Here's the definition of Darboux integration that I'm working with:
A bounded function $f:[a,b] \to \mathbb{R}$ is D-integrable iff $\forall \, \epsilon >0, \exists \, \delta > 0$ such that for every partion $P$ with $||P|| < \delta$ we have $U(P,f) -L(P,f) < \epsilon$
Why do I need every partition to have norm less than $\delta$. Doesn't exhibiting at least one partition do the job? What would go wrong here?
This question is inspired from another definition of D-integration:
A bounded function $f:[a,b] \to \mathbb{R}$ is D-integrable iff for every $\epsilon >0$, there exits a partition $P_{\epsilon}$ such that $U(P_{\epsilon},f) - L(P_{\epsilon},f)< \epsilon$
This definition seems to require that exhibiting at least one partition that satisfies the condition is enough.
Question: Could this have something to do with boundedness for the first definition? For instance, in the first definition if we remove the "function is bounded" requirement, would the definition still be true? You know, just like in the definition of Riemann integration where we don't mention the function is bounded.
For reference: the Riemann integration definition:
A function (bounded or not, we don't mention) $f:[a,b] \to \mathbb{R}$ is R-integrable with integral $I$ iff $\forall \, \epsilon>0, \exists \, \delta >0$ such that for every tagged partion $P^T$ with norm less than $\delta$, we have the $|\sum f(t_i)(x_i - x_{i-1}) - I|< \epsilon$