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Let $(f_n)$ be a sequence of functions $f_n:[a,b]\to\mathbb{R}$, all of which are piecewise continuous. Does $f_n\rightrightarrows f$ imply that $f$ is piecewise continuous?

EDIT:

I wrote a proof that $f$ must be piecewise continuous. However, this proof must be incorrect according to Jahan's answer. Where's the problem?

Suppose $f$ is not piecewise continuous. Then the collection $X$ of points $x\in[a,b]$ at which $f$ is discontinuous is infinite. For each $x_0\in X$ there exists $\epsilon>0$ such that in every neighborhood of $x_0$ there is an $x$ satisfying $|f(x_0)-f(x)|\ge \epsilon$. Suppose $f_n$ is continuous at $x_0$. Then there exists $\delta>0$ such that when $|x-x_0|<\delta$ we have $|f_n(x)-f_n(x_0)|<\epsilon/3$. Since $f_n\rightrightarrows f$, there is some $N$ such that for all $n\ge N$ and all $x\in[a,b]$ we have $|f_n(x)-f(x)|<\epsilon/3$. Now, for $n\ge N$ we have that $$\epsilon\le |f(x)-f(x_0)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|<\epsilon.$$ This is a contradiction. Therefore, for each $x_0$ in $X$, $f_n$ is not continuous at $x_0$ and so its set of discontinuity points is infinite. Therefore $f$ must be piecewise continuous, else some function in the sequence is not piecewise continuous.

Alec Teal
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Ryan
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  • Do the functions necessarily share the same endpoints of their pieces? – Zach Stone Aug 04 '15 at 22:08
  • No. The discontinuity points may not be the same for all $f_n$. They may even never be the same between any two functions in the sequence. – Ryan Aug 04 '15 at 22:16
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    The issue with your proof is that you start with some $f_n$ continuous at $x_0$, and then also try to impose the condition that $|f_n(x) - f(x))| < \varepsilon/3$. What your proof shows is that if $f$ is not continuous at $x_0$, then there exists some $N$ such that $f_n$ is not continuous at $x_0$ for all $n \geq N$. – Alex Zorn Aug 04 '15 at 23:00
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    "Suppose $f_n$ is continuous at $x_0$". As it seems to me by deducing a contradiction from this assumption you only proved that for every discontinuity $x_0 \in X$ of $f$ there has to exist an $N \in \mathbb{N}$ such that $f_N$ is discontinuous at $x_0$. – el_tenedor Aug 04 '15 at 23:02
  • @JahanClaes I don't think that's true. If for for any N you could always find $n > N$ such that $f_n$ were continuous at $x_0$, then you could use the OP's argument. – Alex Zorn Aug 06 '15 at 15:23
  • @AlexZorn my mistake! You are right of course. – Jahan Claes Aug 06 '15 at 20:39

2 Answers2

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Let $f_0$ be a constant function, say $f_0(x)=0$. I'll tell you how to generaete a sequence of functions whose uniform limit isn't piecewise continuous.

From $f_{n-1}$, we'll construct $f_{n}$ in the following manner: Divide your interval $[a,b]$ up evenly into $2^n$ pieces. Starting at the leftmost piece, alternatively raise/lower $f_{n-1}$ by $\frac{1}{2^n}$ in each interval. This new function is $f_n$. You should be able to convince yourself that this function converges uniformly, since $\sup(f_m-f_n)\leq 2\max(\frac{1}{2^n},\frac{1}{2^m})$. You should also be able to convince yourself that the limiting function is discontinuous at every point of the form $a+\frac{(b-a)}{2^n}$.

Essentially, the displacements never "cancel" (i.e., if you raise the function on the left of a point and lower the function on the right of the point, you do the same every subsequent iteration. You'll never lower the function on the right and raise the function on the left). Thus, the function is discontinuous at every point where you broke it apart.

Jahan Claes
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  • Thanks, that example makes sense. While you wrote that example I came up with a (apparently incorrect) proof that $f$ must be piecewise continuous. I can't seem to spot the error in my proof. Care to tell me what it is? I've added it to the question. – Ryan Aug 04 '15 at 22:55
  • Actually, I'm confused as to why $\sup(f_m-f_n)\le\max(2^{-n},2^{-m})$. Shouldn't it be larger? You're adding $2^{-n}$ each time so it seems like should be $2^{-n}+2^{-(n+1)}+2^{-(n+2)}+\cdots+2^{-m}$ – Ryan Aug 04 '15 at 23:03
  • Edited, thanks. – Jahan Claes Aug 05 '15 at 01:00
  • And the comments above pretty much sum up the mistake in your proof, I think. – Jahan Claes Aug 05 '15 at 01:01
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Define $f$ on $[0,1]$ by setting $f = 1/n$ on $(1/(n+1),1/n], n=1,2,\dots , \,f(0) = 0.$ Then $f$ is not piecewise continuous on $[0,1],$ but $f$ is the uniform limit on $[0,1]$ of the sequence $f_n=f\cdot\chi_{(1/n,1]},$ and each $f_n$ is piecewise continuous on $[0,1].$

zhw.
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  • What does $f_{\chi(1/n,1]}$ refer to? – Ryan Aug 04 '15 at 23:23
  • Thanks for the edit, but I still don't know what $\chi_{1/n,1]}$ refers to – Ryan Aug 04 '15 at 23:25
  • I meant to multiply $f$ by $\chi_{(1/n,1]}.$ (I edited it to make it clearer.) Standard notation: $\chi_E$ is the function that is $1$ on $E$ and $0$ everywhere else. I could have also just said $f_n = f$ on $(1/n,1]$ and $f_n=0$ on $[0,1/n].$ – zhw. Aug 04 '15 at 23:26