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I've been trying to prove that If $(f_{n})_{n}$ is cauchy in measure, i.e $\mu(|f_{n}-f_{m}|\geq\delta)<\epsilon$ for $n,m $ relatively big then it converges to a measurable function in measure. Now, if $\mu(\Omega)$ is finite then it is enough to prove that if $(f_{n})_{n}$(which converges cauchy in measure) converges a.e to a measurable function, because in a space of finite measure, convergence almost everywhere implies convergence in measure. This might be a really basic question, but I havent been able to prove it.

Any help would be really appreciated. Thanks guys <3

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Your idea may not work. Consider the probability measure, i.e. $\mu(\Omega)=1$. Define an independent sequence $\{X_n\}$ by $$X_n=\left\{ \begin{array} 1 1 &\text{with probability } n^{-1},\\ 0 &\text{with probability }1-n^{-1}. \end{array} \right. $$ You can check that $X_n\rightarrow 0$ in probability, i.e. in measure but $\{X_n\}$ does not converge almost surely.

For the proof of your problem, see Cauchy in measure implies convergent in measure.

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  • $\begingroup$ But then, whats the use for a finite measure in my problem? o: $\endgroup$ – asd123 Jun 30 at 2:42
  • $\begingroup$ @asd123 I think the assumption of finite measure is redundant. $\endgroup$ – Feng Shao Jun 30 at 3:06
  • $\begingroup$ Thanks for your answers, but then, whats the freakin difference between cauchy in measure and convergence in measure D: I suppose convergence in measure is stronger but no entirely sure haha, i mean the practical difference D: $\endgroup$ – asd123 Jun 30 at 4:49
  • $\begingroup$ @asd123 As far as I know, they’re equivalent. Maybe my last statement is not true in some cases. After all, I’m not very sure about it. My understanding is that the notation of “Cauchy in measure” is a convenience for usage. For example, there are cases where the proof of “Cauchy in measure” is easier than the direct proof of “convergence in measure”. $\endgroup$ – Feng Shao Jun 30 at 8:36

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