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if $f:[0,1] \to \mathbb{R}$ is increasing, show that $f$ is the pointwise limit of a sequence of continuous functions over $[0,1]$

Intuitively this makes sense but I am having trouble with showing why there would be a sequence of continuous functions converging pointwise to $f$.

Clearly there is a sequence converging pointwise to $f$, I can set: $\forall n \in \mathbb{N}, f_n = f$.

How to prove there is at least one which is made up of continuous functions $f_n, \forall n \in \mathbb{N}$ over $[0,1]$ I can't quite figure out the argument.


marked as duplicate by user299912, Claude Leibovici, user91500, Glorfindel, Community Jul 26 '17 at 4:06

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  • $\begingroup$ How about $f_n:= f-1/2^n$? We know $f$ is a.e. continuous, since it is monotone. We define it as above wherever $f$ is continuous and then extend on the countably-many points where it is not? Isn't the set of points of continuity dense in $[0,1]$? $\endgroup$ – gary Jul 25 '17 at 0:59
  • $\begingroup$ @gary Increasing doesn't imply continuous, hence $f - \frac{1}{2^n}$ isn't necessarily continuous. $\endgroup$ – MathematicsStudent1122 Jul 25 '17 at 1:00
  • $\begingroup$ @MathematicsStudent1122: It (meaning monotonous) implies a.e continuous, though, and I am defining it as $f-1/2^n$ at the points where it is continuous, which I believe is dense in $[0,1]$ and then extending from a dense subset into $[0,1]$. $\endgroup$ – gary Jul 25 '17 at 1:03
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    $\begingroup$ @gary Suppose $f(x)=x+H(x-0.5)$. This is continuous on $[0,0.5)\cup(0.5,1]$, so $f_1(x)=f(x)-1$ on $[0,0.5)\cup(0.5,1]$. How do you propose to define $f_1(0.5)$ to make $f_1$ continuous? $\endgroup$ – stewbasic Jul 25 '17 at 1:06
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    $\begingroup$ @gary $H$ is the Heaviside function (en.wikipedia.org/wiki/Heaviside_step_function) $\endgroup$ – stewbasic Jul 25 '17 at 1:15

Let $\mathcal{D}$ be the set of discontinuities of $f$. We know that $\mathcal{D}$ is at most countable, so we enumerate this set by $\mathcal{D} = \{ x_0, x_1, \cdots \}$. Now for each $n \geq 0$, let

$$ \Pi_n = \{ \tfrac{k}{2^n} : 0 \leq k \leq 2^n \} \cup \{ x_0, \cdots, x_n \} $$

and define $f_n : [0, 1] \to \mathbb{R}$ as the linear interpolation of the points $\{ (x, f(x)) : x \in \Pi_n\}$ ordered from left to right. Then

  • It is clear that $f_n$ is continuous and increasing for each $n\geq 0$.

  • If $x \in \cup_{n\geq 0} \Pi_n$, then $x \in \Pi_N$ for some $N$ and hence by construction, $f_n(x) = f(x)$ for all $n \geq N$. So we have $f_n(x) \to f(x)$ as $n\to\infty$.

  • If $x \in [0, 1] \setminus \mathcal{D}$, then for each fixed $m$ there is $a_m, b_m \in \Pi_m$ such that $a_m \leq x \leq b_m$ and $|b_m - a_m| \leq 2^{-m}$. Taking limit as $n\to\infty$ to the inequality $f_n(a_m) \leq f_n(x) \leq f_n(b_m)$, we obtain

    \begin{align*} f(a_m) = \lim_{n\to\infty} f_n (a_m) &\leq \liminf_{n\to\infty} f_n (x) \\ &\leq \limsup_{n\to\infty} f_n (x) \leq \lim_{n\to\infty} f_n (b_m) = f(b_m). \end{align*}

    Taking $m \to \infty$, both $(a_m)$ and $(b_m)$ converge to $x$. Since $x$ is a continuity point of $f$, we have

    $$ f(x) \leq \liminf_{n\to\infty} f_n (x) \leq \limsup_{n\to\infty} f_n (x) \leq f(x) $$

    and hence $f_n(x) \to f(x)$.

Combining altogether, it follows that $f_n \to f$ pointwise on $[0, 1]$ as expected.

  • $\begingroup$ Doesn't a sequence of continuous function on a compact set converge uniformly so that the limit function is continuous? $\endgroup$ – gary Jul 25 '17 at 18:50
  • $\begingroup$ @gary, Continuity is not guaranteed if you take pointwise limit. Recall the archetypal example: $f_n(x) = x^n$ on $[0, 1]$. I suspect that you are confusing this with the statement that a continuous function on a compact subset of $\mathbb{R}$ to $\mathbb{R}$ is uniformly continuous (or with its obvious generalizations). $\endgroup$ – Sangchul Lee Jul 25 '17 at 20:36
  • $\begingroup$ No, I just thought sequences that were defined on a compact space converged uniformly to a continuous function, but you reminded me of that counterexample. $\endgroup$ – gary Jul 25 '17 at 21:03
  • $\begingroup$ @SangchulLee when you say linear interpolation you mean something like the Lagrange polynomial? $\endgroup$ – oliverjones Jul 26 '17 at 4:06
  • $\begingroup$ @oliverjones, It is rather a very straightforward one, joining successive data points by line segments. Check this page to see a picture! $\endgroup$ – Sangchul Lee Jul 26 '17 at 7:03

Note: This answer dose not answer the OP's question. Just see it as a reference.

