Function on half open interval is Riemann-integrable on closed interval I'm having the following question:
Let $f:(a.b] \to \mathbb{R} $ a continuous and bounded function on the open interval $(a, b]$.
Prove that $f$ is Riemann integrable on the closed interval $[a,b]$.
I tried to show that for every $\epsilon > 0$ there exist $\delta > 0$ so that for every partition $p$ that satisfies $\lambda(p) < \delta \Rightarrow U(p,f) - L(p,f) <  \epsilon$, but I don't know how to estimate $U(p,f) - L(p,f)$.
 A: If a function is Riemann integrable it must be bounded. This is a necessary condition and not just a part of the definition of the Riemann integral. Hence, to define the Riemann integral of $f$ over $[a,b]$, any  finite value for $f(a)$ can be assigned so that $f:[a,b] \to \mathbb{R}$ is bounded. Ultimately the value of the integral will not depend upon this choice as the singleton $\{a\}$ has measure $0$.
Since $f$ is bounded on $(a,b]$ we have finite values $M = \sup_{x \in (a,b]}f(x)$ and $m = \inf_{x \in (a,b]} f(x)$.  Define $M' = \max(M,f(a))$ and $m' = \min (m,f(a))$.
Given $\epsilon > 0$, take any point $x_1 \in (a,b)$ such that  $x_1-a < \epsilon/(2(M'-m'))$.  With $x_1$ fixed we have $f$ continuous on $[x_1,b]$ and, therefore, Riemann integrable. Given $\epsilon > 0$ there exists a partition $P': x_1 < x_2 < \ldots < x_n = b$ such that $U(P',f) - L(P',f) < \epsilon/2$.
Extending to a partition $P: a = x_0 < x_1< x_2 < \ldots < x_n = b$ we have
$$U(P,f) - L(P,f) = ( \sup_{x \in [a,x_1]} f(x) - \inf_{x \in [a,x_1]} f(x)) \,(x_1-a) + U(P',f) - L(P',f) \\ \leqslant (M'-m')(x_1-a) +U(P',f) - L(P',f) < \epsilon$$
Therfore, $f$ is Riemann integrable on $[a,b]$ by the Riemann criterion.
