Step 1: Schnirelmann density. For $A \subseteq \mathbb{N}$ with $0 \in A$, define the Schnirelmann density $\sigma(A) = \inf_{n \geq 1} \frac{A(n)}{n}$ where $A(n) = |A \cap [1, n]|$.

Schnirelmann's lemma: If $\sigma(A) \geq 1/2$, then $A + A = \mathbb{N}_0$ (every non-negative integer is a sum of two elements of $A$).

Proof of lemma: For $\sigma(A) \geq 1/2$ and any $n$, consider $A' = A \cap [0, n]$ and $n - A' = \{n - a : a \in A'\}$. We have $|A'| \geq n/2 + 1$ (counting 0), so $|A'| + |n - A'| = 2|A'| > n + 1$, meaning $A' \cap (n - A') \neq \emptyset$ by pigeonhole. Hence $n = a + a'$ for some $a, a' \in A$.

Schnirelmann's sum theorem: If $\sigma(A) + \sigma(B) \geq 1$, then $A + B = \mathbb{N}_0$. More generally, $\sigma(A + B) \geq \sigma(A) + \sigma(B) - \sigma(A)\sigma(B)$.

Step 2: The Goldbach set has positive density. Let $G = \{0, 1\} \cup \{n : n = p + p' \text{ for primes } p, p'\}$. By the Selberg sieve upper bound and explicit computations, the even integers not in $G$ (if any) are rare. More precisely, Vinogradov's theorem (or even earlier results of Schnirelmann using Brun's sieve) shows $\sigma(G) > 0$.

The key input: by Brun's sieve, the number of even integers $\leq 2N$ that are sums of two primes is $\geq cN$ for some $c > 0$. Combined with Goldbach for odd numbers (every odd number $> 1$ is either prime or $2 +$ prime -- wait, this needs more care).

Actually, Schnirelmann proved: $\sigma(\mathcal{P} + \mathcal{P}) > 0$ where $\mathcal{P} = \{0\} \cup \{\text{primes}\}$. This follows from the lower bound: $|\{n \leq N : n = p + p'\}| \geq cN$ for even $n$ (by the circle method or sieve estimates) combined with odd $n = 2 + (n-2)$ where $n - 2$ is handled similarly.

Step 3: Iterating. Since $\sigma(\mathcal{P} + \mathcal{P}) = \alpha > 0$, repeated application of the sum theorem gives $\sigma(k(\mathcal{P} + \mathcal{P})) \to 1$ as $k$ increases. Once $\sigma(k(\mathcal{P}+\mathcal{P})) \geq 1/2$, one more summing gives the full $\mathbb{N}_0$. This shows every $n$ is a sum of $C = 2k + 2$ primes.

More precisely: $\sigma(A + A) \geq 2\sigma(A) - \sigma(A)^2 = 1 - (1-\sigma(A))^2$. By induction: $\sigma(kA) \geq 1 - (1-\sigma(A))^{2^{k-1}}$. For $k \geq \lceil\log_2(1/\sigma(A))\rceil + 1$: $\sigma(kA) \geq 1/2$, and $\sigma((k+1)A) = 1$. Thus $C \leq 2(k+1) = O(\log(1/\alpha))$. $\square$