Suppose interval $I\in[a,b]$ as finitely many limit points, $f:[a,b]\to \mathbb{R} $ is bounded on $[a,b]$ and continuous on $[a,b]\setminus I$. Use the fact that if $f$ is continuous except at finitely many points in an interval, $f$ is integrable on that interval.

I can use Darboux and Reimann integration theorems and definitions.

I believe the correct way to start is to partition the accumulation points into intervals, but I'm not sure the correct way to set this up. I also know that given the hypothesis, there are infinitely many discontinuities on $[a,b]$, but I m also not sure how to prove that. Any help is appreciated, thank you!

  • $\begingroup$ What about the Dirichlet function? On any interval it has finitely many limit points, yet is not Riemann integrable on any interval by Lebesgue. $\endgroup$ – Melody Dec 9 '18 at 19:36
  • $\begingroup$ Part of the hypothesis I have is that $f$ is continuous on $[a,b]\setminus I$. It's a homework question, so I'm assuming it is true! $\endgroup$ – t.perez Dec 11 '18 at 2:53
  • $\begingroup$ If $f$ integral on $[a,b]$, then it is integrable on every subinterval, hence if $f$ so for a counterexample we can assume WLOG $I=[a,b]$. Alternatively, if the subinterval must be proper, just take $f$ to be the Dirichlet function on $I$, and 0 everywhere else. Then $f$ still has finitelt many limit points, but uncountable discontinuities. $\endgroup$ – Melody Dec 11 '18 at 5:08

The wording of the question is a bit confusing. The correct result is the following

Theorem: Let the function $f:[a, b]\to\mathbb{R} $ be bounded on $[a, b] $ and $D$ be the set of its discontinuities on $[a, b] $. If $D$ has a finite number of accumulation points then $f$ is Riemann integrable on $[a, b] $.

For those familiar with basic measure theory note that $D$ is a set of measure zero and hence by Lebesgue's criterion $f$ is Riemann integrable on $[a, b] $.

However the result can be proved using basic theorems on Riemann integration. One can partition the interval $[a, b] $ into a finite number of subintervals such that each subinterval contains only one accumulation point of $D$ and further the accumulation point is an end point of that subinterval.

This reduces the problem to the case when $D$ has only one limit point $a$ (or $b$ and this case is handled similarly). Let $\epsilon >0$ be arbitrary. If $M$ is positive upper bound for $|f|$ on $[a, b] $ then we can choose $c=\min(b, a+(\epsilon/4M))$. Since $a$ is the only limit point of $D$ in $[a, b] $ the interval $[c, b] $ contains only finitely many points of $D$. Then $f$ is Riemann integrable on $[c, b] $ and hence there is a partition $P'$ of $[c, b] $ for which $$U(P',f)-L(P',f)<\frac{\epsilon} {2}$$ Let $P=P'\cup\{a\} $ so that $P$ is a partition of $[a, b] $ and then we have $$U(P, f) - L(P, f) <2M\cdot \frac{\epsilon} {4M}+\frac{\epsilon}{2}=\epsilon$$ and therefore $f$ is Riemann integrable on $[a, b] $.


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