# Proving Riemann integral does not change when finite values of a function is changed.

I know how to prove that the Riemann integral of a function does not change if one point of the function is changed. However, extending that result to a finite set by use of induction is something I have struggled to prove. I just need a hint as to how I should start off the proof. Looking forward to an exchange of ideas. An answer I found for a single point is given below.

#### Solutions Collecting From Web of "Proving Riemann integral does not change when finite values of a function is changed."

Now, let $f_1$ be equal to $f_0$, except at $x_1$; you know that the integral of $f_1$ is the same as that of $f_0$ by the above argument (so then it is also the same as $f$!).

Then, let $f_2$ be equal to $f_1$, except at $x_2$, etc., etc.

Any finite sequence of changes can be decomposed into a sequence of changes one point at a time. The first of these does not change the value of the integral. Then the second doesn’t either by the same theorem.

In other words, you assume that n changes have been made, then because you know that a single point being changed (i.e from n changes to n+1) doesn’t affect the integral, n+1 changes will also have no effect.