Suppose $f : [a, b] → R$ is bounded and satisfies $f ≥ 0$ and
$$\int_{0}^{\overline{1}}f=0$$
Can we conclude it is integrable? Prove it either way
Without considering the non-negativity of $f$, I was able to do the following:
Let $f(x)=0$ when $x \in \mathbb{Q}$ & $f(x)=-1$ when $x \in \mathbb{Q^c}$
Constructing a partition $p$, suppose $p=\{ 0=x_0<x_1<x_2<...<x_n=1 \}$
Then $U(p,f)= \sum_{r=1}^n M_r\delta _r =0$; $infU(p,f)=0$ i.e $\int_{0}^{\overline{1}}f=0$
But $f$ is not riemann integrable as $L(p,f)=\sum_{r=1}^n m_r \delta_r =-1, \ supL(p,f)=-1.$
Since $\int_{0}^{\overline{1}}f \neq {\int_{\underline{0}}^{1}}f$, $f$ is not riemann integrable.
Is there a way I could modify my solution (substitute different values of $f$) or approach this differently to make it valid for $f\ge 0$?
If $f$ is non-negative, then, for each partition $P$ of $[a,b]$, $L(f,P)\geqslant0$. Therefore, $\underline{\int_a^b}f=0$. Since we are assuming that $\overline{\int_a^b}f=0$, $f$ is integrable.