Definition $:$ Let $F$ be a Banach space. A function $f : [a,b] \longrightarrow F$ is said to be Riemann integrable if there exists $\int_{a}^{b} f \in F$ such that for any $\varepsilon \gt 0$ there exists $\delta \gt 0$ such that for any partition $\mathfrak P = \{a = t_0 \lt t_1 \lt t_2 \lt \cdots \lt t_n = b \}$ with $\left \|\mathfrak P \right \| \lt \delta$ and for any choice of $\xi_i \in (t_{i-1},t_i),$ for $i = 1, 2, \cdots, n$ we have $$\left \|\int_{a}^{b} f - \sum\limits_{i=1}^{n} f(\xi_i) (t_i - t_{i-1}) \right \|_{F} \lt \varepsilon.$$ In this case we say that $\displaystyle {\int_{a}^{b} f}$ is the Riemann integral of $f$ on $[a,b].$
Now my question is as follows $:$
In the above definition of Riemann integral will it make any difference if instead of $\xi_i \in (t_{i-1},t_i)$ we stipulate $\xi_i \in [t_{i-1},t_i]\ $? Any suggestion in this regard will be highly appreciated.
Thanks in advance.
It does not hurt anything to allow for $\xi$ to be in closed subintervals. A kind of manual way to see this is to connect the Riemann sum on a tagged partition with tags at endpoints to a Riemann sum on a tagged partition without tags at endpoints. One way to do that is to split into two cases depending on whether two neighboring $\xi$'s are the same or not.
If they are the same, merge those two subintervals; doing this all over results in a partition with mesh $2\delta$ which still goes to zero, and the Riemann sum on this tagged partition is exactly the same as the Riemann sum you already had.
If they are not the same, enlarge the subinterval which had $\xi$ at the endpoint by a little bit, at the expense of its neighbor. As long as the sum of all the enlargement lengths is less than $\varepsilon/4M$ where $M$ is the bound on $f$, the resulting change in the Riemann sum will be less than $\varepsilon$.