Let $a<b\in\mathbb{R}$,$\gamma([a,b])\to\mathbb{C}$ be a piecewise smooth function, $f:\gamma^*\to \mathbb{C}$ be a continouus function. Prove that $\left| \int_\gamma f(z)dz \right|\le \|f \|_\infty\lim\limits_{\|P\|\to 0,P\in\mathcal{P}[a,b]}\lambda(\gamma)$. $P$ is a partition on $[a,b]$. $\lambda(\gamma)$ is the length of $\gamma$.
Everything in this proof is clear to me, except for one thing:
$$\lim\limits_{\|P\|\to 0,P\in\mathcal{P}[a,b]} \sum\limits_{i=1}^n \left| f(\gamma(t_i^*)\gamma'(t_i^*)\right|\Delta t_i\le \|f\|_\infty \lim\limits_{\|P\|\to 0,P\in\mathcal{P}[a,b]} \sum\limits_{i=1}^n \left| \gamma'(t_i^*)\right|\Delta t_i$$
That is,
$$\lim\limits_{\|P\|\to 0,P\in\mathcal{P}[a,b]} \sum\limits_{i=1}^n \left| f(\gamma(t_i^*)\right|\le \|f\|_\infty $$
What is confusing me is the summation sign on the LHS of the last expression. Here we are not just saying that the sup-norm of $f$ is greater than $f$ at any value on $\gamma$, but it is actually greater than the sum over $n$ as $n\to \infty$. How can this be?