Subjects: Measure Theory
Links: Measurable Functions, Measure Spaces and Measurable Spaces, Measures

Th: if $\{f_n\mid n<\omega \}$ is a sequence of extended real values measurable functions on a measurable space $X$, then the following functions are measurable:

• $h(x)=sup\{f_n(x)\mid n<\omega \}$
• $g(x)=inf\{f_n(x)\mid n<\omega \}$
• $f^{\ast }(x)=\underset{n\to \mathrm{\infty }}{lim sup}{f}_n(x)$
• $f_{\ast }(x)=\underset{n\to \mathrm{\infty }}{lim inf}{f}_n(x)$

Cor: if $\{f_n\mid n<\omega \}$ is a sequence of extended real values measurable functions on a measurable space $X$, then the set of convergence of the sequence $\{f_n\mid n<\omega \}$, i.e., the set $$\left{x\in X ; \left\rvert ; \limsup_{n\to \infty} f_n(x) = \liminf_{n\to \infty} f_n(x)\right.\right},$$has a measurable intersection with every measurable set, and consequently, that the function $f$, defined by $f(x)=\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)$ at every $x$ for which the limits exists, is measurable function.

Def: A function $f$, defined on a measurable space $X$, is called simple if there is a finite, disjoint class $\{E_1,\dots ,E_n\}$ me measurable sets and a finte set $\{{\alpha }_1,\dots ,{\alpha }_n\}$ of real numbers such that $$f(x) = \sum_{i = 1}^n \alpha_i\chi_{E_i}(x).$$
Th: Every extended real valued measurable function $f$ is the limit of a sequence $\{f_n\mid n<\omega \}$ of simple functions; if $f$ is non negative, then each $f_n$ may be taken non negative and the sequence $\{f_n\mid n<\omega \}$ may be assumed increasing.

If $f$ is bounded, then the sequence $\{f_n\mid n<\omega \}$ can be made to converge uniformly.

Def: An elementary function is defined the same way as a simple function, the only change being that the number of sets $\{E_n\mid n<\omega \}$ is allowed to be countably infinite.

Prop: Every real valued measurable function $f$ is the limit of uniformly convergent sequence of elementary functions.

Prop: If we disregard the results that requiere order of $\mathbb{R}$, then the results above can be extended to for complex valued sequences of measurable functions.

Convergence

Def: If a certain proposition concerning points of a measure space is true for every point, with exception at most of a set of measure zero, it is customary to say that the proposition is true almost everywhere or a.e.
Def: A function $f$ is called essentially bounded if it is bounded a.e., i.e., if there exists a $M>0$ such that $\mu (\{x\in X\mid |f(x)|>c\})=0$. The infimum of the of the valued of $M$ for which the statement is true is called the essential supremum of $|f|$, abbreviated to $\text{ess sup}(|f|)$.

Def: Let $\{f_n\mid n<\omega \}$ be a sequence of extended real valued functions which converges a.e. on the measure space $X$ to the limit function $f$. This means, of course, that there exists a set $E_0$ of measure zero such that, if $x\notin E_0$ and $\epsilon >0$, then there exists $N\in \omega$, such that

• if $f(x)=-\mathrm{\infty }$, then $f_n(x)<-1/\epsilon$
• If $f(x)\in \mathbb{R}$, then $|f_n(x)-f(x)|<\epsilon$
• if $f(x)=\mathrm{\infty }$ then $f_n(x)<1/\epsilon$
  whenever $n\ge N$.

Def: We shall say that a sequence $\{f_n\mid n<\omega \}$ of real valued function is fundamental a.e. if there exists a set $E_0$ of measure zero such that, if $x\notin E_0$ and $\epsilon >0$, then there's an integer $N<\omega$, with the property if $n,m\ge N$ $$|f_n(x) - f_m(x)|<\varepsilon.$$
Obs: It is clear that if a sequence converges to a finite valued limit function a.e., then it is fundamental a.e., and conversely, that corresponding to a sequence which is fundamental a.e. there always exists a finite valued limit function to which it converges a.e.

Obs: On the space of measurable functions being equal a.e. is an equivalence relation. Let $f,g,h:X\to \mathbb{R}$ measurable functions, then:

• $f=f$ a.e.
• If $f=g$ a.e., then $g=f$ a.e.
• If $f=g$ and $g=h$ a.e., then $f=h$ a.e.

Prop: If $f$ is any real valued, Lebesgue measurable function on the real line, then there exista a Borel measurable function $g$ such that $f=g$ a.e.

Def: The sequence $\{f_n\mid n<\omega \}$ converges to $f$ uniformly a.e. if there's a set $E_0$ of measure zero that, for every $\epsilon >0$, there's an integer $N<\omega$ can be found with the property that for every $x\in X\setminus E_0$, $$|f_n(x) - f(x) |< \varepsilon.$$
Egoroff's Theorem: If $E$ is a measurable set of finite measure and if $\{f_n\mid n<\omega \}$ is a sequence of a.e. finite valued measurable functions which converges a.e. on $E$ to a finite valued measurable function $f$, then, for every $\epsilon >0$, there's exists a measurable subset $F\subseteq E$ such that $\mu (F)<\epsilon$ and such that the sequence $\{f_n\mid n<\omega \}$ converges to $f$ uniformly on $E\setminus F$.

Def: A sequence $\{f_n\mid n<\omega \}$ of a.e. finite valued measurable functions will be said to the measurable function $f$ almost uniformly, if for every $\epsilon >0$, there's a measurable set $F$ such that $\mu (F)<\epsilon$ and such that the sequence $\{f_n\}$ converges uniformly to $f$ uniformly on $X\setminus F$.

