# finite measure and cauchy in measure

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

## 1 Answer

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.

• But then, whats the use for a finite measure in my problem? o: – asd123 Jun 30 at 2:42
• @asd123 I think the assumption of finite measure is redundant. – Feng Shao Jun 30 at 3:06
• 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: – asd123 Jun 30 at 4:49
• @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”. – Feng Shao Jun 30 at 8:36