Prove that a function is Riemannian integrable and its Riemannian integral is 0

Question

Let$f:[a,b]\to \mathbb{R},\forall x_{0}\in[a,b],\lim_{x\to x_0}f(x)=0$,Prove:f(x) is Riemannian integrable on [a,b] and its Riemannian integral$\int_a^bf(x)dx=0$
Using the completeness theorem of real numbers，I have proved that $\forall \varepsilon>0,$there exists at most finite $x\in[a,b]$ such that f(x)$\geq\varepsilon $ .Next step should I prove that f(x) is continuous almost everywhere on [a,b]?I hope someone can help me.Thanks!

Answer 1

I feel like you could just compare with $\int_a^b |f|$. For this integral, you have an obvious bound on the lower Darboux integral, and you have basically already shown that upper Darboux sums can be taken less than arbitrary $\epsilon > 0$.

Answer 2

HINT: There's a quick compactness argument you could employ. Let $\varepsilon > 0$ be given. Then for each $x\in [a,b]$, there is a $\delta_x > 0$ such that $|f(y)|< \varepsilon$ whenever $|y-x|< \delta_x$.

Now, cover $[a,b]$ with the family $\{ (x-\delta_x , x+\delta_x ) : x\in [a,b] \}$. Since $[a,b]$ is compact, we can find $x_1 , \ldots , x_n \in [a,b]$ such that $[a,b]\subset \bigcup_{i=1}^{n} (x_i-\delta_{x_i} , x_i +\delta_{x_i})$.

Now, we may assume that none of the $(x_i-\delta_{x_i} , x_i +\delta_{x_i})$ are contained in the other and furthermore that $x_1 < x_2 < \ldots < x_n$. Form a partition out of the points of the form $a, x_i-\delta_{x_i} , x_i +\delta_{x_i}, b$ and see if you can complete the proof.