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Expose the independent edge indicators of a binomial random graph in a fixed order, and set $M_i=\mathbb{E}[Z\mid \text{the first }i\text{ indicators}]$ for an integrable statistic $Z$. This is a martingale from $\mathbb{E}Z$ to $Z$. If changing one edge changes $Z$ by at most $c$, coupling the unexposed indicators gives $|M_i-M_{i-1}|\le c$, permitting the Azuma-Hoeffding inequality.

We need independent sets in every sufficiently large remaining induced subgraph, rather than just one in the initial random graph. The useful concentration variable is the maximum edge-disjoint clique packing in the complement graph; changing one edge changes that variable by at most one. This avoids the much larger sensitivity of the ordinary clique count in a binomial random graph.

First work in a binomial random graph $H\sim G(m,1/2)$, take $r=r_0(m)$, and let $X$ count its $r$-cliques. Put $\mu =\mathbb{E}X\ge m^{9/5}$ and let $Z$ be its maximum edge-disjoint clique packing size. Form the clique conflict graph whose vertices are these cliques, adjacent when they share an edge. If $Q$ counts unordered conflicting pairs, the Caro-Wei bound gives, for each realization,To verify the bound, randomly order the vertices of a finite graph and retain each vertex preceding all its graph neighbours. These retained vertices are an independent set, with expected size ${\sum }_v(1+\mathrm{deg}v{)}^{-1}\ge X^2/(X+2Q)$ by the Cauchy-Schwarz inequality. Interpret the ratio as zero when $X=0$. A second application of the Cauchy-Schwarz inequality, now to expectations, givesCounting ordered pairs of $r$-sets with $j$ common vertices, including identical sets when $j=r$, yieldsThe shared edges account for the factor $2^{\left(\genfrac{}{}{0ex}{}{j}{2}\right)}$. We claimHere are the estimates, including the large overlaps. For $2\le j\le \lfloor r/2\rfloor$, the probability that a random $r$-set contains any specified $j$ elements is at most $(r/(m-r){)}^j$. Summing over their choices inside the first $r$-set givesThe logarithm of $T_j$ is a convex function, a quadratic in $j$, so the maximum on this interval is at an endpoint. We have $T_2=O(r^4/m^2)$. As $r=(2+o(1)){\log }_{2}m$, the other endpoint satisfiesand is smaller than $T_2$ for sufficiently large $m$. Summing at most $r$ terms gives $O(r^5/m^2)$.

For the other half of the overlaps, write $j=r-s$, where $0\le s\le \lceil r/2\rceil$. The exact identity and an upper bound areThe quantity in parentheses is $m^{-1/2+o(1)}$, which tends to zero. The geometric series is thus at most $2/\mu$ for sufficiently large $m$. This proves the claim. In fact $r^5/m^2=o(m^{-9/5})$ and $1/\mu \le m^{-9/5}$, so eventuallyNow expose the $N=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edge indicators one at a time. The edge-exposure martingale $M_i=\mathbb{E}[Z\mid \text{first }i\text{ indicators}]$ starts at $\mathbb{E}Z$ and ends at $Z$. Deleting one edge destroys at most one member of any edge-disjoint clique packing. Hence changing one indicator changes $Z$ by at most one; coupling the remaining indicators shows $|M_i-M_{i-1}|\le 1$.

Precisely, the Azuma-Hoeffding inequality says that if $(M_i{)}_{i=0}^N$ is a martingale and $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then, for $t>0$,If all $c_i=0$, the martingale is constant. Applying the lower-tail inequality with $c_i=1$ and $t=\mathbb{E}Z$ givesfor sufficiently large $m$. This is the only martingale concentration inequality needed.

Return to the original binomial random graph $G\sim G(n,1/2)$ and set $m=\lceil n/({\log }_{2}n{)}^2\rceil$. For each fixed $m$-element vertex set $U$, the complement graph of $G[U]$ has distribution $G(m,1/2)$. The union bound therefore givessince $m^{8/5}/n\to \infty$. Thus with high probability every vertex set of size at least $m$ contains an independent set of sizeApply greedy colouring by removing independent sets: remove such an independent set, give it a fresh colour, and repeat while at least $m$ vertices remain. Give each of the fewer than $m$ final vertices its own colour. This uses at mostcolours, since $m=o(n/{\log }_{2}n)$. Together with the lower bound, this gives the chromatic number of the half-density binomial random graph: