Showing that g is integrable and $\int^b_a{f}$ = $\int^b_a{g}$

Solution 1:

First of all you may prove the result for the case where there is only one point say $c$ where the f and g have different values.

Note that in that case for any partition of [a,b] the difference between Riemman Sum of $f$ and the Riemman Sum for $g$ is at most $$|f(c)-g(c)|\times \delta$$ where $\delta$ is the maximum length of your subpartition intervals.

Now you pick a partition such that $$ \delta <\frac {\epsilon}{|f(c)-g(c)|}$$

The rest is easy to complete.

Solution 2:

Assuming $f\ne g$ so that $S=\{x: f(x)\ne g(x)\}$ is finite and not empty. Let $M=\max \{|f(x)-g(x)|:x\in S\}.$ Let $|S|$ be the number of members of $S.$

For a partition $P$ let $\|P\|$ be the length of the largest interval belonging to $P$. For any $\epsilon >0$ there exists $\delta_{\epsilon}>0$ such that $\forall P\;(\|P\|<\delta_{\epsilon} \implies U(f,P)-L(f,P)<\epsilon/2).$

So if $\|P\|< \delta_{\epsilon}\cdot \min (1, \epsilon(M|S|)^{-1}/2)$ then $|U(f,P)-U(g,P)|<\epsilon$ and $|L(f,P)-L(g,P)|<\epsilon.$

The idea is simply that, since $S$ is finite, the contribution, ( to a finite sum that approximates the integral of $f$ or of $g$ ), of intervals containing members of $S, $ will be negligible if $\|P\|$ is small enough.

Solution 3:

Put $h (x)=f (x)-g (x) $ and let us prove that $h $ is integrable at $[a,b] $

Assume that $$\forall x\in (a,b] \; h (x)=0$$ and $$h (a)=c>0.$$

Let $\epsilon>0$ enough small given. $h $ is continuous at $[a+\frac {\epsilon}{2c},b]$ and then integrable.

there exist a partition $\sigma_1$ of $[a+\frac {\epsilon}{2c},b] $ such that

$$U (h,\sigma_1)-L (h,\sigma_1)<\frac {\epsilon}{2}$$

put now $$\sigma=\sigma_1\cup \{0\}$$

$\sigma $ is a partition of $[a,b] $

now $$U (h,\sigma)-L (h,\sigma)=$$ $$U (h,\sigma_1)-L (h,\sigma_1)+c\frac {\epsilon}{2c} <\epsilon.$$ This proves that $h $ is integrable at $[a,b] $.

$g=f-h $ is integrable since $f $ and $h $ are integrables at $[a,b] $.