Ask definition of Riemann integral Let $[a,b]$ be an interval and $f$ a function with domain $[a,b]$. We say that the Riemann sums of $f$ tend to a limit $l$ as $m(P)$ tends to $0$ if, for any $\epsilon > 0$, there is a $\delta > 0$ such that, if $P$ is any partition of $[a,b]$ with $m(P) < \delta$, then $|R(f,P)-l| < \epsilon$ for every choice of $s_j \in I_j$.
My question is, I don't know how it is incarnated "as $m(P)$ tends to $0$" in this definition?
Source: Real Analysis and Foundations,
Book by Steven G. Krantz

The interval $[a,b]$ is partitioned as $a = x_0 < x_1 < \ldots < x_{n-1} < x_n = b$ for some $n\in\mathbb N$. In short this partition is denoted by $P$ and $m(P)$ denotes the length of its largest sub-interval. The idea is that, if the largest sub-interval goes to $0$ in length, so must all the others. If the function is Riemann integrable we might write something like
$$ L  := \lim _{m(P) \to 0} \sum f(x_i)\Delta_i =:  \lim _{m(P)\to 0}R(f,P)  $$
where $x_i$ are the knots and $\Delta _i := x_{i}-x_{i-1}$ i.e the length of the respective sub-interval.

So by definition of limit, for every $\varepsilon >0$ there exists $\delta>0$ such that $m(P)<\delta$ implies $|R(f,P)-L|<\varepsilon$.

The limit is taken as $n\to\infty$ i.e the size of a partition grows. Hence, again by definition, at some point, the size of the partition is big enough to guarantee $m(P)<\delta$.
