Prove that $f_n\to f$ in measure on $E$ if and only if given $\varepsilon>0$, there exists $K$ such that |{$x\in E : |f(x)-f_k(x)|>\varepsilon$}|$<\varepsilon$ for $k\ge K$.
The "only if" direction of this is immediate from the definition of convergence in measure, but the other direction is less obvious to me.
Conversely, we suppose that given $\varepsilon >0$, there is a $K$ such that |{$x\in E : |f(x)-f_k(x)|>\varepsilon$}|$<\varepsilon$ for $k\ge K$. My initial thought was to bound the measure of the set in question by, say, $\frac{1}{k}$. But I'm not sure I can do that because $\varepsilon$ not only bounds the measure of the set, but the set also depends on the choice of $\varepsilon$. To show something convergence in measure, I need to show that for every $\varepsilon$ the limit as $k\to\infty$ of the measures of those sets is zero...
[Subquestion: is the use of |$\cdot$| standard for denoting Lebesgue measure? I had never seen it until this course. I had always seen $m(-)$.]
[Sub-subquestion: is there any particular reason that set brackets don't display in math mode? The commands \ { and \ } didn't do anything...]