6

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?

Saaqib Mahmood
  • 26,762
  • 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) – Ben Grossmann Jul 29 '17 at 12:11
  • For part (b), it is notable that "step-functions" are Riemann integrable – Ben Grossmann Jul 29 '17 at 12:15
  • @Omnomnomnom An unbounded can't be Riemann integrable, because you could construct an unbounded sequence of Riemann sums, then. –  Jul 29 '17 at 12:15
  • 1
    @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 $[0,1]$? – Ben Grossmann Jul 29 '17 at 12:16
  • 1
    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. –  Jul 29 '17 at 12:19
  • @Omnomnomnom that's exactly why I've included the links in my post. So please visit the link on Theorem 6.10 in Baby Rudin. There you'll get the relevant definitions. – Saaqib Mahmood Jul 29 '17 at 12:32
  • 1
    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. –  Jul 29 '17 at 12:35
  • @ProfessorVector All right, thanks for clarifying. – Ben Grossmann Jul 29 '17 at 12:36
  • @ProfessorVector can you please also give some reference (i.e. book, webpage, lecture notes, etc. ) where the different notions of integrability included these two have been discussed together, compared and contrasted? – Saaqib Mahmood Jul 29 '17 at 12:40
  • Sure, it's all here: https://en.wikipedia.org/wiki/Darboux_integral –  Jul 29 '17 at 12:42
  • Doesn't the part a) follow from the simple theorem that if $f$ is Riemann integrable over $[a, b] $ then both the function $F, G$ defined by $$F(x) =\int_{a}^{x} f(t) , dt, , G(x) =\int_{x} ^{b} f(t) , dt$$ are continuous on $[a, b] $? – Paramanand Singh Jul 29 '17 at 16:39

1 Answers1

4

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 $x>0.$