$\int_0^1f(x)dx= \lim_{c\rightarrow0}\int_c^1f(x)dx$ I am reading Rudin's "Priciples of Mathematical Analysis", I am trying to solve exercise 6.7 which says:

Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathcal R$ on $[c, 1]$ for every $c > 0$.
Define:
$$\int_0^1f(x)dx= \lim_{c\rightarrow0}\int_c^1f(x)dx$$
if this limit exists (and is finite).
(a) If $f \in \mathcal R$ on $[0, 1]$, show that this definition of the integral agrees with the old one.

My solution is:
Since $f \in \mathcal R$, we know that $f$ is bounded, let $|f| \leq M$.
Let $c < \epsilon / M$.
$$\left| \int_0^1f(x)dx - \int_c^1f(x)dx \right| = \left| \int_0^cf(x)dx \right|$$
$$\leq M.c = \epsilon$$
And we are done.
Is there is a flaw in this solution?
 A: Expanding on the comment, to complete this proof, you also require $f$ to be Riemann Integrable on $[0,1]$. To observe this, we will use the Riemann's Criteria of Integrability.
Let $\epsilon >0$. Choose $c = \frac{\epsilon}{2(M-m)}$, where $M$ and $m$ are the maximum and minimum values of the function $f$ on the interval $[0,1]$.
Note that since $f \in \mathcal R[c,1]$, for this $\epsilon$, $\exists \ P=\{c,x_1,x_2....1\}$ such that
$$U(P,f) - L(P,f) < \frac{\epsilon}{2}$$
Now, consider the partition $P_c = \{0,c,x_1,x_2,...1\}$ of $[0,1]$ (i.e. the partition $P$ with the additional point $0$) Corresponding to this partition we have:
$$U(P_c,f) - L(P_c,f) = \sum_{i=1}^{n} (M_i-m_i)\Delta x_i$$
Suppose the maximum and minimum values of $f$ on $[0,c]$ are $M_1$ and $m_1$ respectively. We can therefore split the summation as:
$$U(P_c,f) - L(P_c,f) = (M_1-m_1)c + \sum_{i=2}^{n} (M_i-m_i)\Delta x_i$$
Note that $M_1-m_1 \leq M-m$. Also, the sum $\sum_{i=2}^{n} (M_i-m_i)\Delta x_i$ is nothing but $U(P,f) - L(P,f)$. Thus,
$$U(P_c,f) - L(P_c,f) < (M-m)c + U(P,f) - L(P,f) < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon$$
Thus, $f \in \mathcal R  \blacksquare$ and you can proceed as you did in your solution.
