I'm reading a proof of the following Thereom:
If $m/n \to \infty$ then $G_{n,m}$ contains a triangle wit high probability.
The first line of the proof states:
Because having a triangle is a monotone increasing property, we can prove the result in $G_{n,p}$ assuming that $np \to \infty$.
I realize that this is probably a very basic technique and there is nothing deep here, but I'm having trouble coming up with a rigorous proof of why it's true. I think an informal justification would go something like this:
For $G_{n,p}$ the expected number of edges is $\binom{n}{2}p \approx n^2p$. If we assume $np\to \infty$, then for $G_{n,m}$ we have $m/n \approx n^2p/n = np \to \infty$.
But I haven't really used the monotone increasing property hypothesis, and I also did a bit of hand waving with the expected number of edges. How do we formalize this?
For the formal justification, being monotone increasing is key to having a very close relationship between probabilities in the two graphs. Fix $m,n,p$ such that $m = \binom n2 p$. Let $M$ be an arbitrary monotone increasing property.
\begin{align} \Pr[G_{n,p} \text{ has $M$}] &= \sum_{k=0}^{\binom n2} \Pr[G_{n,p} \text{ has $k$ edges}] \cdot \Pr[G_{n,k} \text{ has $M$}] \\ &\ge \sum_{k=m}^{\binom n2} \Pr[G_{n,p} \text{ has $k$ edges}] \cdot \Pr[G_{n,k} \text{ has $M$}] \\ &\ge \sum_{k=m}^{\binom n2} \Pr[G_{n,p} \text{ has $k$ edges}] \cdot \Pr[G_{n,m} \text{ has $M$}] \\ &= \Pr[G_{n,m} \text{ has $M$}] \cdot \Pr[G_{n,p} \text{ has $\ge m$ edges}]. \end{align} Similarly, if $M$ were monotone decreasing, we could drop the other half of the sum in the second line, and then get that $\Pr[G_{n,p}\text{ has }M] \ge \Pr[G_{n,m} \text{ has }M] \cdot \Pr[G_{n,p}\text{ has $\le m$ edges}]$.
Either way, the missing probabilities are binomial: they are the probabilities that the random variable $\text{Binomial}(\binom n2, p)$ either exceeds or does not exceed its mean $m$. For a nice symmetric binomial, the probability is at least $\frac12$; in the general case, things get messy, but this paper says it's still at least $\frac14$, provided $1 < m < \binom n2 - 1$. (Assuming $np \to \infty$ lets you upgrade to $\frac13$, but that's immaterial.)
So we get $\Pr[G_{n,p}\text{ has }M] \ge \frac14 \Pr[G_{n,m}\text{ has }M]$. On the other hand, $\neg M$ is also a monotone property, and therefore $\Pr[G_{n,p}\text{ has a}\neg M] \ge \frac14 \Pr[G_{n,m}\text{ has }\neg M]$ which implies $1 - \Pr[G_{n,p}\text{ has }M] \ge \frac14 (1 - \Pr[G_{n,m}\text{ has }M])$. Putting these together, we get the inequality
$$\frac14 \Pr[G_{n,m} \text{ has M}] \le \Pr[G_{n,p} \text{ has M}] \le \frac34 + \frac14 \Pr[G_{n,m} \text{ has M}]$$
This is an awkward inequality with a gap of $\frac34$, but it does have one consequence: if $G_{n,p}$ either has or does not have $M$ with high probability, the same must hold of $G_{n,m}$. In particular, a result about having triangles whp can be proven in $G_{n,p}$ and it will hold in $G_{n,m}$ as well.