# Condition for point wise convergence of a function to be uniform

I've been doing a few practice problems, and I've got some that I am stuck on, one of which is (paraphrasing):

Show that if $f_n$ is a sequence of increasing functions on $[0,1]$ that converges point wise to $f$, and $f$ is continuous then $f_n$ converges uniformly.

It look to me like I can use Egorov's theorem and then examine the points which do not converge uniformly, and use continuity from there, but I can't seem to end the problem. I have (for $y$ in the uniform convergence set and $x$ outside it:

$|f_n(x) - f(x) + f(y) - f(y)| \leq |f_n(x) - f(y)| + \epsilon \leq |f_n(x) - f_n(y)| +\epsilon + |f_n(y) - f(y)|$

Where the last term on the right hand side converges uniformly, my problem is if I try and get the first term to converge uniformly then I seem to just go round in circles. Any advise?

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Are the $f_n$ functions in the sequence continuous as well? –  user35959 Dec 12 '12 at 3:54
The question does not say so. If they were then we would be done (as the RHS of what I wrote above would converge) –  Fredo Baron Dec 12 '12 at 3:59
I could be way off, but I'm moderately suspicious. Let f = 1 and consider $f_n = \chi_{[0,\frac{1}{2}-\frac{1}{2n}] \cup [\frac{1}{2},1]}$. Then, $f_n$ is increasing, pointwise convergent to $f$ , but not uniformly convergent. (Note, $\chi_{A}$ denotes the characteristic function that is 1 for $x \in A$ and zero outside. –  user35959 Dec 12 '12 at 4:22
I think you are right, meaning that we must have $f_n$ being continuous, meaning we are done. Thanks –  Fredo Baron Dec 12 '12 at 4:27

With the additional assumption that the $f_n$ are continuous, try this out: Let $E_n = \{x: |f_n(x)-f(x)| <\epsilon \}$. By construction, these are open sets. Now, since $f_n$ is increasing, what can we say about the $E_n$?. Next, since $f_n \to f$ pointwise, we get another useful property of the $E_n$. Lastly, note that $[0,1]$ is a compact set. Use these facts and you should be able to get that there exists an $N$ such that for $n > N$, $|f_n(x) - f(x)| < \epsilon$ for all $x \in [0,1]$ (i.e., $f_n$ uniformly converges).