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Prove or disprove: If $f$ is non-decreasing real valued function on $[0,1]$ then there is a sequence $f_n$ of continuous function on $[0,1]$ such that for each x $ \in [0,1] $, we have $f_n(x)$ converges to $f(x)$

I am thinking of If assume f is continuous then we can find sequence of polynomial $p_n(x)$ which converges to $f$ even uniformly. but f is not given to be continuous then how do we do? any suggestions and hints are welcomed.

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  • $\begingroup$ how ? could provide me some hints @ Daniel 9 $\endgroup$ – user145993 Jan 3 '16 at 16:31
  • $\begingroup$ Take spline functions generated from a function value table with step size $2^{-n}$. The only problem being the limits at jump points. $\endgroup$ – LutzL Jan 3 '16 at 16:36
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The answer is "yes". Let $B$ denote the set of functions which are the pointwise limit of continuous functions (this set is usually called the set of Baire class 1 functions).

Let $S$ denote the set of step functions over $[0,1]$. You can prove the following facts:

  • $S$ is a subset of $B$

  • every non-decreasing function is the uniform limit of a sequence of functions in $S$

  • $B$ is closed under uniform limits

Putting these together yields an answer to your question.

Note that, since the set of points of continuity of Baire class 1 functions is dense, the set of points of continuity of any monotonic function is dense !

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You note that $X=C([0,1])$ with the sup norm $||\cdot ||_{sup}$ is a metric space, then it is first countable. Your function is an accumulation point for $X$, then there is sequence $\lbrace f_n\rbrace$ of $X$ such that it converges uniformly to $f$. This result is simply obtained for all accumulation points in first countable topological spaces.

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