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Suppose $f : [ 0, 1 ] → \mathbb{R}$ and $\lim\limits_{x \rightarrow c} f ( x )$ exists for all $c \in [ a , b ]$. Show that $f$ is Riemann integrable on $[ a , b ]$.

I can show this $f$ is bounded on $[a,b]$ since the limit exists at every point. But I'm not sure how to proceed. Any hints or help would be great.

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  • $\begingroup$ You should replace $[a,b]$ with $[-1,1]$ or vice versa. $\endgroup$ – Haris Gusic Feb 20 at 4:07
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Show that such $f$ with a limit that exists on every point of interval $[a,b]$ is discontinuous in an at most countable set (it is proved here on this site). The main idea is showing that the set $\{x : f(x) \neq g(x)\}$, where $g(y) = \lim_{y \rightarrow x} f(y)$, is at most countable.

Then, apply the Lebesgue criterion to show that the function is integrable since countable sets have measure zero.

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  • $\begingroup$ The question asks for Riemann integrability, so the Lebesgue criterion doesn't apply. $\endgroup$ – TonyK Feb 21 at 21:34
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    $\begingroup$ @TonyK, the Lebesgue criterion is for Riemann integrability. See en.wikipedia.org/wiki/Riemann_integral middle of page. $\endgroup$ – twnly Feb 21 at 21:41
  • $\begingroup$ You are right, of course. Sorry. $\endgroup$ – TonyK Feb 21 at 22:08
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Sketch of a more elementary approach: Let $\epsilon > 0.$ For every $x\in [0,1]$ there is an interval $I_x$ centered at $x$ where

$$\sup_{I\setminus \{x\}} f - \inf_{I\setminus \{x\}}< \epsilon/2.$$

The reason we need to remove $x$ from $I_x$ is because $\lim_{\,t\to x} f(t)$ pays no attention to the value of $f$ at $x.$

Because $[0,1]$ is compact, finitely many of these intervals cover $[0,1].$ Let's label them $I_{x_k},$ $k=1,2,\dots n.$ The $x_k$ form a partition of $[0,1],$ but we need to tweak this partition because the values of the $f(x_k)$ could be a bit wild. However, as you noted, $|f|$ is bounded by some $M.$ So for each $k$ we can choose $y_k<x_k<z_k$ very close to $x_k$ and then let $P$ be the partition formed by throwing in all the $x_k,y_k,z_k.$ For this partition, we will have

$$U(f,P)-L(f,P)<\epsilon,$$

which implies $f$ is Riemann integrable on $[0,1].$

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  • $\begingroup$ Can you explain a bit more why we need to omit $x \in I$. I used a similar argument to prove $f$ is not bounded, but I did not omit $x$. Also, I'm not quite sure where your first inequality came from @zhw. $\endgroup$ – user439126 Feb 25 at 22:15
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    $\begingroup$ It's possible for $f$ to have limit $0$ at $x$ while $f(x)=10.$ In other words, the value $f(x)$ is irrelevant in the limit process. In my first inequality, I am using the fact that the limit exists at $x.$ All values in $I_x\setminus \{0\}$ are close to the limit, hence close to each other. $\endgroup$ – zhw. Feb 26 at 23:39
  • $\begingroup$ I see, thank you. Now I am stuck on $\sup f(x) - \inf f(x) < \frac{\varepsilon}{2}$ in $[y_k, x_k]$ or $[x_k,z_k]$. How does $f$ being bounded and $y_k$ and $z_k$ being very close to $x_k$ help? @zhw. $\endgroup$ – user439126 Feb 26 at 23:50
  • $\begingroup$ Also, when you say all values in $I_x \setminus \{x\}$ are close to the limit, do you mean all $f(y)$ for $y \in I_x \setminus \{x\}$ are close to the limit? @zhw. $\endgroup$ – user439126 Feb 26 at 23:56
  • $\begingroup$ For the first question, note that $\inf (L-\epsilon,L+\epsilon)=L-\epsilon $,$\sup (L-\epsilon,L+\epsilon)=L+\epsilon .$ So $\sup - \inf = 2\epsilon.$ Yes on the second question. $\endgroup$ – zhw. Feb 28 at 17:16

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