Proving that $\left| \int_\gamma f(z)dz \right|\le \|f \|_\infty\lim\limits_{\|P\|\to 0,P\in\mathcal{P}[a,b]}\lambda(\gamma)$

Question

Let $a

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?