Proving integral with finite number of values greater than any small number is $0$ I was given this question and think it is mostly wrong,there were also more mistakes than what I am going to write, but I would like to know how wrong it is.So would there be any validity to this approach, or is it completely incorrect?
Question Suppose that $f:[0,1]\rightarrow [0,1]$ satisfies the following properties:given any $\epsilon>0,f(x)<\epsilon$ except at finitely many points, prove $\int_{0}^{1}f=0$
My attempt:
Let $T=\{y_1,...,y_m\}$ be the set of points such that $f(x) \geq \epsilon$ let $P=\{x_0,...,x_n\}$ be a partition of $[0,1]$.
Let $A=\{k|[x_{k-1},x_k] \cap T\neq \emptyset\}$
Let $B=\{k|[x_{k-1},x_k]\cap T=\emptyset\}$
Then there are at most $2m$ intervals having endpoints with indices in $A$ containing points in $T$. Let $P*$ be a partition containing these intervals such that
$\Delta x_k\leq\frac{\epsilon}{4m}$
Then $U(f,P*)=\sum_{k \in A}\text{sup}\{f(x)|x \in [x_{k-1},x_k]\}\Delta x_k<\frac{\epsilon}{2}$
similarly $L(f,P*)=\sum_{k \in A}\text{inf}\{f(x)|x \in [x_{k-1},x_k]\}\Delta x_k<\frac{\epsilon}{2}$
Now consider a partition $Q$ of intervals with endpoints with indices in $B$.
Let $\Delta x_k<\frac{1}{2(n-2m)}$ Then since $f(x)<\epsilon$ for $f$ on intervals having endpoints with indices in $B$
$U(f,Q)<\frac{\epsilon}{2}$ and $L(f,Q)<\frac{\epsilon}{2}$
Then $L(f,Q)+L(f,P*)<\epsilon$ and $U(f,Q)+U(f,P*)<\epsilon$
My question is was this on the right track of a correct approach?Is there a different way to approach this? When I saw the question I felt a similarity between this question and the proof that thomae's function is integrable, but I do not think I studied/understood that proof well enough, to successfully come up for a proof for this. And I realized in that proof, that didn't come up with separate partitions for intervals having endpoints with indices in $A,B$ like I did.
 A: For each $n$ define $s_n=(x,f(x)\ge \frac{1}{n})$.  Hypothesis $s_n$ is finite.  Therefore $S=\cup_ns_n$ consisting of all points where $f(x) \gt 0 $ is countable.  Countable sets have measure zero, so the measure of the set where $f(x)=0$ is one and the integral $=0$.
Without measure theory: Since $S$ is countable, order the list $(x_k)$ and for each $k$ bracket the point by an interval $(x_k-\frac{\epsilon}{2^k},x_k+\frac{\epsilon}{2^k})$  Union of covering intervals has length $\lt 2\epsilon$. Let $\epsilon \to 0$, so integral over non-zero points $=0$.
A: Your idea is correct and simple. Just deal with $U(P, f) $.
Let $k$ be the number of points where $f(x) \geq\epsilon$. And choose norm of partition less than $\epsilon/2k$. The upper sum on intervals which don't contain these points does not exceed $\epsilon$. And on the intervals which contain these points the upper sum does not exceed $2k(\epsilon /2k)=\epsilon$. Hence the upper sum $U(P, f) $ does not exceed $2\epsilon $.
I am not sure why you chose different $\Delta x_k$ for these different types of intervals. This probably makes your proof a bit difficult to comprehend (at least for me).
You can see that we don't need to deal with lower sum separately because it is already non negative and does not exceed any upper sum.
