A Propositon of Riemann Integral

Question

• Let $f$ be integrable on $[a, b],$ and suppose $g$ is a function on $[a, b]$ such that

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

except for finitely many $x$ in $[a, b] .$ Show that $g$ is integrable and

$$ \int_{a}^{b} f=\int_{a}^{b} g $$.

How can I show, can you help?

There is a solution here (but I couldn't understand, the solution is very complicated.).

Answer 1

Because $f-g=0$ a.e., $$ \int (f-g) = 0 \text{.} $$ Then $$ \int f = \int (f-0) = \int (f - (f-g)) = \int g \text{,} $$ showing $g$ is integrable by showing to what it integrates.

Answer 2

Suppose $f$ is integrable and $f(x) = g(x)$ except at one point $c$.

Since $f$ is integrable, it is bounded by some $B$. For simplicity, assume that $c$ is in the interior of the interval, the edge cases are similar.

Let $B'=\max(B,|g(c)|)$.

Pick some $\epsilon>0$ and choose points $c_1,c_2$ in the interior of the interval such that $c_1 < c < c_2$ and $2B'(c_2-c_1) < \epsilon$. Now choose a partition $P$ such that $U(f,P)-L(f,P) < \epsilon$, then let $P'$ be the $P$ partition with the points $c_1,c_2$ added. We note that $|L(f,P')-L(g,P')| < \epsilon$, $|U(f,P')-U(g,P')| < \epsilon$ and $U(f,P')-L(f,P') < \epsilon$. Hence we have \begin{eqnarray} U(g,P')-L(g,P') & \le & U(g,P')-U(f,P') + U(f,P')-L(g,P') \\ & < &\epsilon + U(f,P')-L(g,P') \\ & = &\epsilon + U(f,P')-L(f,P')+L(f,P')-L(g,P') \\ & < & 3 \epsilon \end{eqnarray} It follows that $g$ is integrable. Furthermore, since we can always refine a partition $P$ such that $|L(f,P')-L(g,P')|$ is arbitrarily small, it follows that $\int g = \sup_P L(g,P) = \sup_P L(f,P) = \int f$.

If $f,g$ differ at a finite number of points we can use the above analysis (changing $f$ one point at a time) to show that $\int f = \int g$.