Improper Riemann integral of bounded function is proper integral

Let $f:[a,b) \rightarrow \mathbb R$ be Riemann integrable on each compact subinterval of $[a,b)$ and bounded on $[a,b)$. Let $g:[a,b] \rightarrow \mathbb R$ be arbitrary extension $f$ ( i.e. $g|_{[a,b)}=f$). Why $g$ is Riemann integrable on $[a,b]$?

Solution 1:

We need to use the following theorem by Lebesgue.

Theorem (Lebesgue) A bounded $ \mathbb{R} $-valued function $ f $ defined on a closed interval is Riemann-integrable if and only if the set of discontinuities of $ f $ has measure $ 0 $.

Let $ D^{f} $ denote the set of discontinuities of $ f $ in $ [a,b) $. Next, let $ (b_{n})_{n \in \mathbb{N}} $ be a strictly increasing sequence in $ [a,b) $ such that $ \displaystyle \lim_{n \to \infty} b_{n} = b $. For each $ n \in \mathbb{N} $, define $$ D^{f}_{n} \stackrel{\text{def}}{=} \{ x \in [a,b_{n}] ~|~ \text{$ f|_{[a,b_{n}]} $ is discontinuous at $ x $} \}. $$

For each $ n \in \mathbb{N} $, as we have assumed $ f $ to be Riemann-integrable on $ [a,b_{n}] $, Lebesgue’s Theorem yields $ \mu(D^{f}_{n}) = 0 $. Hence, $$ 0 \leq \mu(D^{f}) = \mu \left( \bigcup_{n=1}^{\infty} D^{f}_{n} \right) \leq \sum_{n=1}^{\infty} \mu(D^{f}_{n}) = \sum_{n=1}^{\infty} 0 = 0, $$ which gives us $ \mu(D^{f}) = 0 $.

Now, extend $ f: [a,b) \to \mathbb{R} $ to $ g: [a,b] \to \mathbb{R} $, which is bounded. It is not difficult to see that the set of discontinuities of $ g $ is the set $ D^{f} $ plus possibly the point $ b $ itself. As such, the set of discontinuities of $ g $ has measure $ 0 $. By applying Lebesgue’s Theorem once more, we conclude that $ g $ is Riemann-integrable on $ [a,b] $.

Solution 2:

There follows what was going to be the text of a new question, "Bounded function with improper integral on finite interval also has proper integral?" until I discovered belatedly that it was an exact duplicate of this one. (I could have sworn I had Googled thoroughly before consulting all my analysis textbooks! Never mind.)

I'm gambling that it's less of a breach of etiquette to have added the irrelevant 'proof-verification' tag to this old question than it would have been to have asked an entirely new question. If not, please excuse me - I'm a newbie, trying hard not to screw up here!

I just want to check my reasoning on this subtle point, which I unconsciously glossed over when trying to answer another recent question, and which also isn't mentioned in any of the many standard textbooks that I subsequently consulted. $\newcommand{\abs}[1]{\left\lvert#1\right\rvert}$ $\newcommand{\R}{\mathbb{R}}$ $\renewcommand{\phi}{\varphi}$

Suppose that (i) the function $f: [a, b] \to \R$ is bounded, and (ii) the improper Riemann integral $\int_{a+}^b f = \lim_{\epsilon \to 0+} \int_{a + \epsilon}^b f$ exists. My claim is that the proper Riemann integral $\int_a^b f$ also exists, and it is equal to the improper integral $\int_{a+}^b f$.

(The change of variables $x \mapsto a + b - x$ yields the corollary that the existence of $\int_a^{b-} f = \lim_{\epsilon \to 0+} \int_a^{b - \epsilon} f$ implies the existence of $\int_a^b f$, with the same value.)

The closest thing to any mention of this fact is in Exercise 6.7(a) of Rudin's Principles of Mathematical Analysis (3rd ed.): but that is much easier, because he hypothesises Riemann-integrability, whereas I only hypothesise boundedness.

Using Lebesgue's criterion for Riemann-integrability to reduce the problem to the case considered by Rudin (the set of discontinuities of $f$ in $[a, b]$ is the union of the sets of discontinuities of $f$ in $[a + \frac{b - a}{n}, b]$, and so has measure zero) would constitute excessive force, I feel. A proof from first principles is desirable.

[That was written in ignorance of the existence of this thread, and not as a dig at the other answer!]

Here is such a proof, I think:

Let $M$ be any upper bound of $\{\abs{f(x)}: a \leqslant x \leqslant b\}$ such that $2M(b - a) > 1$. For $n = 1, 2, \ldots$, let $\epsilon_n = \frac{1}{2nM}$, and let $P_n$ be a partition of $[a + \epsilon_n, b]$ on which the upper and lower Darboux sums of $f$ both differ from $\int_{a + \epsilon_n}^b f$ by less than $\frac{1}{2n}$. The upper and lower Darboux sums of $f$ on $\{a\} \cup P_n$ both differ from $\int_{a + \epsilon_n}^b f$ by less than $\frac{1}{n}$, so $f$ has a sequence of upper Darboux sums over $[a, b]$ that converges to $\int_{a+}^b f$, and also a sequence of lower Darboux sums over $[a, b]$ that converges to $\int_{a+}^b f$. Hence, $f$ is Riemann integrable on $[a, b]$, and $\int_a^b f = \int_{a+}^b f$. Q.E.D.

Application (a lemma for which I needed the above as a subsidiary lemma):

Let $g: \R \to \R$ be a bounded function whose improper Riemann integral $\int_{-\infty}^\infty g$ exists, let $\phi: (a, b) \to \R$ be a continuously differentiable increasing bijection, and let the function $F: [a, b] \to \R$ be defined by $F(y) = g(\phi(y))\phi'(y)$ ($a < y < b$), $F(a)$ and $F(b)$ being given arbitrary values. Then $\int_{-\infty}^\infty g = \int_a^b F$.

Proof:

By the theorem on change of variable in a Riemann integral (see e.g. Rudin, op. cit., Theorem 6.19), $F$ is Riemann-integrable on any closed subinterval $[c, d]$ of $(a, b)$, and $\int_c^d F = \int_{\phi(c)}^{\phi(d)} g$. Therefore, the improper Riemann integral $\int_{a+}^{b-} F$ exists, and equals the improper Riemann integral $\int_{-\infty}^\infty g$. But $F$ is bounded on $[a, b]$, so the previous lemma implies that $\int_a^b F$ exists and equals $\int_{-\infty}^\infty g$. Q.E.D.

Are these proofs valid? Are there shorter or clearer proofs? Are these proofs indeed a song and dance about nothing?

Update:

The second 'proof' isn't valid. In the original context, I had devoted a lot of effort to ensuring that $F$ was bounded on $[a, b]$. Indeed, that was the crux of the argument (about the improper integral over $\R$ of a function of moderate decrease). Of course, one can't just baldly assert the same thing in the general context! So, scrub the assertion that this is also a 'lemma'. I'm glad the first lemma is OK.