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 $[a,b]$?  
 A: 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] $.
A: 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.
