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If $\mu(A) <\infty $, then from almost everywhere convergence follows the convergence in the measure . I don't understand what the "convergence in the measure" means. Waiting for your explanation.

I am trying to understand proposition 2 from the following link :


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Please provide some context, where is this statement taken from? etc. – Asaf Karagila Jun 8 '12 at 10:41
Wikipedia is waiting for you! – Sangchul Lee Jun 8 '12 at 10:41
@AsafKaragila : I have posted the reference. Would like to hear some explanation. – Theorem Jun 8 '12 at 10:50

2 Answers 2

up vote 3 down vote accepted

The proposition says that if $(X,\mu)$ is a measure space, and $\mu(X)$ is finite, then:

Whenever $f_n$ is a sequence of real valued functions such that for almost all $x\in X$ we have $\lim f_n(x)=f(x)$ for some $f$,

Then for every $\varepsilon>0$ the measure of the sets $E_n=\left\{x\in X: |f_n(x)-f(x)|>\varepsilon\right\}$ approach zero, that is $\lim\limits_n\mu(E_n)=0$.

The first type of convergence is called "almost everywhere convergence" and the second type is called "convergence in measure". The proposition tells us that for finite-measure spaces one implies the other.

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We say that a sequence of functions $f_{n}:A\to \mathbb{R}$ converges to a function $f:A\to\mathbb{R}$ in measure, if for every $\varepsilon>0$ we have \begin{equation*} \lim_{n\to\infty}\mu(\{x\in A:|f_{n}(x)-f(x)|>\varepsilon\})=0. \end{equation*} So the statement says that if $\mu(A)<\infty$ then a.e. convergence implies convergence in measure as defined above.

Edit: I didn't look in every detail the proof that you gave, but here's a simple way to see it by using Fatou's reverse Lemma for measurable sets, which requires the finiteness of $\mu(A)$ since its proof uses convergence of measure for decreasing sequence of sets. Note that \begin{align*} \{x\in A:\lim_{n\to\infty}f_{n}(x)=f(x)\} &=\{x\in A:\forall n\in\mathbb{N}\exists m\in\mathbb{N}\,\,s.t.\,\,\forall i\geq m\,\,|f_{i}(x)-f(x)|\leq \frac{1}{n}\}\\ &=\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}\bigcap_{i\geq m}\{x\in A:|f_{i}(x)-f(x)|\leq \frac{1}{n}\}. \end{align*} So if $f_{n}\to f$ a.e. then the complement of the above set has measure zero, whence by de-morgan's law \begin{align*} 0 &=\mu\Big(\bigcup_{n=1}^{\infty}\bigcap_{m=1}^{\infty}\bigcup_{i\geq m}\{x\in A:|f_{i}(x)-f(x)|> \frac{1}{n}\}\Big) \overset{\forall n\in\mathbb{N}}{\geq}\mu\Big(\bigcap_{m=1}^{\infty}\bigcup_{i\geq m}\{x\in A:|f_{i}(x)-f(x)|> \frac{1}{n}\}\Big) \\ &=\mu\Big(\limsup_{i\to\infty}\{x\in A:|f_{i}(x)-f(x)|> \frac{1}{n}\}\Big)\overset{Fatou}{\geq} \limsup_{i\to\infty}\,\mu(\{x\in A:|f_{i}(x)-f(x)|> \frac{1}{n}\}) \\ &\geq \liminf_{i\to\infty}\,\mu(\{x\in A:|f_{i}(x)-f(x)|> \frac{1}{n}\})\geq 0. \end{align*} So there in fact holds an equality everywhere, and in particular \begin{equation*} \lim_{i\to\infty}\mu(\{x\in A:|f_{i}(x)-f(x)|>\frac{1}{n}\})=0 \end{equation*} for all $n\in\mathbb{N}$, from which it follows that $f_{n}\to f$ in measure.

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Thank you very much. Can you explain me how u changed from $limsup$ to $liminf$ towards the last step ? – Theorem Jun 8 '12 at 14:10
@Ananda: for any sequence $(a_{n})_{n=1}^{\infty}$ of real numbers we have $\limsup\, a_{n}\geq \liminf\, a_{n}$, so I just applied this to the above line. – T. Eskin Jun 8 '12 at 15:03

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