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Let a function $f:A \rightarrow \mathbb{R}$, defined on a set $A\subset \mathbb{R}$ without isolated points, has a finite limit $g(x)$ in each point $x\in A$.

I wish to show that $g: A \rightarrow \mathbb{R}$, $g(x)=\lim_{t \rightarrow x} f(t)$, is continuous.

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Define $\tilde{f}$ as follows: $\tilde{f}(x)=f(x)$ if $f$ is continuous at $x$; $\tilde{f}(x)=\lim_{t \to x} f(x)$ otherwise. Can you compare $\lim_{t \to x} g(x)$ to $\lim_{t \to x}\tilde{f}(x)$? – Siminore Aug 25 '12 at 10:31
up vote 1 down vote accepted

Let $x$ in $A$ and $\varepsilon\gt0$. By definition of $g(x)$ as a limit, there exists $\eta\gt0$ such that every $z$ in $A$ such that $|z-x|\lt2\eta$ and $z\ne x$ is such that $|f(z)-g(x)|\leqslant\varepsilon$. Let $x'$ in $A$ such that $|x'-x|\lt\eta$ and $x'\ne x$. Then every $z$ in $A$ such that $|z-x'|\lt\min\{|x'-x|,\eta\}$ and $z\ne x'$ is such that $z\ne x$ and $|z-x|\lt2\eta$, hence $|f(z)-g(x)|\leqslant\varepsilon$. In particular, the limit $g(x')$ of $f(z)$ when $z\to x'$ with $z$ in $A$ and $z\ne x'$, is such that $|g(x')-g(x)|\leqslant\varepsilon$.

This is valid for every $x'$ in $A$ such that $|x'-x|\lt\eta$ and $x'\ne x$, hence $g$ is continuous at $x$.

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Suppose $g$ is discontinuous at $a\in A$. Let $a_k$ be a sequence in $A$ such that $d(a_k, a) < 1/k$ and such that for some constant $c>0$ and for all all $k$, $d(g(a_k),g(a))> c$. For each $k$ let $b_k$ have $d(b_k, a_k) < 1/k$ and $d(f(b_k),g(a_k)) < c/2$. Then I claim that $b_k$ converges to $a$ yet $f(b_k)$ does not converge to $g(a)$. This is a contradiction.

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