5
$\begingroup$

Here is Prob. 7, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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.

(b) Construct a function $f$ such that the above limit exists, although it fails to exist with $\lvert f \rvert$ in place of $f$.

Here I'll only attempt Part (a):

My Attempt:

Here is the link to a post of mine here on Math SE where I've copied the definition of the Riemann and Riemann-Stieltjes integral that Rudin uses (i.e. Definitions 6.1 and 6.2 in Baby Rudin, 3rd edition):

Theorem 6.10 in Baby Rudin: If $f$ is bounded on $[a, b]$ with only finitely many points of discontinuity at which $\alpha$ is continuous, then

As $f \in \mathscr{R}$ on $[0, 1]$, so $\int_0^1 f(x) \ \mathrm{d} x$ exists in $\mathbb{R}$.

According to the statement of the problem, we only need to show that $$ \lim_{c \to 0+} \int_c^1 f(x) \ \mathrm{d} x = \int_0^1 f(x) \ \mathrm{d} x. \tag{0}$$

Let $\varepsilon > 0$ be given. We need to find a real number $\delta> 0$ such that $$ \left\lvert \int_c^1 f(x) \ \mathrm{d} x \ - \ \int_0^1 f(x) \ \mathrm{d} x \right\rvert < \varepsilon \tag{1} $$ for any real number $c$ such that $0 < c < \delta$.

Now let's choose a real number $\delta_0 \in (0, 1)$, and let us choose $c$ such that $0 < c < \delta_0$.

Then as $f \in \mathscr{R}$ on $[0, 1]$ and as $c \in (0, 1)$, so by Theorem 6.12 (c) in Baby Rudin $f \in \mathscr{R}$ on $[0, c]$ and on $[c, 1]$, and $$ \int_0^c f(x) \ \mathrm{d} x \ + \ \int_c^1 f(x) \ \mathrm{d} x = \int_0^1 f(x) \ \mathrm{d} x. \tag{2} $$

Here is the link to my Math SE post on Theorem 6.12 (c) in Baby Rudin, 3rd edition:

Theorem 6.12 (c) in Baby Rudin: If $f\in\mathscr{R}(\alpha)$ on $[a, b]$ and $a<c<b$, then $f\in\mathscr{R}(\alpha)$ on $[a, c]$ and $[c, b]$

In the light of (1) and (2), we can conclude that we now only need to show that there exists a real number $\delta > 0$ such that $$ \left\lvert \int_0^c f(x) \ \mathrm{d} x \right\rvert < \varepsilon \tag{3} $$ for any real number $c$ such that $0 < c < \delta$, and we now also know that $ 0 < c < \delta_0 < 1$.

As $f \in \mathscr{R}$ on $[0, 1]$, so $f$ is also bounded on $[0, 1]$ and hence also on $[0, c]$. Let $M \colon= \sup \{ \ f(x) \ \colon \ 0 \leq x \leq c \ \}$.

Then by Theorem 6.12 (d) in Baby Rudin, we have $$ \left\lvert \int_0^c f(x) \ \mathrm{d} x \right\rvert \leq M c. \tag{4} $$

Here is the link to my Math SE post on Theorem 6.12 (d) in Baby Rudin, 3rd edition:

Theorem 6.12 (d) in Baby Rudin: If $\lvert f(x) \rvert \leq M$ on $[a, b]$, then $\lvert \int_a^b f d\alpha \rvert \leq \ldots$

So if we choose our $\delta$ such that $$0 < \delta < \min \left\{ \ \delta_0, \frac{\varepsilon}{M+1} \ \right\}, $$ then, for any real number $c$ such that $0 < c < \delta$, we have $0 < c < \delta_0$ so that $c \in (0, 1)$ and from (4) we also have $$ \left\lvert \int_0^c f(x) \ \mathrm{d} x \right\rvert \leq M c \leq \frac{M \varepsilon}{M+1} < \varepsilon, $$ which by virtue of (3) implies that (1) holds.

