Proving the equivalence of an alternative definition of the Riemann integral (Baby Rudin)

Question

This is Exercise 7, Chapter 6 in 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.

I think to show that to show that this new definition above "agrees with the old one," one would need to show that this new definition above implies that the old definition holds AND that the old definition implies that this new definition holds.

I want to inquire above about the former; how does one show that this new definition above implies that the old definition holds?

My attempt/idea for this proof is pretty limited: I was thinking of proving that assuming the new definition above, it follows that $$ \int_{-0}^1 f(x)\, dx = \int_{0}^{-1} f(x)\, dx.$$ where $\int_{-0}^1 f(x)\, dx$ is the lower Riemann integral and $\int_{0}^{-1} f(x)\, dx$ is the upper Riemann integral.

To show this, I thought of showing

$$\int_0^1 f(x)\, dx = \int_{-0}^1 f(x)\, dx$$ and $$\int_0^1 f(x)\, dx = \int_{0}^{-1} f(x)\, dx$$ which I now think would be incorrect since I would then be using the old definiton itself. My latest thought on this is that I would need to show something of the form $$ \textrm{[something]} \implies \left|\int_{-0}^1 f(x)\, dx - \int_{0}^{-1} f(x)\, dx\right| < \epsilon$$ and the desired result will follow because $\epsilon$ is arbitrary.

Am I even on the right track? How can the desired result be proven?

Answer 1

What you must show is that whenever $f \in \mathscr{R}$ on $[0, 1]$, we have

$\lim\limits_{c \to 0+} \int\limits_c^1 f(x) dx = \int\limits_0^1 f(x) dx$

Or equivalently,

$\lim\limits_{c \to 0+} (\int\limits_c^1 f(x) dx - \int\limits_0^1 f(x) dx) = 0$

Now we have $\int\limits_0^1 f(x) dx = \int\limits_0^c f(x) dx + \int\limits_c^1 f(x) dx$ by Theorem 6.12(c) in Baby Rudin. Then we have $\int\limits_c^1 f(x) dx - \int\limits_0^1 f(x) dx = \int\limits_0^c f(x) dx$. Thus, it suffices to show that

$\lim\limits_{c \to 0+} \int\limits_0^c f(x) dx = 0$

Since $f \in \mathscr{R}$, $f$ is bounded on $[0, 1]$. Let $B$ be an upper bound of $|f|$; WLOG, let $B$ be positive. Now suppose we have $\epsilon > 0$. WLOG, take $\epsilon < B$. Then for all $c$ s.t. $0 < c < \frac{\epsilon}{B}$, we have $|\int\limits_0^c f(x) dx| \leq B (c - 0) = Bc < \epsilon$ by Theorem 6.12(d) in Baby Rudin. Thus, we have shown that $\lim\limits_{c \to 0+} \int\limits_0^c f(x) dx = 0$ as required.