# Prove $f$ is integrable on $[a,b]$ if $f$ has finitely many accumulation points on $[a,b]$.

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!

• What about the Dirichlet function? On any interval it has finitely many limit points, yet is not Riemann integrable on any interval by Lebesgue. – Melody Dec 9 '18 at 19:36
• 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! – t.perez Dec 11 '18 at 2:53
• 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. – Melody Dec 11 '18 at 5:08

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 $$[a, b]$$.