# Proof that Riemann-integrability is preserved by changing the function at a converging sequence of points

Let $$(x_n)^{\infty}_{n=1}\in[a,b]$$ such that $$\lim\limits_{n\to\infty}x_n=l$$. Let $$g:[a,b]\to \mathbb{R}$$ be bounded function. Assume that there exists a Riemann integrable function $$f:[a,b]\to \mathbb{R}$$ with the following property

$$f(x)=g(x)$$

for every $$x\in[a,b]$$, except maybe $$x_1,x_2,x_3,...$$

Then, $$g$$ is a Riemann integrable function.

Any help will be appreciated.

• Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. Apr 27, 2013 at 11:31
• @Lord_Farin, i give you a privilege editing my "question". Apr 27, 2013 at 11:51
• Show that in fact that $\int_a^b f(x)\,dx=\int_a^b g(x)\,dx$. Towards this end, let $\epsilon>0$. Construct a Riemann sum for $g$ that differs from $\int_a^b f(x)\, dx$ by at most $\epsilon$. Towards this end, choose a subinterval $I$ of the partition to contain $l$ and have small width. Then choose subintervals of small width for the finitely many $x_n$ that do not belong to $I$. Fill in the rest of the partition with subintervals of small width. If you make those small widths small enough, the Riemann sum will be within $\epsilon$ of $\int_a^b f(x)\,dx$. Apr 27, 2013 at 12:00
• @Sandra As far as I can improve the question, I have. But I can't provide the context it was encountered in and your attempts. Nobody but you can provide those. Apr 27, 2013 at 12:20
• @Lord_Farin: You are awesome! Apr 27, 2013 at 12:27

## 1 Answer

By Lebesgue theorem, a bounded function $f$ on a segment is a Riemann integrable iff the set of the discontinuity points of $f$ has the Lebesgue measure $0$.