Proving that, if a function $f:[a,b] \rightarrow \mathbb{R}$ be zero except at finitely many points, it is Riemann-integratable with integral $0$ :
Let the function attain non-zero values at $x_i
$, where $i={1, 2, 3,…n}$, where $x_i<x_{i+1}$, with respective values being $a_i$.
Let us now split the domain into finite number almost disjoint intervals, $I_1, I_2, …, I_{\omega}$ , thus forming a finite partition of $[a,b]$ . Let us now assume a finite subsequence of intervals, $I_{n_1}, I_{n_2},…, I_{n_k}$, containing the respective non zero values. It is to be noted, with proper "refinement" of partitions, we can make $max\{k\}=n$ [as $n$ is finite].
Further, assume the length of the respective subintervals, containing the numbers be $ \delta_i$, where $i$ runs from $1$ to $k$.
When $k<n$, we have $sup(f)_{I_{n_i}}=max\{a_{t_i}\}=\alpha_{t_i}$ [ where $a_{t_i} \in I_{n_i}$ ] and $inf(f)_{I_{n_i}}=0$. For this type of partition modes, we have $U(f;P)= \sum_{i=1}^{k} \delta_i \alpha_{t_i} $ and $L(f;P)=0$.
As the subintervals containing the points containing $x_i$ becomes fine enough, we can have $x_i \notin I_{n_a} \cap I_{n_b}$ for any $i, a$ and $b$.
Now, we have $U(f;P')= \sum_{i=1}^{n} \delta_i x_i \leq U(f;P) $ with $L(f;P')=0$ as usual.
Now, as the norm of the partition $P'$ approaches $0$, i.e. $ ||P'| | \rightarrow 0$, $\delta_i \rightarrow 0 $. Hence $ \mathbf{inf} \ U(f; P')=0$. So,
$ \mathbf{inf} \ U(f; P')=0= \mathbf{sup} \ L(f; P')$.
QED
Is the above proof adequate and valid?
Best Answer
It seems that you are assuming that $f$ is nonnegative in order to conclude that $L(f,P)=0$ for any partition $P$.
Under that assumption, I interpret $P'$ as a refinement of $P$ and $U(f; P')= \sum_{i=1}^n \delta_i x_i $ should be $U(f; P')= \sum_{i=1}^n \delta_i a_i .$ Also, we have used the notation $\delta_i$ to refer to the width of intervals in both $P$ and $P'$.
A possible direction for this question is to prove that if $g$ is integrable on $[a,b]$ and it differs from $h$ by $1$ point, then $h$ is integrable and share the same value. To handle the case of finitely many points, we just have to use induction.