# 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

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.