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

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!

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 $$\epsilon > 0$$.

– Community Bot
Commented Mar 17 at 3:51
• Thank you!I understand. Commented Mar 21 at 4:51

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.

• Thanks!I have completed the proof! Commented Mar 21 at 4:52
• Did my hint help or did you use some other approach?
– ash
Commented Mar 21 at 16:24