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Hi can you help me with the following:

$\{f_n\}$ a sequence of increasing functions with $f_n\to f$ in measure on $[a,b]$. Show that $f_n(x)\to f(x)$ at every $x$ where $f(x)$ is continuous.

Thanks a lot!!

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What have you tried? –  Davide Giraudo Sep 8 '12 at 12:42
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This looks like homework; you should read meta.math.stackexchange.com/questions/1803/…;, edit accordingly, and add the "homework" tag. –  Nate Eldredge Sep 8 '12 at 12:57
    
Do you mean the sequence is increasing? Or do you mean you have a sequence of strictly increasing functions? The first case I think is false. Consider $f = 3$, and $f_n = 3$ everywhere except at $\frac{a+b}{2}$ and $\frac{a+2b}{2}$. At $\frac{a+b}{2}$, $f_n = 0$. At $\frac{a+2b}{2}$, $f_n = 3 - \frac{1}{n}$. This is an increasing sequence of functions that certainly converges in measure but $f_n(\frac{a+b}{2}) \not\to f(\frac{a+b}{2})$. However, $f$ is clearly continuous at $\frac{a+b}{2}$. –  Shankara Pailoor Sep 8 '12 at 18:45

1 Answer 1

I believe that even if you say the sequence consists of non-decreasing functions, the statement is false. Let $f(x) = 3$ on $[a,b]$. Take, $$\{f_n\} = \left\{ \begin{array}{lr} 3 & \text{if } x \neq a\\ 0 & \text{if } x = a \end{array} \right.$$

$f_n$ is clearly non-decreasing and converges to $f$ in measure. However, $\lim_n{f_n(0)} = 0 \neq f(0)$.

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I have to add f_n is continous and each f_n is non-decreasing. –  Salih Ucan Sep 8 '12 at 19:12
    
But f_n's are NOT uniformly continuous for every n?? –  Salih Ucan Sep 8 '12 at 20:56
    
|fn(x)−fn(z)|<ϵ this does not hold for every n?? –  Salih Ucan Sep 8 '12 at 20:57
    
you are right, we can just show for a particular $n$ that $|f_n(x) - f_n(z)| < \epsilon$. Sorry about that. –  Shankara Pailoor Sep 8 '12 at 21:04
    
One more question...is there no condition that says $f_n(x) \leq f_{n+1}(x)$? –  Shankara Pailoor Sep 8 '12 at 23:05

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