Since $\varepsilon > 0$ was arbitrary, therefore (0) holds as well, as required.

Is this proof correct and rigorous enough for Rudin? If not, then where is it lacking?

Is this proof the same as the proof asked for by Rudin?

$\endgroup$
  • $\begingroup$ It seems that you've taken $f \in \mathscr R$ on $[0,1]$ and concluded that $|f| \leq M$ for some $M \geq 0$. Does one really follow from the other? What exactly is the definition of $\mathscr R$ in this context? (I don't remember Rudin's conventions here) $\endgroup$ – Omnomnomnom Jul 29 '17 at 12:11
  • $\begingroup$ For part (b), it is notable that "step-functions" are Riemann integrable $\endgroup$ – Omnomnomnom Jul 29 '17 at 12:15
  • 1
    $\begingroup$ @ProfessorVector are you saying that a function such as $$ f(x) = \begin{cases} 1/\sqrt{x} & x \neq 0\\ 0 & x = 0 \end{cases} $$ would not be considered Riemann integrable on $[0,1]$? $\endgroup$ – Omnomnomnom Jul 29 '17 at 12:16
  • 1
    $\begingroup$ There's nothing to be "considered", it just isn't, by definition of the Riemann integral. It exists only as an improper integral, i.e. a limit. This exercise is exactly about the difference between the two notions. $\endgroup$ – Professor Vector Jul 29 '17 at 12:19
  • 1
    $\begingroup$ Strictly speaking, that definition is called Darboux integral, it's different from the original definition of the Riemann integral, but it's easy to show that both definitions are equivalent. $\endgroup$ – Professor Vector Jul 29 '17 at 12:35
4
$\begingroup$

It's correct, just a bit too long for my taste, that distracts from the main ideas. These bits would be sufficient:

Then as $f \in \mathscr{R}$ on $[0, 1]$ and as $c \in (0, 1)$, so by Theorem 6.12 (c) in Baby Rudin $f \in \mathscr{R}$ on $[0, c]$ and on $[c, 1]$, and $$ \int_0^c f(x) \ \mathrm{d} x \ + \ \int_c^1 f(x) \ \mathrm{d} x = \int_0^1 f(x) \ \mathrm{d} x. $$ As $f \in \mathscr{R}$ on $[0, 1]$, so $f$ is also bounded on $[0, 1]$ and hence also on $[0, c]$: An unbounded function can't be Riemann integrable, because one could construct an unbounded sequence of Riemann sums, then.
Let $M \colon= \sup \{ \ f(x) \ \colon \ 0 \leq x \leq c \ \}$.
Then by Theorem 6.12 (d) in Baby Rudin, we have $$ \left\lvert \int_0^c f(x) \ \mathrm{d} x \right\rvert \leq M c. $$

It's clear that the latter converges to $0$ as $c\rightarrow0.$
As for b), a simple example would be $f(x)=\frac1x\sin\frac1x$ for $x>0.$

$\endgroup$
  • $\begingroup$ Why does $f \in \mathscr{R}$ on $[0, 1]$ imply that $f$ is bounded on $[0, 1]$? $\endgroup$ – Omnomnomnom Jul 29 '17 at 12:13
  • $\begingroup$ @Omnomnomnom it is by virtue of Definitions 6.1 and 6.2 in Rudin. It is only for bounded functions that he has defined Riemann-integrability. $\endgroup$ – Saaqib Mahmood Jul 29 '17 at 12:28
  • $\begingroup$ @SaaqibMahmuud Thank you, everything makes sense then $\endgroup$ – Omnomnomnom Jul 29 '17 at 12:35
  • $\begingroup$ @Omnomnomnom you're welcome. Can we connect off Math SE too? My profile has my contact info. $\endgroup$ – Saaqib Mahmood Jul 29 '17 at 12:37
  • $\begingroup$ @ProfessorVector I aim to write my proofs in sufficient enough detail for a student taking their very first course in analysis to be able to understand them. $\endgroup$ – Saaqib Mahmood Jul 29 '17 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.