An argument in approximating Riemann-type sums: assuming without loss of generality a given partition is contained in all partitions. I am trying to understand a simple argument from the proof of 17.6 below. So the proof proceeds by taking as $\Delta$ a common refinement of the partitions for the step processes $\tau$ and $c$. Then it says that, without loss of generality we may assume that $\Delta$ is contained in all the partitions $\Pi$ of $[0,T]$.
I cannot see, however, why this assumption can be made. How does proving the result in this scenario imply that given all partitions $\Pi$, which may not contain $\Delta$, if the mesh of $\Pi$ tends to $0$, we still get the result?


 A: In this case we can say that $\Delta \subset \Pi$ without loss of generality because, as the mesh-size $|\Pi|\rightarrow 0$ the partition $\Delta$ must lie in $\Pi$ at some point.  Since we're interested in the behaviour at the limit of the mesh-size, that's enough.
In slightly more analytic terms we can say that $\exists N \in {\mathbb N}$ such that $\Delta \subset \Pi$ whenever $|\Pi|<1/N$.  To see this, first note that since the mesh-size shrinks by increasing the number of points in $\Pi$ it is clear that:

*

*if $N_1 < N_2$ then $\Delta \subset \Pi$ whenever $|\Pi|<1/N_1 \Rightarrow \Delta \subset \Pi$ whenever $|\Pi|<1/N_2$

*whenever a point of $\Delta$ appears in $\Pi$ it remains in there.

So we only need to show that no point of $\Delta$ is omitted from $\Pi$ as $|\Pi|\rightarrow 0$.  Let us suppose otherwise to derive a contradiction.
Let $\Delta = \{0=t_0 < t_1 < \cdots <t_M = T \}$ and suppose that $t_j$ appears nowhere in $\Pi$ as $|\Pi| \rightarrow 0$.  Let $\{\Pi_N \}_{N\in {\mathbb N}}$ be subsequent refinements of $\Pi$ so that $\Pi_N \subset \Pi_{N+1}$ and $|\Pi_N|<1/N$ for each $N$.  Since $t_j \not\in \Pi_N$ for all $N$ we can find points $p_{N,1} < t_j < p_{N,2}$ in each partition $\Pi_N$ with $|p_{N,1}-p_{N,2}|<1/N$ and moreover, $p_{N,1} < p_{N+1,1} < \cdots < t_j$.  So $|p_{N+k,1} - p_{N+k+1,1}| < 1/(N+k+1) \rightarrow 0$ as $k \rightarrow \infty$.
So as $|\Pi| \rightarrow 0$ we have $|p_{N,1}-t_j| \rightarrow 0$ and since the real numbers are complete, this contradicts our claim that $t_j$ is omitted from $\Pi$.
