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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.

oliverjones
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  • 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 $[0,1]$? – gary Jul 25 '17 at 00:59
  • @gary Increasing doesn't imply continuous, hence $f - \frac{1}{2^n}$ isn't necessarily continuous. – MathematicsStudent1122 Jul 25 '17 at 01:00
  • @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 $[0,1]$. – gary Jul 25 '17 at 01:03
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    @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? – stewbasic Jul 25 '17 at 01:06
  • @gary at any discontinuities of $f$ wouldn't $f - \frac{1}{2^n}$ have a discontinuity at the point as well? – oliverjones Jul 25 '17 at 01:07
  • @stewbasic: What is $H$? And wouldn't $f(0.5)=0.5$ make $f$ continuous? – gary Jul 25 '17 at 01:08
  • It's a collary of Lusin's theorem. – C. Ding Jul 25 '17 at 01:08
  • @oliverjones: $f$ is defined that way at points of continuity. And if $f$ is not continuous , unless I am missing something, it can be extended. – gary Jul 25 '17 at 01:09
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    @gary $H$ is the Heaviside function (https://en.wikipedia.org/wiki/Heaviside_step_function) – stewbasic Jul 25 '17 at 01:15
  • @gary you are missing something. Being continuous a.e. doesn't mean it can be extended to a continuous function. A function with a single jump discontinuity is a counterexample. – spaceisdarkgreen Jul 25 '17 at 01:23
  • @spaceisdarkgreen: But this seems to contradict the result that every unifomly continuous function defined on a dense subset can be extended continuously. $f$ is a.e. continuous on a compact subset, therefore it is a.e. uniformly continuous by compactness.What am I missing? – gary Jul 25 '17 at 01:31
  • @gary I think the gap in your argument is that although continuity on a compact set implies uniform continuity, the same isn't true for a.e. continuity. In any case your conclusion is clearly false for a function with a step discontinuity (eg see my link). – stewbasic Jul 25 '17 at 01:33
  • @stewbasic: D'oh myself: the set of points where f is continuous is not necessarily compact. – gary Jul 25 '17 at 01:35
  • @stewbasic I was given the hint: prove for any interval of $[0,1]$ that the characteristic function is the pointwise limit of a sequence of continuous functions. Which I believe I can then define $f_n(x) = \max(0, 1 - n\chi_I)$ where $I \subseteq [0,1]$. But I don't see how this helps. – oliverjones Jul 25 '17 at 07:02
  • Why do you consider the pointwise limit? – C. Ding Jul 25 '17 at 07:03
  • @C.Ding I don't understand your question. – oliverjones Jul 25 '17 at 07:04
  • Why do you need the pointwise-limit other than a.e.-limit? – C. Ding Jul 25 '17 at 07:08
  • @C.Ding because that is the question I was asked. a.e.-limit is not what was asked for so I can't just use that. – oliverjones Jul 25 '17 at 07:09
  • "A function whose continuity points are everywhere dense in every nonempty perfect set can be represented as a limit of continuous functions." This statement is taken from Boas, A primer of real functions, page 125. I suspect that this could be useful, since a monotone function is continuous axcept at countably many points where one-sided limits exist. – Siminore Jul 25 '17 at 07:46

4 Answers4

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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.

Sangchul Lee
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  • Doesn't a sequence of continuous function on a compact set converge uniformly so that the limit function is continuous? – gary Jul 25 '17 at 18:50
  • @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). – Sangchul Lee Jul 25 '17 at 20:36
  • 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. – gary Jul 25 '17 at 21:03
  • @SangchulLee when you say linear interpolation you mean something like the Lagrange polynomial? – oliverjones Jul 26 '17 at 04:06
  • @oliverjones, It is rather a very straightforward one, joining successive data points by line segments. Check this page to see a picture! – Sangchul Lee Jul 26 '17 at 07:03
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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.

C. Ding
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  • 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. – stewbasic Jul 25 '17 at 01:30
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    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. – stewbasic Jul 25 '17 at 01:31
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    I need the sequence of functions to converge at every point (pointwise) and I don't currently have that theorem at my disposal. – oliverjones Jul 25 '17 at 01:37
  • By the same token I can use my argument, since a Real-valued function can only have countably-many jumps. – gary Jul 25 '17 at 01:40
  • @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? – C. Ding Jul 25 '17 at 06:42
  • @C.Ding I'm not sure, but I'd be interested in seeing how that follows; would you mind giving more detail? – stewbasic Jul 25 '17 at 06:53
  • @stewbasic I'm also not sure , and we may consider a sigular function here: https://www.encyclopediaofmath.org/index.php/Cantor_ternary_function – C. Ding Jul 25 '17 at 07:01
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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.

charmd
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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.

Siminore
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