Obs: We see that Egoroff's theorem assets that on a set of finite measure convergence a.e. implies almost uniform convergence.

Cor: If $E$ is a measurable set of positive finite measure, and if $\{f_n\mid n<\omega \}$ is a sequence of a.e. finite valued measurable functions which is fundamental a.e., then there exist $M>0$ and an $F\subseteq E$ of positive measure such that for every $n<\omega$, and every $x\in F$ $|f(x)|\le c$.

Cor: If $E$ is a measurable set of a $\sigma$-finite measure, if $\{f_n\mid n<\omega \}$ is a sequence of a.e. finite valued measurable functions which converges a.e. on $E$ to a finite valued measurable function, then there exists a sequence of $\{E_m\mid m<\omega \}$ of measurable sets such that $\mu (E\setminus \bigcup _{m<\omega }{E}_n)=0$ and such that the sequence $\{f_n\}$ converges uniformly on each $E_m$.

Prop: If $\{f_n\mid n<\omega \}$ is a sequence of measurable functions which converges to $f$ almost uniformly, then $\{f_n\mid n<\omega \}$ to $f$ a.e.

Convergence in Measure

Th: Suppose that $f$ and $f_n$, $n<\omega$ are real valued measurable functions on a set $E$ of finite measure, and write for every $\epsilon >0$, $$E_n(\varepsilon) := {x \in X \mid |f_n(x) - f(x)| \ge \varepsilon}, \qquad n < \omega.$$The sequence $\{f_n\mid n<\omega \}$ converges to $f$ a.e. on $E$ iff $$\lim_{n \to \infty} \mu\left(E\cap \bigcup_{m = n}^\infty E_m(\varepsilon)\right) = 0$$for every $\epsilon >0$.

Def: A sequence $\{f_n\mid n<\omega \}$ of a.e. finite valued, measurable functions converges in measure to the measurable function $f$ if, for every $\epsilon >0$, $\underset{n\to \mathrm{\infty }}{lim}\mu (\{x\in X\mid |f_n(x)-f(x)|\ge \epsilon \})=0$. In accordance with our general comment on different kinds of convergence, we shall say that a sequence $\{f_n\mid n<\omega \}$ of a.e. finite valued measurable functions is fundamental in measure if, for every $\epsilon >0$, $$\lim_{n, m \to \infty} \mu({x\in X\mid |f_n(x) - f_m(x)| \ge \varepsilon}) = 0.$$
Obs: Every subsequence of a sequence which is fundamental in measure is fundamental in measure.

Obs: If a sequence of finite valued measurable functions converges a.e. to a finite limit (or is fundamental a.e.) on a set $E$ of finite measure, then it converges in measure (or is fundamental in measure) on $E$.

Obs: If $X=\mathbb{N}$ is the set of all integers, $\mathcal{S}=P(\mathbb{N})$, and for every $E\in \mathcal{S}$, $\mu (E):=|E|$, then, for the measure space $(\mathbb{N},P(\mathbb{N}),\mu )$, convergence in measure is equivalent to uniform convergence everywhere.

Obs: On a set of infinite measure convergence a.e. implies convergence in measure.

Th: Almost uniform convergence implies convergence in measure.

Prop: Every subsequence of a sequence which is fundamental in measure is fundamental in measure.

Prop: If $\{f_n\mid n<\omega \}$ converges in measure to $f$, then $\{f_n\mid n<\omega \}$ is fundamental in measure. If $\{f_n\mid n<\omega \}$ converges in measure to $g$ then $f=g$ a.e.

Obs: There are the measure space $(X,\mathcal{S},\mu )$ that are totally finite, and sequences of functions $\{f_n\mid n<\omega \}$ that converge in measure but there are no points that converges.

Def: a sequence of measurable functions is uniformly fundamental in measure if for every $\epsilon >0$ and $\delta >0$, there's a $N<\omega$ such that for all $m,n\ge N$, $$\mu({x \in X \mid |f_n(x) - f_m(x)|\ge \varepsilon})<\delta. $$
Th: If $\{f_n\mid n<\omega \}$ is a sequence of measurable functions which is fundamental in measure, then some subsequence $\{f_{n_k}\mid k<\omega \}$ is almost uniformly fundamental.

Th: If $\{f_n\mid n<\omega \}$ is a sequence of measurable functions which is fundamental in measure, then there exists a measurable function $f$ such that $\{f_n\mid n<\omega \}$ converges in measure to $f$.

Prop: Suppose the measure space $(X,\mathcal{S},\mu )$ is totally finite, and let $\{f_n\mid n<\omega \}$ and $\{g_n\mid n<\omega \}$ be sequence of finite valued measurable functions converging in measure to $f$ and $g$ respectively. Then the following are true:

• If $\alpha ,\beta \in \mathbb{R}$, then $\{\alpha f_n+\beta g_n\mid n<\omega \}$ converges in measure to $\alpha f+\beta g$.
• $\{|f_n|\mid n<\omega \}$ converges to $|f|$.
• If $f=0$ a.e, then $\{f_n^2\mid n<\omega \}$ converges in measure to $f^2$.
• The sequence $\{f_{n}g\mid n<\omega \}$ converges in measure to $fg$.
• The sequence $\{f_n^2\mid n<\omega \}$ converges in measure to $f^2$.
• The sequence $\{f_n{g}_n\mid n<\omega \}$ converges in measure to $fg$.