Let $f(x)$ be a function on the closed interval $[a,b]$. We define the limit of Riemann Sums $S_{P}=\sum_{k=1}^{n} f(c_{k})\Delta x_{k}$ on $[a,b]$ as the mesh $\mu{(P)}\rightarrow 0$ to be the number $I$ if
Given $\epsilon>0$, $\exists\delta$ such that for every partition $P$ of the interval $[a,b]$
$$\mu{(P)}<\delta \Longrightarrow |S_{P}-I|<\epsilon$$
for any $c_{k}$ we choose in the subintervals $[x_{k-1},x_{k}]$.}
\noindent If the definition holds, then the limit of a Riemann Sum would exist and we can write
$$\lim_{\mu{(P)}\to 0} \sum_{k=1}^{n} f(c_{k})\Delta x_{k}=I$$
How am I supposed to interpret the definition of the definite above? The norm/mesh $\mu{(P)}$ of the partition approaches zero, and we get a finer partition. But wouldn't that create a new partition with its own Riemann Sum? Am I supposed to interpret this as a sequence of partitions where the norm/mesh $\mu{(P)}$ is approaching zero?
Thanks.
The definition is the following: We write $\int_a^b f(x)\,dx=I$ if for every $\epsilon>0$ there exists a $\delta>0$ such that for every partition $P=\{x_0,x_1,\dots x_n\}$ of $[a,b]$ with $\mu(P)<\delta,$ we have
$$|\sum_{k=1}^{n}f(c_k)\Delta x_k - I|<\epsilon$$
for all choices of $c_k\in [x_{k-1},x_k], k= 1,\dots, n.$