Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...]

share|improve this question
    
@Bey: You are missing conditions; presumably, "...there exists $K$ such that for all $k\geq K$, ..." –  Arturo Magidin Nov 3 '10 at 16:58
1  
@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
1  
@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
show 1 more comment

1 Answer

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 $|\{x\in E : |f(x)-f_k(x)|>\min(\varepsilon,\delta)\}|<\min(\varepsilon,\delta)$.

share|improve this answer
    
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
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.