$\newcommand{\u}{\mathfrak{U}}$ Definition. For a divisor $\u$ on $M$, we set a $\Bbb C$-vector space $$L(\u) = \{f\in\mathcal{K}(M);(f)\geq \u\},$$ where $\mathcal{K}(M)$ denotes a field of meromorphic functions on $M$. $$r(\u) =\dim_{\Bbb C}L(\u).$$ Similarly, we define a $\Bbb C$-vector space $$\Omega(\u) = \{\omega;\omega\ \text{is a meromorphic differential with}\ (\omega)\geq\u\}.,$$ and $$i(\u) = \dim\Omega(\u).$$
Theorem (Riemann-Roch). Let $M$ be a compact Riemann surface of genus $g$ and $\u$ an integral divisor on $M$. Then $$r(\u^{-1}) = \deg\u-g+1+i(\u).$$
Let $D$ be a divisor on $M$ which is a compact Riemann surface of genus $g$. Then $i(D) = 0$ provided $\deg D>2g-2$.
This is a statement in Farkas & Kra Riemann surfaces without much explanation. Why this is true?
Take any meromorphic form $\alpha$ on $X$. Let $K=(\alpha)$ (canonical divisor), Then $i(D)\cong r(D-K)$ by sending $\omega$ to $\omega/\alpha$. Then we can get another form of Riemann-Roch: $$ r(-D)-r(D-K)=1-g+deg D $$ Let $D=0$ and $D=K$, we can get $deg K=2g-2$. So if $\omega\in i(D)$ and $deg (D)\geq 2g-1$, then $(\omega)-(\alpha)\geq D-K$. Then $$ deg\left(\frac{\omega}{\alpha}\right)\geq 1 $$ But deg $(f)$ should be $0$ by Residue Theorem. So $\omega=0$.
We actually proved that all meromorphic forms have the same degree $2g-2$.