Why this is true using Riemann-Roch theorem Let $C$ be a curve of genus $g$ over an algebraically closed field $k$ and $K=k(C)$ the field of rational functions of $C$. Consider $P$ a point at $C$.
What I know:
For each $r\in \mathbb N$, we have $l(rP)\le l((r+1)P)$, because we have the following inequality betweeen these divisors: $rP\le(r+1)P$.
I'm trying to understand this claim:

l(0)=1 and by Riemann-Roch theorem we have $l\big((2g-1)P\big)=g$. So
  if $N_r\doteqdot N_r(P)=l(rP)$, then we have $1=N_0\le N_1\le\ldots\le
 N_{2g-1}=g$.

What I didn't understand is why $N_{2g-1}=g$, i.e., why $l\big((2g-1)P\big)=g$.
Riemann-Roch theorem:
Let $W$ be a canonical divisor over a curve $C$ of genus $g$. Then, for every divisor $D$, we have 
$$l(D)=\deg(D)+1-g+l(W-D)$$
I really need help!
Thanks in advance.

Notice that it follows from Riemann-Roch Theorem that $$l(D) = deg(D) + 1 - g$$ if $deg(D) \geq 2g-1$.
Skecthing the proof, if you have $h \in l(W-D)-\lbrace 0 \rbrace$ then $(h) \geq D-W$ and it follows that $$0 = deg((h)) \geq deg(D-W) > 0$$
which is a contradiction.  
Take $D = (2g -1)P$ and you will have the result. 
