I want to follow up on my previous question. Based on the comments and responses I got for my previous question, I developed a new proof for Baby Rudin Ch. 6. Ex. 7a. The exercise is:
Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c, 1]$ for every $c>0$. Define \begin{equation}\tag{7.0} \int_0^1 f(x) \, dx = \lim_{c \to 0} \int_{c}^1 f(x) \, dx \end{equation} if this limit exists (and is finite). If $f \in \mathscr{R}$ on $[0, 1]$, show that this definition of the integral agrees with the old one.
My latest attempt:
Firstly, we state a modified version of Theorem 6.12 (c) in the text; call it Theorem 6.12 (c$^*$). Theorem 6.12 (c) states that:
If $f \in \mathscr{R}(\alpha)$ on $[a, b]$ and if $a<c<b$, then $f \in \mathscr{R}(\alpha)$ on $[a, c]$ and on $[c, b]$ and \begin{equation*} \int_a^c f \, d\alpha + \int_c^b f \, d\alpha = \int_a^b f \, d\alpha \end{equation*}
Now, if we allow for the possibility of $c=b$ in Theorem 6.12 (c), we obtain Theorem 6.12 (c$^*$). To wit, take $c=b$ in Theorem 6.12 (c). Then,
If $f \in \mathscr{R}(\alpha)$ on $[a, b]$ and if $a<c\le b$, then $f \in \mathscr{R}(\alpha)$ on $[a, c]$ and on $[c, b]$ and \begin{equation*} \int_a^c f \, d\alpha + \int_c^b f \, d\alpha = \int_a^b f \, d\alpha + \int_b^b f \, d\alpha = \int_a^b f \, d\alpha \end{equation*}
In other words, we can safely replace $a<c<b$ with $a<c\le b$ in Th. 6.12. Now, we prove the original question. Suppose $f$ is a real function and is Riemann integrable on $[0, 1]$ in the sense of the definition of Riemann integrals given on Pg. 121 in Rudin. Let $0<c\le1$ and that the limit on the right-hand side of (7.0) exists.
By Theorem 6.12 (c$^*$), $f$ is Riemann integrable on $[c, 1]$. We want to show that \begin{equation} \lim_{c \to 0} \int_{c}^1 f(x) \, dx = \int_0^1 f(x) \, dx = \lim_{c \to 0} \int_{0}^1 f(x) \, dx \end{equation}
Let $\epsilon> 0$ be given. It suffices to that there exists a $\delta>0$ such that
\begin{equation}\tag{7.1}
0< |c-0| < \delta \implies \left|\int_{c}^1 f(x) \, dx- \int_{0}^1 f(x) \, dx \right| < \epsilon
\end{equation}
By Theorem 6.12 (c$^*$), we know that $f$ is Riemann integrable on $[0, c]$ and
\begin{equation}\tag{7.2}
\int_0^c f(x) \, dx = \int_0^1 f(x) \, dx - \int_c^1 f(x) \, dx
\end{equation}
Since $f$ is Riemann integrable on $[0, c]$, it must also be bounded on $[0, c]$, for if $f$ is not bounded on $[0, c]$, then it cannot be Riemann integrable on $[0, c]$. Put $M = \sup\limits_{0 \, \le x \, \le c} |f(x)|$. Clearly, $M \ge 0$. If $M =0$, then $f(x) = 0 \, \forall x$, which implies
\begin{equation}
\int_0^1 f(x) \, dx =0= \int_{c}^1 f(x) \, dx
\end{equation}
which means that the implication in (7.1) will hold for any $\delta$. So, assume WLOG that $0 < \epsilon\le M$. By Theorem 6.12 (d),we have
\begin{equation*}
\left| \int_0^c f(x) \, dx\right| \le M(c-0) = Mc \stackrel{(7.2)}{\implies} \left| \int_0^1 f(x) \, dx - \int_c^1 f(x) \, dx\right| \le Mc
\end{equation*}
Put $\delta = \dfrac{\epsilon}{M}$. Notice that $c\in(0, 1]$ and $\dfrac{\epsilon}{M}\in(0, 1]$. If $c=\dfrac{\epsilon}{M}, Mc=\epsilon$ and
\begin{equation*}
\left| \int_0^1 f(x) \, dx - \int_c^1 f(x) \, dx\right| \le Mc = \epsilon
\end{equation*}
If $c<\dfrac{\epsilon}{M}$, we have
\begin{equation*}
\left| \int_0^1 f(x) \, dx - \int_c^1 f(x) \, dx\right| \le Mc < M \cdot \frac{\epsilon}{M} = \epsilon
\end{equation*}
If $c>\dfrac{\epsilon}{M}, \dots$
My question:
Is Theorem 6.12(c$^*$) actually correct? If my proof correct, or is there something in the proof that can be improved? Can someone please suggest how the proof can be completed? Thanks!
