TITLE: Proof of equivalence of conformal and complex structures on a Riemann surface.
QUESTION [8 upvotes]: I am trying to understand the fundamentals of Riemann surface theory and so far I have the following:
--Definition 1.
A conformal structure on a Riemann surface $\Sigma$ is an equivalence class of metrics
$$ [g]=\{e^{2u}g \colon u\in \mathcal{C}^\infty(\Sigma)\} $$
--Definition 2. A complex structure on a Riemann surface $\Sigma$ is an equivalence class of complex atlases, where two atlases are considered equivalent iff their union forms a new complex atlas.
Note 1. Each complex structure has a canonical representative given by the maximal atlas
---Riemann's Uniformization Theorem. In any given conformal structure, there exists a unique metric with constant curvature of either $1,0$ or $-1$. 
Note 2 This gives a means of choosing a canonical representative for each conformal structure.
My questions are these:
(I) Which theorem tells us that there is a bijection between these two definitions of conformal and complex structures? 
(II) What is meant by the "Riemann moduli space"?
Many Thanks,
A.
EDIT: In establishing the claim (I), it appears to be neccessary to take as assumption that a surface is orientable.

REPLY [3 votes]: As Dmitry Zaitsev correctly noted, the proof in Part (1) of the accepted answer is incomplete: The (complex) isomorphisms $T_pM\to {\mathbb C}$ are well-defined only pointwise: Unless the tangent bundle $TM$ is trivial, you cannot even make choices of these isomorphisms continuous with respect to $p$. What an (almost) complex structure on the tangent bundle $TM$ gives you is the orientation on $M$ and the notion of an angle between tangent vectors. It does not give a well-defined notion of length for tangent vectors. The standard way to deal with this problem is as follows:
Let $\{(U_\alpha, \phi_\alpha)\}_{\alpha\in A}$ be a locally finite system of holomorphic coordinates on the given Riemann surface $X$, i.e. the cover ${\mathcal U}= \{U_\alpha\}_{\alpha\in A}$  of $X$ is locally finite (every point is covered by finitely many charts). Why such a cover exists is a tricky issue: I will simply assume that $X$ is paracompact. (It is a theorem due to Rado that every Riemann surface is paracompact. See the discussion here.) Paracompactness means that every open cover admits a locally finite subcover. 
Now, given this, we proceed as follows. Let $\{\eta_\alpha: \alpha\in A\}$ be a partition of unity subordinate to the open cover ${\mathcal U}$. 
For each $\alpha\in A$ equip $U_\alpha$ with the pull-back metric $g_\alpha= \phi_\alpha^*(|dz|^2)$, obtained by the pull-back of the Euclidean metric from ${\mathbb C}$ via $\phi_\alpha$. Set $h_\alpha= \eta_\alpha g_\alpha$ and extend $h_\alpha$ by zero to the rest of $X$. The result is a semi-Riemannian metric $h_\alpha$ on $X$ (it is only positive semidefinite rather than definite on tangent spaces). 
Lastly, set
$$
h=\sum_{\alpha\in A} h_\alpha.
$$ 
Since ${\mathcal U}$ is locally finite, this sum is a smooth and semi-Riemannian; it is a Riemannian metric since  $\{\eta_\alpha: \alpha\in A\}$ is a partition of unity. Conformality of this metric on $X$ follows from the following observation: 
Observation. Let $g=\rho(z)|dz|^2$ be a conformal Riemannian metric on an open subset $U\subset {\mathbb C}$. Let $f: V\to U$, a biholomorphic map from an open subset $V\subset {\mathbb C}$. Then the pull-back metric $f^*(g)$ equals
$$
\rho(f(w))|f'(w)|^2 |dw|^2
$$
and, hence, is again conformal. Thus, conformality of a Riemannian metric on a Riemann surface is independent of the local holomorphic chart. In particular, a finite sum of conformal metrics is again conformal. 
Lastly, here is an amusing and little-known fact (due to Robert Gunning and Raghavan Narasimhan). Suppose that $X$ is a connected noncompact Riemann surface. Then $X$ admits a conformal Riemannian metric of zero curvature. (Such a metric is typically incomplete.)