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On an independent Bernoulli product measure, the Harris-Kleitman inequality states that increasing events $E,F$ obeyThe same holds for two decreasing events; an increasing and a decreasing event have the reversed inequality.

Prove the stronger function version by induction on the number of coordinates. For increasing bounded functions $f,g$, condition on the last coordinate $Z$, with $\mathrm{Pr}(Z=1)=p$. The law of total covariance givesThe first term is nonnegative by induction, and the other factors are nonnegative by monotonicity. The one-coordinate case is the same formula without the first term. Event indicators prove the assertion. Two decreasing functions follow by replacing both by their negatives; replacing only one gives negative correlation of increasing and decreasing events.

Here is a precise Janson inequality and its lower-tail concentration form. Let $I_i$ indicate containment of a specified set of independent Bernoulli coordinates, set $X={\sum }_i{I}_i$, $\mu =\mathbb{E}X$, and letThis Janson dependency sum is ordered and excludes the diagonal. ThenA zero denominator means the trivial case $X=0$ almost surely.

To prove the avoidance bound, expose the avoidance events in order. For a fixed $i$, separate earlier indices into those whose supports are disjoint from $S_i$ and those overlapping it. Write $C$ for avoidance of the first group and $D$ for avoidance of the second. The event $C$ is independent of $I_i$, while $C$ is decreasing and $\{I_i=I_j=1\}$ is increasing. The correlation inequality therefore givesThe first inequality divides a nonnegative numerator by $\mathrm{Pr}(D\mid C)\le 1$; the second uses a union bound and negative correlation with $C$. Conditioning events of zero probability already make full avoidance impossible. Multiply the resulting conditional avoidance bounds to obtain the Janson sequential product lemma. Taking logarithms and using $\log (1-u)\le -u$ gives $-\mu +\Delta /2$, since each unordered dependency pair is counted once in the sequential sum.

For the Janson lower-tail bound by independent thinning, attach independent selectors of probability $t$ to the events. Their total expectation and dependency sum are $t\mu ,t^2\Delta$, so avoidance of the selected events givesSet $t=1-e^{-s}$ and apply the exponential Markov bound to $e^{-sX}$. Using $s-s^2/2\le 1-e^{-s}\le s$ yieldsChoosing $s=a/(\mu +\Delta )$ proves the concentration assertion. For avoidance alone, optimizing $-t\mu +t^2\Delta /2$ also gives $e^{-{\mu }^2/(2\Delta )}$ when $\Delta \ge \mu$, and the original bound gives $e^{-\mu /2}$ when $\Delta \le \mu$.

To outline the sharp chromatic application, view an independent set of $G(n,p)$ as a clique of its complement, whose edge probability is $q=1-p$. Fix $0<\epsilon <1$, let $m=\lceil n/(\log n{)}^2\rceil$ and $r=\lfloor (2-\epsilon ){\log }_{b}n\rfloor$. For a particular $m$-vertex set, let $X$ count independent $r$-sets. Then $\mu =\left(\genfrac{}{}{0ex}{}{m}{r}\right)q^{\left(\genfrac{}{}{0ex}{}{r}{2}\right)}$, and the overlap calculation givesOverlap of only one vertex shares no edge, hence is independent. For completeness, the summand is at most $(r^2/(m-r){)}^{\ell }{b}^{\left(\genfrac{}{}{0ex}{}{\ell }{2}\right)}/\ell !$. Since $r\le (2-\epsilon /2){\log }_{b}m$ eventually, large overlaps are bounded by $m^{-c_{\epsilon }\ell }$; finitely many smaller overlaps are dominated by $\ell =2$. Also $\log \mu ={\Omega }_{p,\epsilon }((\log m{)}^2)$, so the diagonal contribution $1/\mu$ is negligible. The concentration bound consequently givesA union bound over at most $2^n$ vertex sets still tends to zero. Thus independent sets in every large subset of a dense random graph are available uniformly, including adaptively obtained remainders. Repeatedly remove an independent $r$-set and give it a new colour, then colour the fewer than $m$ leftover vertices individually. This uses at most $n/r+m$ colours. Combining with the first-moment lower bound and letting fixed $\epsilon$ be arbitrarily small gives the accurate leading asymptoticThe uniform subset estimate, rather than merely the existence of one large independent set, is what makes the sharp upper bound possible.

The parameter relations give $\mu =\log (n/c)$ andFor each fixed integer $r\ge 1$, take an $r$-vertex set $S$ and let $X_S$ count triangles meeting $S$. The event that every vertex of $S$ is counted by $T$ is exactly $X_S=0$.

The number of such triangles isThus their total expected count is ${\mu }_S=N_S{p}^3=r\mu +O_r(n{p}^3)=r\mu +o(1)$.

Distinct triangles have dependent containment events only when they share an edge. If that common edge meets $S$, there are $O_r(n)$ choices for it and $O(n^2)$ choices for the two other vertices. If the common edge avoids $S$, both other vertices must lie in $S$, giving only $O_r(n^2)$ pairs. Each pair needs five edges. Consequently its ordered Janson dependency sum isJanson inequality gives the upper bound $\mathrm{Pr}(X_S=0)\le \exp (-{\mu }_S+{\Delta }_S/2)$. For the lower bound, the avoidance of each triangle is a decreasing event. Repeated application of Harris' inequality givesThe error $n^2{p}^6=O((\log n{)}^2/n^2)$ also tends to zero. The bounds match multiplicatively:Summing over ordered distinct vertices yields the factorial moments

Here is an explicit justification that these moments determine the limit. For every integer $t\ge 0$, the finite alternating sumsare upper bounds for $\mathrm{Pr}(T=t)$ when $L$ is even and lower bounds when $L$ is odd. This follows by applying the Bonferroni inequalities to $\left(\genfrac{}{}{0ex}{}{T}{t}\right){\sum }_j(-1{)}^j(\genfrac{}{}{0ex}{}{T-t}{j})$; the full sum is precisely the indicator of $T=t$. First let $n\to \infty$ for a fixed truncation and use the factorial-moment limits. Then let the even and odd truncations grow. Both bounds tend toHence every point probability has the Poisson distribution limit, proving the Poisson threshold for vertices lying in no triangle:This is also the factorial-moment criterion for Poisson convergence, here justified rather than left as an unproved implication.

Let $X$ count the triangles of the binomial random graph $G(n,p)$. ThenTwo distinct triangle indicators are dependent only when the triangles share an edge. Hence their ordered Janson dependency sum is

If $0<p\le n^{-1/2}$, then $\Delta =O(\mu )$. The first Janson inequality givesFor the reverse bound, the events that individual triangles are absent are decreasing, so Harris' inequality giveswhere $p<1/2$ keeps the logarithmic estimate uniform. Thus the probability is $e^{-\Theta (p^3{n}^3)}$.

If $n^{-1/2}\le p<1/2$, the extended Janson bound givesFor a lower bound, fix a balanced bipartition of the vertices and require every edge inside either part to be absent. The resulting graph is bipartite, hence triangle-free, and this event has probabilityCombining the bounds proves the second regime.

For a fixed set of $r$ vertices, exclude all triangles meeting that set. Their total expectation is $r\log (n/c)+o(1)$. Distinct triangle pairs sharing an edge contribute $O_r(n^3{p}^5)=o(1)$ to the Janson dependency sum. Harris' inequality and Janson inequality bound the avoidance probability between matching exponentials, giving $(c/n{)}^r(1+o(1))$. The factorial-moment criterion for Poisson convergence then gives the displayed law.