# Cauchy in measure implies convergent in measure (given converging in measure subsequence)

Suppose that there is a sequence $f_n$ which is Cauchy in measure, and suppose further that there exists some subsequence $f_{n_k}$ which converges in measure to $f$. Prove that $f_n$ converges in measure to $f$.

Attempt:

We know that $f_n$ is Cauchy in measure, so $\forall\epsilon>0$ we have that

$$m(|f_n-f_m|\geq \epsilon)<\epsilon, \forall m,n\geq N\in\mathbb{N}$$

We also know that the subsequence $f_{n_k}$ converges in measure which means that

$$m(|f_{n_k}-f|\geq \epsilon)<\epsilon, \forall k\geq N_1\in\mathbb{N}$$

I was thinking of somehow combining the two inequalities above using the Triangle inequality, with something along the lines of :

$$m(|f_n-f\geq \epsilon)|\leq\ldots\leq m(|f_n-f_m+f_{n_k}-f|\geq \epsilon)\leq m(|f_n-f_m|\geq \epsilon)+m(|f_{n_k}-f|\geq \epsilon)<\epsilon+\epsilon=2\epsilon$$

Ultimately I aim to show that $$m(|f_n-f|\geq \epsilon)<\epsilon$$

I am having trouble justifying the "$\ldots$" aspect of the inequality (if this is the correct way to go about the problem).

Hint:

Given measurable functions $f,g$ and $\alpha>0$, then

$$\{ |f-g|\ge\alpha\}\subset \{|f|\ge \alpha/2\}\cup\{|g|\ge\alpha/2\}$$

To see this, note that if $|f(x)|<\alpha/2$ and $|g(x)|<\alpha/2$, then by the triangle inequality, $|f(x)-g(x)|<\alpha/2$.

Let $\alpha>0$ be given.

From $|f(x)-f_{m}(x)|\leq |f(x)-f_{n_{k}}(x)|+|f_{n_{k}}(x)-f_{m}(x)|$

we have

$\{x\in X : |f(x)-f_{m}(x)|\geq \alpha\} \subseteq \{x\in X : |f(x)-f_{n_{k}}(x)|\geq \frac{\alpha}{2}\} \cup \{x\in X : |f_{n_{k}}(x)-f_{m}(x)|\geq \frac{\alpha}{2} \}$.

Since,

$\mu ( \{x\in X : |f(x)-f_{n_{k}}(x)|\geq \frac{\alpha}{2} \}) \to 0, k\to \infty$ and $\mu(\{x\in X : |f_{n_{k}}(x)-f_{m}(x)|\geq \frac{\alpha}{2} \})\to 0,k,m\to \infty$, we have that $(f_{n})_{n\in N}$ converges to $f$ in measure.