real analysis - 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

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.