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Say we have a sequence of functions $(f_n):(0,1)\rightarrow \mathbb{R}$ which convergence point-wise to a function $f$, and converge uniformly to $f$ on every compact sub-interval of $(0,1)$. Is there anything about the sequence of functions $(f_n)$ that is not preserved in the limit that would be if $(f_n)$ converged uniformly to $f$ on all of $(0,1)$?

Edit: I should clarify, I'm also interested in if the limit can be 'passed' through the integral (assuming it exists) $\int f_n$, incase that wasn't clear.

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    Well, you can, what I'm interested in is if you do assume they are, does the limit pass through the integral?2012-03-06

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Consider the function $f_n$ whose graph coincides with the $x$-axis on $[0,1-{1/ n }] $. On the interval $[1-1/n,1]$ the graph is comprised of the straight line segments from $[1-{1 / n },0]$ to $[1-{1 /(2 n) },2n]$ and from $[1-{1 /(2 n) },2n]$ to $[1,0]$ (over $[1-{1 / n },1]$, the graph is a "triangular spike" of height $2n$).

Then the sequence $(f_n)$ converges pointwise to $f=0$ on $[0,1]$ and converges uniformly on $[0,1-\epsilon]$ for each $1>\epsilon>0$. But, $\int_0^1 f_n=1$ for each $n$ while $\int_0^1 f=0$.

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    Actually I misspoke, I was concerned with if the limit of your$f_n$was well defined, not the limit of their antiderivatives, but now I see I was wrong in either case since in your limit the tip of your triangle drops to zero and does not go to infinity, mea culpa.2012-03-06
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A similar answer, based on what David said: consider the sequence of functions $(f_n):[0,1]\rightarrow\mathbb{R}$ defined as: $f_n(x)=n^2x(1-x)^n$ It's easy to check $f_n$ converges pointwise to $f=0$, but: $1=\lim_{n\to\infty}\int_0^1 f_n(x)\,\mathrm dx\neq\int_0^1 \left(\lim_{n\to\infty}f_n(x)\right)\,\mathrm dx=0$