Cauchy in Measure Trick Question 
I am refering to the above question in Wheedon's Real Analysis book..
The definition of converging in measure given is $$\lim_{k\to\infty}|\{x\in E: |f(x)-f_k(x)|>\epsilon\}|=0$$ for every $\epsilon>0$.
What the question asks us to prove seems to follow directly from the definition. I am worried that I may be missing something since the other questions in the book are usually not this trivial.
Is there a trick to this?
I see one potential issue in the same $\epsilon$ in the question. Perhaps using the definition only implies $|\{f-f_k|>\epsilon_1\}|<\epsilon_2$, we are supposed to prove that it works using the same epsilon?
Thanks for any help.
 A: Convergence in measure implies that given $\epsilon>0$, there exists $K$ s.t. for all $k>K$, we have $\mu(\{x:|f_k(x)-f(x)|>\epsilon\})<\epsilon$. This part is therefore easy.
Now suppose it is given that for any $\epsilon>0$, there exists $K$ s.t. for any $k>K$, we have $\mu(\{x:|f_k(x)-f(x)|>\epsilon\})<\epsilon$. Let $\epsilon_1$ and $\epsilon_2$ be given. 
We are to find $K$ s.t. for any $k\geq K$, $\mu(\{x:|f_k(x)-f(x)|>\epsilon_1\})<\epsilon_2$.
Suppose $\epsilon_1\geq \epsilon_2$. Then by hypothesis, we know there exists $K$ s.t. for all $k\geq K$, $\mu(\{x:|f_k(x)-f(x)|>\epsilon_2)<\epsilon_2$. Now since $\epsilon_1\geq \epsilon_2$, we have $\{x:|f_k(x)-f(x)|>\epsilon_1\})\subseteq\{x:|f_k(x)-f(x)|>\epsilon_2\}$. Thus, $\mu(\{x:|f_k(x)-f(x)|>\epsilon_1\})\leq \mu(\{x:|f_k(x)-f(x)|>\epsilon_2\})<\epsilon_2$ for all $k\geq K$. And we are done.
If however $\epsilon_1<\epsilon_2$, then we know that there exists $K$ s.t. for all $k>K$, we have $\mu(\{x:|f_k(x)-f(x)|>\epsilon_1\})<\epsilon_1<\epsilon_2$ (the second inequality follows as $\epsilon_1<\epsilon_2$). Thus we are done again.
