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Let $f : [a, b] \to \mathbb{R}$ be a Riemann integrable function. If $g : [a, b] \to \mathbb{R}$ is another function and $S = \lbrace x : f(x) \neq g(x)\rbrace$ contains exactly $n$ points, show that $g$ is also Riemann integrable from any of the equivalent definitions of Riemann integrability. Can anyone give me some hints?

Theo Bendit
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Walls
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    One of the equivalent definitions is that a bounded function is Riemann integrable iff it is continuous almost everywhere, i.e., everywhere except on a set of measure zero. – Davey Jul 18 '18 at 21:02
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    The laziest could be Lebesgue's criterion. You only need to realize that $S$, being finite, is of measure zero. Therefore, the points of discontinuity of $g$ are those of $f$ changed by possibly a subset of $S$, which results in a set of measure zero. –  Jul 18 '18 at 21:02
  • These n points have to be discontinuous, and also then each function can be thought of as covering same area except since point discontinuities don't affect area over x-axis...hope this helps. – Pi_die_die Jul 18 '18 at 21:03
  • One of my concerns is that what dose the set S imply? – Walls Jul 18 '18 at 21:07

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Let $S:=\{x_{1},x_{2},\cdots,x_{n}\}$. I will suppose below that $a<x_{1}$ and $x_{n}<b$. Clearly, $S$ is isolated. Thus, we can find $\delta>0$ such that $x_{k+1}-x_{k}>2\delta$. We may also suppose that $x_{1}-a>\delta$ and $b-x_{n}>\delta$. Now, consider the following. \begin{align} \int_{a}^{b}f(u)\mathrm{d}u ={}&\int_{a}^{x_{1}-\delta}f(u)\mathrm{d}u+\int_{x_{1}+\delta}^{x_{2}-\delta}f(u)\mathrm{d}u+\cdots+\int_{x_{n-1}+\delta}^{x_{n}-\delta}f(u)\mathrm{d}u+\int_{x_{n}+\delta}^{b}f(u)\mathrm{d}u\\ {}&+\int_{x_{1}-\delta}^{x_{1}+\delta}f(u)\mathrm{d}u+\cdots+\int_{x_{n}-\delta}^{x_{n}+\delta}f(u)\mathrm{d}u\\ ={}&\int_{a}^{x_{1}-\delta}g(u)\mathrm{d}u+\int_{x_{1}+\delta}^{x_{2}-\delta}g(u)\mathrm{d}u+\cdots+\int_{x_{n-1}+\delta}^{x_{n}-\delta}g(u)\mathrm{d}u+\int_{x_{n}+\delta}^{b}g(u)\mathrm{d}u\tag{1}\label{eq1}\\ {}&+\int_{x_{1}-\delta}^{x_{1}+\delta}f(u)\mathrm{d}u+\cdots+\int_{x_{n}-\delta}^{x_{n}+\delta}f(u)\mathrm{d}u.\tag{2}\label{eq2} \end{align} Note that as $\delta\to0$, each term in \eqref{eq2} tends to $0$, and \eqref{eq1} becomes $\int_{a}^{b}g(u)\mathrm{d}u$.

bkarpuz
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