Riemann integral over $[a,b), (a,b)$ or $(a,b]$ Couple days ago I finished the reading of chapter 6 Riemann-Stiltjes Integral from PMA Rudin. And I noted the he defines Riemann Integral only for bounded functions on $[a,b]$. 
But what would be if we consider Riemann integral for $[a,b), (a,b)$ or $(a,b]$? Does sums $U(P,f)$ and $L(P,f)$ changes in these cases? Since we have no information about values $f(a)$ or $f(b)$ because we use them in these above sums. 
Let's take a look at this example.
1. Define $f(x)=1$ on $(0,1)$. How to evaluate $\int \limits_{0}^{1}f(x)dx$? In this case we don't nothing about $f(0)$ and $f(1)$. Can anyone show accurately and clearly how it can be evaluated? And prompt how to be in other analogous examples?
I would be really greatful for your help!
 A: As long as $f$ is "integrable", you can redefine any finite number of points of $f$ without changing the integral: That is, if $g(x)=f(x)$ everywhere in some interval $I$ except perhaps on the set $\{x_1,x_2,...,x_n\}$, then $\int_Ig(x)dx=\int_If(x)dx$. This fact is not hard to prove directly using Riemann sums.
In particular, if we have $f$ defined only on $(0,1)$, giving values to $f$ at the endpoints will change nothing in evaluating the integral. (More specifically, to evaluate $\int_0^1fdx$ where $f\colon(0,1)\rightarrow\mathbb{R}$ is bounded, just set $f(0)=f(1)=0$)
A: Generally, definite integrals over closed intervals are considered proper and integrals over open or half-open intervals are improper, defined as the limit of proper integrals. 
For instance, over the the interval $[0,1)$, the integral is defined as
$$\lim_{x\to 1^+} \int_0^x f(t) \ dt$$
Over the interval open interval $(0,1)$, the integral is defined as a double limit
$$\lim_{x\to 0^-} \lim_{y\to 1^+} \int_x^y f(t) \ dt$$
A: A Riemann integral over $[a,b]$ has a definition: $$\lim_{\max\Delta x_i\to0}\sum f(x^*_i)\,\Delta x_i$$ where the sum is over the subintervals from some partition of $[a,b]$, with $x^*_i$ coming from a subinterval, and $\Delta x_i$ being the length of that subinterval. The limit is over any sequence of partitions where the maximum subinterval length is eventually arbitrarily small (if you are far enough along in the sequence of partitions). If $f$ is continuous, this limit is guaranteed to exist (meaning, it takes a finite value), by a theorem from calculus.
You could try applying the same definition to an interval with at least one open end. But then the limit is not guaranteed to exist. Examples come in the form of functions that blow up near the open endpoint. For example, $\int\limits_{(0,1]}\frac1x\,dx$ just doesn't exist. They also come in the form of functions with weird behavior near the open endpoint: $\int\limits_{(0,1]}\sin\mathopen{}\left(\frac1x\right)\mathclose{}\,dx$ just doesn't exist either. But as long as $f$ is continuous on $[a,b]$, $\int\limits_{[a,b]}f(x)\,dx$ will always exist.
