Exercise 6.7 of Baby Rudin 
Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c, 1]$ for every $c > 0$. Define
$$ \int_0^1 f(x) \ \mathrm{d} x = \lim_{c \to 0} \int_c^1 f(x) \ \mathrm{d} x $$
if this limit exists (and is finite).
(a) If $f \in \mathscr{R}$ on $[0, 1]$, show that this definition of the integral agrees with the old one.

After some clarification Clarification On Exercise 7 of Chapter 6 of Baby Rudin.
My attempt: To prove: If $f \in \mathscr{R}$ on $[a,b]$, then $ \int_0^1 f(x) \ \mathrm{d} x = \lim_{c \to 0} \int_c^1 f(x) \ \mathrm{d} x $. Since $f \in \mathscr{R}$ on $[0,1]$, $f \in \mathscr{R}$ on $[0,c]$ and $[c,1]$, $\forall c \in [0,1]$. Let $F(c)= \int_c^1 f(x) \ \mathrm{d} x $, for $c \in [0,1]$. By theorem 6.20, $F(c)$ is continuous on $c \in [0,1]$. Since it (domain of $F$) is closed, it contains all of it’s limit point. By theorem 4.6, $\lim_{c \to 0} F(c)= \lim_{c \to 0} \int_c^1 f \ \mathrm{d} x = F(0)= \int_0^1 f \ \mathrm{d} x$. Now by definition 4.25, $F(0)=F(0+)= \lim_{c \to 0^+} \int_c^1 f \ \mathrm{d} x= \int_0^1 f \ \mathrm{d} x$. Our desired equality.
Is this proof correct?
 A: A) Suppose $f \in \mathcal{R}$ on $[0,1]$. Let $\epsilon >0$. Let $M= \sup\{ |f(x)|: 0 \leq x \leq 1\}$. Fix $c \in (0, \frac{\epsilon}{4M})$, and consider any partition of $[0,1]$ that contains $c$, where the lower and upper Riemann sums are $\sum M_j (t_j - t_{j-1})$ and $\sum m_j (t_j - t_{j-1})$ of $f$ differ by less than $\frac \epsilon 4$. Then the partition of $[c,1]$ formed by the points of this partition that lie in the interval surely has the property that its upper and lower Riemann sum $\sum M'_j(t_j - t_{j-1}) $ and $\sum m'_j (t_j - t_{j-1})$ differ by less than $\frac \epsilon 4$. From above we have
$$\sum M_j (t_j - t_{j-1}) \frac \epsilon 4 < \int_0^1 f(x) dx < \sum m_j (t_j - t_{j-1}) + \frac \epsilon 4$$
and
$$\sum M'_j (t_j - t_{j-1}) \frac \epsilon 4 < \int_c^1 f(x) dx < \sum m'_j (t_j - t_{j-1}) + \frac \epsilon 4$$
Moreover, we have
$$\left|\sum M_j (t_j - t_{j-1}) - \sum M'_j (t_j - t_{j-1}) \right| < \frac \epsilon 4$$
and
$$\left|\sum m_j (t_j - t_{j-1}) - \sum m'_j (t_j - t_{j-1}) \right| < \frac \epsilon 4.$$
From these two inequalities you can see that
$$\left|\int_0^1 f(x) dx - \int_c^1 f(x)dx \right|< \epsilon.$$
B)- let
$$f(x) = (-1)^n (n+1).$$
For $\frac{1}{n+1} < x\leq \frac{1}{n}, n=1,2,...$ Then if $\frac{1}{N+1} \leq \leq \frac{1}{N}$ we have
$$\int_c^1 f(x) dx = (-1)^N (N+1) (\frac 1 N -c) + \sum_{k=1}^{k-1} \frac{(-1)^k}{k}.$$
Since $0\leq \frac{1}{N} -c \leq  \frac{1}{N} - \frac{1}{N+1} = \frac{1}{N (N+1)}$, the first term on the right-hand side tends to zero as c goes to zero, while the sum converges to $\ln 2$. However,
$$\int_c^1 |f(x)| dx = (N+1) (\frac{1}{N}) + \sum_{k=1}^{N-1} \frac{1}{k},$$
although in this case the first term in the right-hand side goes to zero, in contrast, the summation goes to infinity.