Since $f:[0,1]\to \mathbb{R}$ is increasing, it is a bounded Borel measurable function. By Lusin's theorem, there is a sequence $\{f_n\}$ of continuous functions such that $f_n=f$ on a Borel set $E_n$ which satisfies $m\{[0,1]\backslash E_n\}<\frac{1}{2^n}$.

Since $\sum_{n=1}^{\infty}m\{[0,1]\backslash E_n\}<\infty$, for almost every point in $[0,1]$, it is contained in finte sets of $\{[0,1]\backslash E_n\}$ and therefore $f_n$ converges to $f$ almost everywhere.

  • $\begingroup$ If $E_n=[0,1]\setminus\{0.5\}$ then $m([0,1]\setminus E_n)=m(\{0.5\})=0$ but $0.5$ is in infinitely many $[0,1]\setminus E_n$? I think it's not enough to know that $f_n$ approximates $f$ well almost everywhere, because you need different sequences depending on whether $f$ is left or right continuous at a jump discontinuity. $\endgroup$ – stewbasic Jul 25 '17 at 1:30
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    $\begingroup$ Ah sorry I missed "almost" in your last sentence. The question asks for $f_n$ to converge to $f$ at every point, not just almost every point. $\endgroup$ – stewbasic Jul 25 '17 at 1:31
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    $\begingroup$ I need the sequence of functions to converge at every point (pointwise) and I don't currently have that theorem at my disposal. $\endgroup$ – oliverjones Jul 25 '17 at 1:37
  • $\begingroup$ By the same token I can use my argument, since a Real-valued function can only have countably-many jumps. $\endgroup$ – gary Jul 25 '17 at 1:40
  • $\begingroup$ @stewbasic If every incresing function is the pointwise limit of a sequence of continuous functions, then so is every bounded variation function. Is this true? $\endgroup$ – C.Ding Jul 25 '17 at 6:42

Here is another answer, which main idea is : if $f$ is right or left continuous, there is an "explicit" solution (see the integrals below).

Claim : Every increasing function $f$ can be written as a sum of an increasing right-continuous function and a left-continuous function.

Proof : the set of points at which $f$ is discontinuous is at most countable : it will be denoted $(a_n)_{n \in \mathbb{N}}$. Now we take $g : x \mapsto \sum \limits_{\substack{n \in \mathbb{N}\\ a_n \le x}} f(a_n)-f(a_n^-)$ and $h = f-g$. It is easy to show that $g$ is right-continuous. If $x \notin \{a_n\}$, $f$ and $g$ are continuous at $x$, so $h$ is continuous at $x$. If $x=a_n$, $g(t) \underset{t \to x^-}{\longrightarrow} g(x)-\big( f(a_n)-f(a_n^-) \big)$, so $h(t) \underset{t \to x^-}{\longrightarrow} f(a_n^-)-g(x)+\big( f(a_n)-f(a_n^-) \big)=h(x)$. Hence $f=g+h$ with $f$ increasing and right-continuous, and $h$ left-continuous.

$ $

If we write $f=g+h$ with $g$, $h$ as in the claim, both $f$ and $g$ are increasing, thus locally bounded and with at most countably many discontinuity points. Hence $f$ and $g$ are Riemann integrable, and so is $h$. For $n \ge 1$, we can thus consider : $$ g_n : x \mapsto n \displaystyle{\int_x^{x+\frac{1}{n}}} g(t)dt,\quad \ h_n : x \mapsto n \displaystyle{\int_{x-\frac{1}{n}}^x} h(t)dt, \quad \ f_n : x \mapsto g_n(x)+h_n(x).$$

As $g$ (resp. $h$) is right (resp. left)-continuous, it is easy to prove that for all $x$, $g_n(x) \underset{n \to +\infty}{\longrightarrow} g(x)$ and $h_n(x) \underset{n \to +\infty}{\longrightarrow} h(x)$, so $\big(f_n\big)_{n \ge 1}$ converges pointwise to $f$. As $g$ and $h$ are locally bounded, $g_n$ and $h_n$ are continuous for all $n$, and thus $\big( f_n \big)_{n \ge 1}$ is a sequence of continuous functions.


So: monotone functions are Baire-one functions, as you can see here and here. This means, by definition, that elements of this class are pointwise limits of continuous functions, see here. Hence the statement is proved.


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