Lef $f$ be a nonnegative real-valued function on $[0, 1]$ that is Riemann integrable on $[0, 1]$ with integral $I \in \mathbb R$. Show that if $f=0$ on $[0, 1]\backslash D$ for some countable subset $D$ of $[0, 1]$, then $I=0$.
Note: Do NOT use Lebesgue integration here. I already know that $f=0$ a.e. here hence $I=0$ is the straightforward result. What I want to know is how to prove the same result using the definition and properties of Riemann integrability.
My attempt: I tried a special case when $D= \left\{\dfrac 1n : n\in \mathbb R\right\}$ as follows:
Since $f$ is Riemann integrable, $f$ is bounded and let $M$ be an upper bound of $f$. Now consider a partition $P=\{0=x_0, x_1, ..., x_n=1\}$. Then for given $\epsilon>0$ and $x_1<\epsilon/2M$, all but finitely many elements of $D$ belong to the interval $[x_0, x_1]$. Then $M_1 = \mathrm {sup}\{f(x): x\in [x_0, x_1]\}\le M$ and $m_1=\mathrm{inf}\{f(x): x\in [x_0, x_1]\}=0$. Now $1, 1/2, ..., 1/N$ be $N$ elements which are not in $[x_0, x_1]$. Then construct a partition $P=\{0=x_0, x_1, ..., x_n=1\}$ such that $||P||=\sup_i\{x_{i+1}-x_i\}<\epsilon/2MN$ and $x_i \neq \dfrac1m$ for any $i$ and $m$. Then there are exactly $N$ number of $i$'s such that $\dfrac 1m\in [x_i, x_i+1]$. Therefore,\begin{align*}U(f, P)-L(f, P) &=M_1 \times \dfrac{\epsilon}{2M}+ \sum_{i=1}^{n}(M_{i+1}-m_{i+1})(x_{i+1}-x_{i})\\&<M \times \dfrac{\epsilon}{2M}+\sum_{i=1}^nM \dfrac{\epsilon}{2MN}\\&= \frac{\epsilon}2+ MN\dfrac{\epsilon}{2MN}\\&=\epsilon\end{align*}
Therefore, $f$ is Riemann integrable.
However, I have no idea how to generalize this proof. In this case we have only one limit point, $x=0$, but we cannot guarantee that there are only finitely many limit points of $D$ in general. Does anyone have other ideas?
Thank you for your help!
hint
Let $\epsilon>0$ given.
$f $ is integrable at $[0,1] \implies \exists \delta>0 \;:$
for any partition $P=(x_0,x_1,...x_n)$ with $\|P\|=\max_{0\le i \le n-1} (x_{i+1}-x_i)<\delta $
all the Riemann sums $S=\sum_{i=0}^{n-1}f (\xi_i)(x_{i+1}-x_i) $, satisfy
$$|\int_0^1 f-S|<\epsilon. $$
You can choose $\xi_i \in [x_{i+1},x_i]\backslash D$ such that $f (\xi_i)=0$.
hence $|\int_0^1f -0|<\epsilon $.
You are nearly done.