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Given a function $f :[a,b]\to \mathbb{R}$, let $g :[a,b] \to \mathbb{R}$ be a function that differ from $f$ at finitely many points. We need to show that the Riemann integral of $g$ is $\int_a^b f$.

I understand the proof for when there is only a single point in which the two function differ. I don't know how to use induction to prove for the case of more than 1 point in which $f$ and $g$ differ.

Alejandro Tolcachier
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user 6663629
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    Divide the interval of integration into finitely many segments, each of which contains one of the points where they differ. – saulspatz Jan 26 '20 at 20:21

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Let $\{a_1,\ldots,a_n\}$ the points that $g$ differs to $f$. I'll assume you know the result when there is only one point of difference.

Let $f_1=\begin{cases} f(x)& x\neq a_1 \\ g(x) & x=a_1 \end{cases}$, then the integrals of $f_1$ and $f$ coincides because they differ at only one point.

Let $f_2=\begin{cases} f(x)& x\neq a_1,a_2\\ g(x)& x=a_1, a_2\end{cases}$, then the integrals of $f_1$ and $f_2$ coincides.

In general, if $f_i=\begin{cases} f(x)& x\neq a_1,\ldots, a_i\\ g(x)& x=a_1,\ldots, a_i\end{cases}$ then the integrals of $f_i$ and $f_{i-1}$ coincides. As $g=f_n$ we are done.

Alejandro Tolcachier
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