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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 :

http://medvegyev.uni-corvinus.hu/CeuAdv2.pdf.

Thanks.

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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.

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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