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The following problem is Exercise 7.K from the book van Rooij-Schikhof: A Second Course on Real Functions and it is very close to a question which was recently discussed in chat.

So I thought that sharing this interesting problem with other MSE users could be useful. Here's the problem:

Let $L$ be the set of all functions $f\colon [0,1]\to\mathbb R$ that have the property that $\lim\limits_{x\to a} f(x)$ exists for all $a \in [0, 1]$. Show that:
(i) $L$ is a vector space. Each $f \in L$ is bounded.
(ii) For each $f \in L$, define $f^c(x): = \lim\limits_{y\to x} f(y)$ ($x \in [0, 1]$). $f^c$ is continuous.
(iii) '$f^c =0$' is equivalent to 'there exist $x_1, x_2, \dots$ in $[0,1]$ and $a_1, a_2,\dots$ in $U$ with $\lim\limits_{n\to\infty} a_n = 0$, such that $f(x_n) = a_n$ for every $n$, and $f=0$ elsewhere'.
(iv) Describe the general form of an element of $L$. Show that every $f\in L$ is Riemann integrable.

The original question in the chat was about functions $\mathbb R\to\mathbb R$, but it does not change much in the parts (iii) and (iv).

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    "$\lim f(x) $ exists for all $a$" is the same as saying that $f$ is continuous, right?2012-06-30
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    @MattN. No. For example $f=\chi_{\{0\}}$. Continuity would be: Limit exists at each point and it is equal to the value of the function: $\lim\limits_{x\to a} f(x)=f(a)$.2012-06-30
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    Martin, I do not understand your $\chi_{(0)}$ example. What about the sequence $a_n=0$ if $n$ is even and $a_n=1/n$ otherwise? This _is_ a sequence converging to $0$, is'nt it?2012-06-30
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    @Thomas [Limit of a function](http://en.wikipedia.org/wiki/Limit_of_a_function#Functions_on_the_real_line) at a point $a$ is defined using [punctured neighborhoods](http://en.wikipedia.org/wiki/Punctured_neighborhood#Punctured_neighbourhood) of this point. Or, if you want sequential definition, you have to take sequences which does not contain the value $a$ as one of their terms.2012-06-30
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    @MartinSleziak Thanks (especially for your second comment to Thomas). Bottom-line (yet again): "Know your definitions." (<- talking to myself)2012-06-30
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    @Martin: but if you use punctured neighborhood, what about the function $f(x) = 1/x$ for $x>0$, and $f(0) = 0$? Are the limits also required to be finite?2012-06-30
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    Also, I wonder if the following characterization is right: "$f$ belongs to $L$ if and only if there exist a compact set $K \subset [0,1]$ and continuous functions $g : [0,1] \to \mathbb{R}$ and $h: K \to \mathbb{R}$ such that $h = 0$ on $K'$ and $f = g+h$".2012-06-30
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    @D.Thomine You are right that if we allow the limit to be $\pm\infty$, we get also some unbounded functions. But most of the rest should remain the same, since $[-\infty,\infty]$ is homeomorphic to $[0,1]$.\\ I am not sure I understand your second comment. What do you denote by $K'$. Derived set (limit points)? Complement?2012-06-30
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    In my precedent comment, $K'$ is the derived set.2012-06-30
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    @D.Thomine The set $K=\{0\}\cup\{1/n; n\in\mathbb N\}$ is compact and $K'=\{0\}$. The function defined by $f(x)=1$ for $x\in\{1/n; n\in\mathbb N\}$ and $f(x)=0$ otherwise fulfills $f|_{K'}=0$, but it does not have limit at $0$. I don't know about the other implication.2012-07-01
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    @Martin: This function is not continuous on $K$.2012-07-01
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    Related question: http://math.stackexchange.com/questions/980022/a-function-having-limit-at-every-point-but-continuous-nowhere2014-10-21

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