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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.