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Let $f$ be a function with the property that every point of discontinuity is a removable discontinuity, i.e., $\lim_{y \to x}f(y)$ exists for all $x$, but $f$ may be discontinuous at some (even infinitely many) numbers $x$. Define $g(x)= \lim_{y\to x}f(y)$. Prove that $g$ is continuous.

Since $g(a)= \lim_{y\to a}f(y)$, by definition, it follows that for any $\epsilon \gt 0$ there is a $\delta \gt 0$ such that $|f(y)-g(a)|\lt \epsilon$ for $0\lt |y-a| \lt \delta$. This means that

$$g(a)-\epsilon \lt f(y) \lt g(a)+\epsilon$$. for $0\lt |y-a| \lt \delta$.

What I need to show is that if $|x-a| \lt \delta$, we have

$$g(a)-\epsilon \lt \lim_{y\to x}f(y) \lt g(a)+\epsilon$$.

I'm having trouble reaching this conclusion with the information I have. I'd appreciate any solutions or suggestions.

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  • $\begingroup$ A point of discontinuity is just a point at which the function is not continuous. The function g is continuous iff g is continuous at each point. $\endgroup$ – Jacob Wakem Jun 28 '17 at 14:04
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We are trying to show that for any $a$ and any $\epsilon \gt 0$ there is some $\delta \gt 0$ such that $|g(x)-g(a)| \lt \epsilon$ for all $|x-a| \lt \delta$
From the definition of $g$ we know that for $\frac{\epsilon}{2}$ there is $\delta$ such that for all $0 \lt |x-a| \lt \delta$ we have $|f(x)-g(a)| \lt \frac{\epsilon}{2}$
Now if $f$ is continuous at $x$ then $f(x)=g(x)$ and the inequality follows.

Let $x_0,\ \ 0 \lt |x_0-a| \lt \delta$ be a discontinuity point of $f$. Then we know that $x_0$ is removable and that $g(x_0) = \lim_{x \to x_0}f(x)$ and so there is $\delta_2$ such that for $0 \lt |x-x_0| \lt \delta_2$ we have $|f(x)-g(x_0)| \lt \frac{\epsilon}{2}$.
Let $x_1$ be a point such that $|x_1-a| \lt \delta$ and $|x_1-x_0| \lt \delta_2$ therefore: $$ |g(x_0) - g(a)| \le |g(x_0)-f(x_1)| + |f(x_1)-g(a)| \lt \epsilon $$

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    $\begingroup$ I'm sorry but I don't really understand how this helps. Could you explain it a bit more? $\endgroup$ – takecare Feb 22 '15 at 14:50
  • $\begingroup$ For $0 \lt |y-a| \lt \delta$, $f$ is continuous in $y$ and therefore $f(y)=g(y) \implies |f(y)-g(a)| = |g(y)-g(a)| \lt \epsilon$ $\endgroup$ – benji Feb 22 '15 at 15:43
  • $\begingroup$ @benji: Why must $f$ be continuous in a punctured neighbourhood of $a$? $\endgroup$ – Martin R Feb 22 '15 at 18:49
  • $\begingroup$ @user135204 please see my update. earlier I missed the part about infinitely many discontinuities... $\endgroup$ – benji Feb 23 '15 at 23:17
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The Let $a$ be in the domain of $g$ and $\epsilon > 0$. From $g(a)= \lim_{y\to a}f(y)$ is follows – as you already said – that there is a $\delta > 0$ such that

$$ g(a)-\epsilon \lt f(y) \lt g(a)+\epsilon \text{ for all } y \in B_\delta(a) \setminus \{ a \} $$ (where $B_\delta(a) := \{ y \mid |y - a| \lt \epsilon\} $ is the "ball" with radius $\delta$ and center $a$).

Now for each $x \in B_\delta(a) \setminus \{ a \}$ there is a $r > 0$ such that $B_r(x) \subset B_\delta(a) \setminus \{ a \}$ and therefore

$$ g(a)-\epsilon \lt f(y) \lt g(a)+\epsilon \text{ for all } y \in B_r(x)\, . \tag 1 $$ By taking the limit $y \to x$ in $(1)$ it follows that $$ g(a)-\epsilon \le g(x) \le g(a)+\epsilon \, . \tag{2} $$

The last relations are trivially true for $x = a$. So for each $\epsilon > 0$ we have found a $\delta > 0$ such that $(2)$ holds for all $x \in B_\delta(a) $. This proves the continuity of $g$ at $a$.

