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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: You are missing conditions; presumably, "...there exists $K$ such that for all $k\geq K$, ..." – Arturo Magidin Nov 3 '10 at 16:58
@Bey: this is a problem of the LaTeX interpreter. To get the brackets to display, use a double backslash, \\{ and \\}; the first acts as an escape character telling the interprer that the next backslash is the LaTeX command character. – Arturo Magidin Nov 3 '10 at 17:03
@Arturo: I had some issues with my formatting that caused part of my post to disappear. Should be correct now – Bey Nov 3 '10 at 17:03
[deleted obsolete correction] Answer to subquestion: it is standard in some books, not in others. I've also seen $\lambda(\cdot)$. – Jonas Meyer Nov 3 '10 at 17:14
@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. – Arturo Magidin Nov 3 '10 at 17:19
up vote 4 down vote accepted

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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Ah! And we know that {x in E : |f(x)-f_k(x)|>e} is contained in {x in E : |f(x)-f_k(x)|>min(e,d)}, so the measure of the former is less than/equal to the measure of the latter, which is less than min(e,d) < d. Is my thinking correct? – Bey Nov 5 '10 at 0:42
@Bey: Yes it is. – Jonas Meyer Nov 5 '10 at 3:11
Wonderful! Thank you so much, Jonas – Bey Nov 5 '10 at 3:38
@Bey: You're welcome. – Jonas Meyer Nov 5 '10 at 4:03

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