$f$ is a bounded function, $f:[a,b]\rightarrow R$ Prove that if $f$ is continuous on (a,b], then $f$ is integrable on [a,b] $f$ is a bounded function, $f:[a,b]\rightarrow R$ Prove that if $f$ is continuous on (a,b], then $f$ is integrable on [a,b]
My question is: Is there any way to prove this other than using a $\delta-\epsilon$ with Darboux/Riemann sums? And if so, could I get a hint on what to use to fulfill this proof? 
 A: A proper proof does not seem to be possible without the use of $\epsilon, \delta$ arguments. I will try to provide a proof using Darboux sums.
First we note the following condition of integrability:
A function $f$ bounded on $[a, b]$ is Riemann integrable on $[a, b]$ if and only if for any $\epsilon > 0$ there is a partition $P$ of $[a, b]$ such that $$U(P, f) - L(P, f) < \epsilon$$ where $L(P, f), U(P, f)$ are lower and upper darboux sums for $f$ over partition $P$.
We are given that $f$ is bounded on $[a, b]$ and thus let $M, m$  be the supremum and infimum of $f$ on $[a, b]$. Now given any $\epsilon > 0$ it is possible to choose a positive number $h$ such that $(M - m)h < \epsilon / 2$. Also let $h$ be small enough to satisfy $h < b - a$ and consider $c = a + h$. We are given that $f$ is continuous on $(a, b]$ and hence $f$ is continuous on $[c, b]$ and therefore Riemann integrable on $[c, b]$. Thus there is a partition $P'$ of $[c, b]$ such that $$U(P', f) - L(P', f) < \frac{\epsilon}{2}$$ Consider the partition $P$ of $[a, b]$ obtained by adding point $a$ in $P'$ i.e. $P = P' \cup \{a\}$. We can see that if $M', m'$ are supremum and infimum of $f$ on $[a, c]$ then $$U(P, f) - L(P, f) = (M' - m')(c - a) + U(P', f) - L(P', f)$$ Noting that $c = a + h$ and the fact that $M' - m' \leq M - m$ we get $$U(P, f) - L(P, f) < (M - m)h + \frac{\epsilon}{2} < \epsilon $$ It now follows that $f$ is Riemann integrable over $[a, b]$.
What we see here is that discontinuity at possibly an end point of the interval does not impact the integrability of $f$. By splitting interval $[a, b]$ into a finite number of sub-intervals it is possible to extend the above result to get the following:
If $f$ is bounded on $[a, b]$ and $f$ is continuous except for a finite number of discontinuities in $[a, b]$ then $f$ is Riemann integrable on $[a, b]$.
However this is not the final story. It is possible that $f$ can have infinite number of discontinuities and yet be integrable (for example monotone functions are Riemann integrable but can possess at max a countable number of discontinuities, also see this answer). There are functions which have an uncountable number of discontinuities and are yet Riemann integrable. In fact counting the number of discontinuities of $f$ to infer Riemann integrability is not the right approach, but rather one needs to figure out how much space these points of discontinuities take up from the interval $[a, b]$ and this view was championed by Henri Lebesgue to develop his theory of measure and integration.
A: This is a simple consequence of the Lebesgue's theorem:

Bounded function $f\colon [a,b]\to \mathbb R$ is R-integrable if and only if it is
  continuous almost everywhere (in the sense of Lebesgue measure)

Since you have a bounded function on $[a,b]$, which is continuous everywhere except at $\{a\}$, R-interability is obvious by the above theorem.
