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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...]

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    @Bey: `<` and `>` also cause problems, because they are interpreted as HTML markup; use the backslash as an escape character `\<` and `\>`, or use `\lt` and `\gt`.2010-11-03

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Convergence in measure means that for all $\varepsilon\gt0$, |\{x\in E : |f(x)-f_k(x)|>\varepsilon\}| goes to $0$, which means that for all $\varepsilon\gt0$, for all $\delta\gt0$, there exists $K$ such that |\{x\in E : |f(x)-f_k(x)|>\varepsilon\}|<\delta for all $k\geq K$. Taking $\delta=\varepsilon$ gives the "only if". To see "if", given positive $\varepsilon$ and $\delta$, take $K$ such that $k\geq K$ implies that |\{x\in E : |f(x)-f_k(x)|>\min(\varepsilon,\delta)\}|<\min(\varepsilon,\delta).

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    @Bey: You're welcome.2010-11-05