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Let $V_t$ be the total variation of $X$ on $[0,t]$, and set ${\tau }_n=\inf \{t:V_t\ge n\}$. Adaptation and continuity of $V$ make these stopping times; finite variation on every compact interval gives ${\tau }_n\uparrow \infty$. The stopped process $U=X^{{\tau }_n}$ satisfies $|U_t|\le n$ and has total variation at most $n$. It is a martingale by the bounded local martingale criterion, and is square-integrable.

Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_m=t$ with mesh tending to zero. The martingale-difference orthogonality of the increments givesPathwise, however,by uniform continuity on $[0,t]$. These sums are bounded by $n^2$, so the dominated convergence theorem gives $\mathbb{E}{U}_t^2=0$. Hence $U_t=0$ almost surely. Take all rational $t$ on a common probability-one event and use continuity to obtain $U\equiv 0$. Then let $n\to \infty$. This proves the continuous finite-variation local martingale is constant result in its zero-starting form:The proof does not assume the uniqueness of quadratic variation that is to be established next.

The martingale property gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$. Both $X_m$ and $X_n$ are square-integrable, so their product is integrable by the Cauchy-Schwarz inequality. Using the ${\mathcal{F}}_m$-measurability of $X_m$ in conditional expectation givesThe pull-out step for the possibly unbounded $X_m$ can be justified by first truncating it, then passing to the limit in the L2 norm and using the Cauchy-Schwarz inequality. Expanding the square now yieldsThe left side is nonnegative. Thus $a_n=\mathbb{E}[X_n^2]$ is nondecreasing and bounded above by $M$, and so $a_n\to a\le M$. For $m\le n$,By symmetry of the norm difference this proves that $(X_n)$ is a Cauchy sequence in the L2 space. The completeness established in part (a) gives an $X_{\infty }\in L^2$ withThis is the L2-bounded martingale convergence theorem, derived from martingale-difference orthogonality and completeness. The same sequence also converges almost surely by part (b), since $\mathbb{E}|X_n|\le \sqrt{M}$; uniqueness of a limit in probability identifies that limit with this $X_{\infty }$.

Use the representation from part (a), and put ${\Delta }_{j}M=M_{T_{j+1}}-M_{T_j}$, $U_j=h_j{\Delta }_{j}M$ and $N=H\cdot M$. The optional stopping theorem in the form needed here says that for a uniformly integrable martingale $L$ and stopping times $\sigma \le \tau$, including terminal infinite times, $\mathbb{E}[L_{\tau }\mid {\mathcal{F}}_{\sigma }]=L_{\sigma }$. An L2-bounded continuous martingale is uniformly integrable and has the required terminal value.

Consequently $\mathbb{E}[{\Delta }_{j}M\mid {\mathcal{F}}_{T_j}]=0$. Conditioning on ${\mathcal{F}}_t$ and distinguishing whether $T_j\le t$ givesThus $N_t=\mathbb{E}[{\sum }_j{U}_j\mid {\mathcal{F}}_t]$ is a martingale. Its paths are continuous, since each stopped increment is continuous and zero until its left endpoint.

For $i<j$, $U_i$ is ${\mathcal{F}}_{T_j}$-measurable, so the same conditional identity yields $\mathbb{E}(U_i{U}_j)=0$. These martingale-difference orthogonality relations also hold for the unweighted increments. ThereforeHere $N_{\infty }={\sum }_j{U}_j\in L^2$. By conditional Jensen, $\mathbb{E}{N}_t^2\le \mathbb{E}{N}_{\infty }^2$ for every $t$, soThis proves the norm bound without presupposing the Itô isometry requested next.

No quadratic variation argument is needed. For deterministic $u<v$, the martingale property and square integrability giveConsequentlyApply this identity separately to each consecutive grid increment. The second moments telescope, givingThis uniform bound follows directly from martingale-difference orthogonality, including the final interval specified by the ceiling.

Write $\mu =1+\epsilon$, $p=\mu /n$, $q=1-p$, and let $T$ be the total progeny of a binomial branching process. The preceding binomial branching survival correction gives $\rho =(2+o(1))\epsilon$. For a branching process conditioned on extinction, the offspring probability generating function is $f((1-\rho )s)/(1-\rho )$, with meanfor sufficiently large $n$. Summing expected generation sizes therefore gives $\mathbb{E}[T;T<\infty ]=(1-\rho )/(1-f^{\prime }(1-\rho ))=O(1/\epsilon )$.

Put $K=\lceil {\epsilon }^{-3}\rceil$ and let $N_{\ge K}$ count vertices in graph components of order at least $K$. A breadth-first exploration of a binomial random graph is dominated by $T$. The Markov inequality applied to finite total progeny givesAlways $L_1\le K+N_{\ge K}$. Since $n{\epsilon }^4\ge n^{1/3}\to \infty$, we have $K=o(\epsilon n)$ and $n/(\epsilon K)=O(n{\epsilon }^2)=o(\epsilon n)$. Hence $\mathbb{E}{L}_1\le (2+o(1))\epsilon n$. This bound controls the expected value directly, including rare large graph components.

For the lower bound, use the breadth-first exploration of a binomial random graph with $U_t$ unseen vertices, $A_t$ active vertices and $t$ explored vertices, starting from $U_0=n$, $A_0=0$. Let $b_t=1_{\{A_{t-1}=0\}}$ indicate a new root. Conditional on the past, the newly discovered count $Z_t$ has a binomial distribution $\mathrm{Bin}(U_{t-1}-b_t,p)$. