What is the meaning of $\lim_{\Delta(P) \to 0} F(P) = L$ for partitions Let $[a,b]$ be an interval, and denote by $\mathcal P[a,b]$ the family of all partitions of $[a,b]$, i.e. sets $P = \{ a = x_0 < x_1 < \ldots < x_n = b \}$. 
For some $P \in \mathcal P[a,b]$ set $\Delta(P) := \max_{i=1,\ldots, n} (x_i - x_{i-1})$, i.e. the mesh size. If $F : \mathcal P[a,b] \to \mathbb R$ is some function, what does the expression
$$
 \lim_{\Delta(P) \to 0} F(P) = L
$$
mean? Does it mean 
i) that for each $\varepsilon > 0$ there exists some partition $P$ such that for each partition $P'$ with $P' \supseteq P$ (i.e. $P'$ refines $P$) we have
$$
 |F(P') - L| < \varepsilon 
$$
or,
ii) that for each $\varepsilon > 0$ there exists some $\delta > 0$ such that for each partition $P$ with $\Delta(P) < \delta$ we have
$$
 |F(P') - L| < \varepsilon
$$
or,
iii) that for each sequence of partitions $\{ P_n \}$ with $\lim_{n\to \infty} \Delta(P_n) = 0$ we have
$$
 \lim_{n\to \infty} F(P_n) = L
$$
where in the last case we have limits over the reals, defined as usual.
So what does it mean? And are all notions written above equivalent?
 A: It means (ii), which is clearly equivalent to (iii).
It's also clear that (ii) implies (i): Given $\delta>0$,
if $P$ has mesh less than $\delta$ and $P'$ is a refinement
of $P$ then $P'$ has mesh less than $\delta$.
But in this generality (i) does not imply (ii). Silly example:
Say $[a,b]=[-1,1]$. Define $F(P)$ by saying that $F(P)=1$ if
$0\in P$, $F(P)=0$ otherwise. Then (i) holds with $L=1$, but (ii)
does not hold; there's no $\delta>0$ such that $\Delta(P)<\delta$ implies that $0\in P$.
(By "$0\in P$" I mean that, in the notation at the top of the OP, there exists $j$ with $x_j=0$.)

EDIT: Of course this led to the real question, namely whether (i) implies (ii) for Riemann sums. Of course in that case we need to modify the whole setup somewhat, because a Riemann sum depends on a partition and a choice of points in each interval. A very informal proof that (i) implies (ii) for Riemann sums:
Say $f$ is a bounded function on $[a,b]$, say $|f|\le M$. Suppose $\epsilon>0$, and let $P$ be as in (i).
Given an interval $I$, let $S_I$ and $s_I$ denote the sup and inf of $f$ on $I$. Let $|I|$ denote the length of $I$. 
Say $P=(a=x_0<\dots<x_n=b)$. Define $I_j=[x_{j-1},x_j]$. Now if (i) holds we must have $$\sum_{j=1}^n(S_{I_j}-s_{I_j})|I_j|\le\epsilon.$$
Now say we have another partition $P'=(a=y_0<\dots<y_m=b)$ with mesh less than some small number $\delta$. Let $J_k=[y_{k-1},y_k]$. For $j=1,\dots,n$ let $$E_j=\{k:J_k\subset I_j\},$$and let $$F=\{1,2,\dots,m\}\setminus\bigcup_{j=1}^nE_j.$$Now if $k\in E_j$ then $S_{J_k}\le S_{I_j}$ and $s_{J_k}\ge s_{I_j}$; hence $$\sum_{k\in E_j}(S_{J_k}-s_{J_k})|J_k|\le (S_{I_j}-s_{I_j})|I_j|,$$hence $$\sum_{j=1}^n\sum_{k\in E_j}(S_{J_k}-s_{J_k})|J_k|\le \epsilon.$$On the other hand if $k\in F$ then there exists $j$ so that $x_j$ lies in the interior of $J_k$. Hence $F$ has at most $n$ points, and so if $\delta$ is small enough then $$\sum_{k\in F}(S_{J_k}-s_{J_k})|J_k|\le n(2M)\delta\le\epsilon.$$Add it all up and you get $$\sum_{k=1}^m(S_{J_k}-s_{J_k})|J_k|\le2\epsilon.$$
