If $f$ is integrable on $[a,b]$ and $g$ is a function on $[a,b]$ . Show $g$ is integrable and $\int_a^b f(x) = \int_a^b g(x)$.

Question

Problem:

If $f$ is integrable on $[a,b]$ and $g$ is a function on $[a,b]]$ so that $f(x) = g(x)$ except for finitely many $x \in [a,b]$. Show $g$ is integrable and $\int_a^b f(x) = \int_a^b g(x)$.

I have looked at the solution from this page.

Where they use induction to prove this, such as this:

However my question is how do they arrive at $$t_k - t_{k-1}= \frac{\epsilon}{12B}$$ and $$|U(g,p) - U(f,p)| \le 2[B-(-B)]*\max \{ t_k - t_{k-1}\} < \frac{\epsilon}{3} $$

Thank you very much.

Answer 1

For $|U(g,p) - U(f,p)| \le 2[B-(-B)]*\max \{ t_k - t_{k-1}\} < \frac{\epsilon}{3} $:

Remember that $f=g$ except for exactly one point. The worst case is that precisely this point is one of the points of the partition $P$. In this case, by the definition of upper sum, $U(f,P)$ and $U(g,P)$ only differes on what hapepens in these two intervals, so $U(f,P)-U(g,P)$ calcels except on AT MOST, two (consecutive) intervals. So

$$U(f,P)-U(g,P)=(M_k(f)-M_k(g))(t_k-t_{k-1})+(M_{k+1}(f)-M_{k+1}(g))(t_{k+1}-t_{k}), $$ where $M_j(f)$ is the max of $f$ in the $j$-th interval (and the same for $g$).

Can you finish from here?