prove that function is integrable on uncountable discontinous interval 
Prove that if f is bounded on [0, 1] and continuous on [0, 1] \ {$  \frac{1}{n}: n \in \mathbb{N}  $}, then f is integrable on [0, 1]

I choose N such that $\frac{1}{N} $ < $\epsilon $. Then f is integrable on [1/N, 1] because [1/N, 1] has countable discontinuous interval. But I don't know how to prove integrable on  the interval between [0, 1/N]. 
can anybody give me a hint?
 A: Take $\varepsilon>0$. Let $L>0$ be such that $(\forall x\in[0,1]):\bigl|f(x)\bigr|<L$. In $\left[\frac\varepsilon{4L},1\right]$, $f$ is integrable since it has finitely many points of discontinuity. Therefore, there is a partition $P$ of $\left[\frac\varepsilon{2L},1\right]$ such that$$\overline\sum\left(f|_{\left[\frac\varepsilon{4L},1\right]},P\right)-\underline\sum\left(f|_{\left[\frac\varepsilon{4L},1\right]},P\right)<\frac\varepsilon2.$$So, if $P^*=\{0\}\cup P$, $\overline\sum(f,P^*)-\underline\sum(f,P^*)<\varepsilon$.
A: Using Lebesgue integral
$f$ has a countable set $S = \{1/n ; n \in \mathbb N\}$ of discontinuities and vanishes elsewhere, hence $S$ Lebesgue measure is vanishing and $f$ is Lebesgue integrable with $\int_0^1 f = 0$.
Using Riemann integral
The functions
$$s_k(x) = \begin{cases}
0 & 0 \le x < 1/k \\
1 & 1/ k \le x \le 1 \cap S\\
0 & 1/ k \le x \le 1 \cap [0,1] \setminus S
\end{cases}$$
and
$$S_k(x) = \begin{cases}
1 & 0 \le x < 1/k \\
1 & 1/ k \le x \le 1 \cap S\\
0 & 1/ k \le x \le 1 \cap [0,1] \setminus S
\end{cases}$$
are step functions such that $s_k \le f \le S_k$ for all $k \in \mathbb N$. Moreover $0 \le \int_0^1 \left(S_k-s_k \right) \le 1/k$ converges to $0$ with $k$.
Therefore $f$ is Riemann integrable and $\int_0^1 f = 0$ as $\lim\limits_{k \to \infty} \int_0^1 s_k = \lim\limits_{k \to \infty} \int_0^1 S_k = 0$.
