Is any continuous bounded function on $(a,b)$ Riemann integrable? Is any continuous bounded function on $(a,b)$ Riemann integrable? Traditionally, we don't discuss the Riemann integrability when the domain is not a closed and bounded interval. When dealing with other domain, such as $[a,\infty)$, we sometimes refer it to the improper integral. However, if we use the definition from the multivariable Riemann integration case, we can talk about such integral on $(a,b)$. So is any continuous bounded function on $(a,b)$ Riemann integrable under this meaning? If so, how can we prove it? Is their a way to avoid digging into $\epsilon-\delta$?
 A: Apostol in his Mathematical Analysis describes the multiple integrals on compact intervals (closed and bounded intervals which are product of many one dimensional closed intervals) and later extends the definition to Jordan measurable sets via the use of characteristic functions.
Thus the same argument applies to the one dimensional case also. The set $(a, b) $ is Jordan measurable and if $f$ is defined on $(a, b) $ and we create another function $g:[a, b] \to\mathbb{R} $ via $g(x) =f(x), x\in(a, b) $ and $g(a) =g(b) =0$ then we say that $f$ is Riemann integrable on $(a, b) $ if and only if $g$ is Riemann integrable on $[a, b] $ and then their integrals are equal. 
It should now be obvious that if $f$ is bounded and continuous on $(a, b) $ then $f$ is Riemann integrable on $(a, b) $.

I don't know what is gained from the definition of multiple integrals given in Wikipedia which involves half open intervals. The approach by Apostol seems less clumsy and is general enough. 
A: A function is Riemann integrable on $[a,b]$ iff it is bounded and continuous save on a set of Lebesgue measure zero. Here the discontinuities
of $f$ are within the finite set $\{a,b\}$ and so $f$ is Riemann integrable.
