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$\{f_n\}$ are absolutely continuous functions on $[0,1]$, we know that if $f_n$ are uniformly convergent to a function $f$, then $f$ is continuous.

The question is: is the function $f$ absolutely continuous?

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Recall the Weierstrass Approximation Theorem: for every continuous function $f: [0,1] \rightarrow \mathbb{R}$, there is a sequence of polynomials $P_n$ such that $P_n$ converges uniformly to $f$ on $[0,1]$.

Therefore if $P$ is any property of a function $f: [0,1] \rightarrow \mathbb{R}$ possessed by all polynomial functions, then any continuous function $f: [0,1] \rightarrow \mathbb{R}$ is a uniform limit of functions satisfying property $P$.

Try this out with $P$ being the property of absolute continuity: if

(1) Every polynomial function is absolutely continuous on $[0,1]$ and
(2) Every uniform limit of absolutely continuous functions is absolutely continuous, then
(3) Every continuous function on $[0,1]$ would be absolutely continuous.

I leave it to you to follow up on this syllogism and solve the problem.

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    @Nate: I didn't say that absolutely continuous functions are uniformly dense in all continuous functions. Perhaps the OP will draw that conclusion from what I said -- perhaps correctly! -- but I didn't say it. :)2011-09-20
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I have a counterexample about (3)Every continuous function on [0,1] would be absolutely continuous: Cantor functions is continuous on [0,1] and not absolutely continuous on [0,1].