A possible mistake in a proof of lower-semicontinuity of length

Question

In the book "A Course in Metric Geometry" (By Dmitri Burago, Yuri Burago Sergei Ivanov), there is a short proof of lower semicontinuity of length induced by a metric: (Prop 2.3.4, pg 35)

Let $\gamma_j:[a,b] \to (X,d)$ converge pointwise to $\gamma$. Fix $\epsilon >0$.Let $Y=\{y_1,...y_N \}$ be a partition of $\gamma$ such that $\Sigma(Y) \ge L(\gamma)-\epsilon$, where $$\Sigma(Y)=\sum_i d(\gamma(y_i),\gamma(y_{i+1})).$$ Define $\Sigma_j(Y)=\sum_i d(\gamma_j(y_i),\gamma_j(y_{i+1}))$, and choose $j$ large enough so that the inequality $d(\gamma(y_i),\gamma_j(y_i))< \epsilon$ hols for all $y_i \in Y$. Then:

$$ L(\gamma) \le \Sigma(Y) + \epsilon \le \Sigma_j(Y) + 2N\epsilon +\epsilon \le L(\gamma_j)+(2N+1)\epsilon$$ where the $2N\epsilon$ factor comes from going through $\gamma(y_i) \to \gamma_j(y_i) \to \gamma_j(y_{i+1}) \to \gamma_j(y_i) \to \gamma_(y_{i+1})$.

The authors then say we take $\epsilon \to 0$ and finish.

My problem:

$N$ is in fact a function of $\epsilon$. How can I be sure that $N(\epsilon)\epsilon \to 0$ when $\epsilon \to 0$.

More formally, we can think of $N(\epsilon)$ as the minimal size of a partition which induces an $\epsilon$-approximation of $\gamma$. clearly this quantity increases with $\epsilon$, and without any knowledge on how "wild" is $\gamma$ I see no way how to gain control over it.

Answer 1

Yes, there is a mistake; See errata here.

The solution is to choose $j$ large enough, so that $d(\gamma(y_i),\gamma_j(y_i))< \frac{\epsilon}{N}$.

Answer 2

This is not an answer to your specific question about the proof in the book. It seems worthwhile to point out that the lower semicontinuity of arc length is much more trivial than that proof makes it appear - the proof you sketch seems to me to be sort of missing the point.

Since $\sum_j(Y)$ is just a finite sum it is clear that $\sum(Y)=\lim_j\sum_j(Y)$. And $\sum_j(Y)\le L(\gamma_j)$ by definition. So $$\sum(Y)=\liminf_j\sum_j(Y)\le\liminf_j L(\gamma_j).$$Hence the definition of $L$ shows that $$L(\gamma)=\sup_Y\sum(Y)\le\liminf_j L(\gamma_j).$$