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    $\begingroup$ I don't fully understand the last inequality. Intuitively it's clear that taking the limit $y\to x$ we would get the inequality, but I don't know how to show rigorously that $g(a)-\epsilon \le \lim_{y\to x}f(y) \le g(a)+\epsilon$. Could you please explain this part more clearly? $\endgroup$ – takecare Feb 24 '15 at 12:07
  • $\begingroup$ @user135204: From $f(y) < K$ for all $y$ in a neighbourhood of $x$ is follows that $\lim_{y \to x} f(x) \le K$. It is the same as with sequences: If $a_n < K$ for all $n$ then $\lim_{n \to \infty} a_n \le K$ (if the limit exists). – (You can of course prove that with $\epsilon$ and $\delta$, but I don't want to repeat what benji added to his answer.) $\endgroup$ – Martin R Feb 24 '15 at 12:10
  • $\begingroup$ The last inequality follows from the limit inequality theorem. milefoot.com/math/calculus/limits/LimitInequalityProofs05.htm $\endgroup$ – The Cryptic Cat Jun 30 '17 at 5:23
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Here is what we know:

$$ g(x)= \lim_{y\to x}f(y) \tag{1} $$ $$ g(a)-\epsilon \lt f(y) \lt g(a)+\epsilon \text{ for } 0\lt |y-a| \lt \delta \tag{2} $$ $$ |x-a| \lt \delta \tag{3} $$

What we want to show:

$$ g(a)-\epsilon \lt \lim_{y\to x}f(y) \lt g(a)+\epsilon \tag{4} $$

From $(1)$ it follows that for any $\epsilon \gt 0$ there is a $\delta \gt 0$ such that $|f(y)-g(x)|\lt \epsilon$ for $0\lt |y-x| \lt \delta$. But $|f(y)-g(x)| = |g(x)-f(y)|$. This means that:

$$ f(y)-\epsilon \lt g(x) \lt f(y)+\epsilon \text{ for } 0\lt |y-x| \lt \delta \tag{5} $$

From $(5)$, by $(1)$, it follows that:

$$ f(y)-\epsilon \lt \lim_{y\to x}f(y) \lt f(y)+\epsilon \text{ for } 0\lt |y-x| \lt \delta \tag{6} $$

Now where $f$ is continuous we have $\lim_{y\to x}f(y) = f(x) = g(x) \implies g(x) \le f(x) \le g(x)$, respectively $g(x)-\epsilon \le f(x) -\epsilon$ and $f(x) +\epsilon \le g(x)+\epsilon$. By this, in conjunction with $(2)$ and $(6)$, it follows that for $|y-a| \lt \delta$ and $0\lt |y-x| \lt \delta$:

$$ g(a)-\epsilon \le f(y)-\epsilon \lt \lim_{y\to x}f(y) \lt f(y)+\epsilon \le g(a)+\epsilon \tag{7} $$

Which implies the desired result, $(4)$.

By $(5)$, given $(3)$, this also implies that:

$$ g(a)-\epsilon \le g(x) \le g(a)+\epsilon \tag{8} $$

which shows that $|g(x)-g(a)| \le \epsilon$ for all $x$ satisfying $|x-a| \lt \delta$. Thus $g$ is continuous.

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Assume for simplicity that $f$ is defined on an open set $\Omega\subset{\mathbb R}^n$, and consider a point $a\in \Omega$. Let an $\epsilon>0$ be given. By definition of $g$ there is a $\delta>0$ such that $$|f(x)-g(a)|<\epsilon\qquad\forall\>x\in \dot U_\delta(a)\tag{1}\ ,$$ whereby $U_\delta(a)$ denotes the $\delta$-neighborhood of $a$, and the dot removes the point $a$ from this neighborhood. Consider an arbitrary point $x_0\in\dot U_\delta(a)$. By definition of $g$ there is a $\delta'>0$ such that $$|f(x)-g(x_0)|<\epsilon\qquad \forall\>x\in \dot U_{\delta'}(x_0)\ .\tag{2}$$ Choose a point $x$ on the open segment $\>]a,x_0[\>$ in $(2)$ and then obtain, using $(1)$, that $$|g(x_0)-g(a)|<2\epsilon\ .$$ Since this is true for all $x_0\in\dot U_\delta(a)$ we are done.

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  • $\begingroup$ Could you elaborate on how $|g(x)-g(a)|< 2\epsilon$ can be obtained from $(2)$ using $(1)$? I suppose you somehow derive the inequality $|g(x) - g(a)| \le |f(x)-g(a)| + |f(t)-g(x)| \lt 2\epsilon$ (as in the answer by user @benji). But I don't see how it holds in all cases. P.S. This is problem 17.d. in chapter 6 of Spivak's Calculus. Further, I believe to understand the notation $U_\delta(a)$ but would nonetheless appreciate a link that explains it in more detail. Spivak does not use it (at least not until chapter 7). $\endgroup$ – user245312 Jun 28 '17 at 16:41
  • $\begingroup$ @user245312: See my edit. $\endgroup$ – Christian Blatter Jun 28 '17 at 18:05

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