitsch: Measure and probability theory . An introduction. 2nd, revised and expanded edition. Springer-Verlag, Berlin Heidelberg 2014, ISBN 978-3-642-45386-1 , doi : 10.1007 / 978-3-642-45387-8 .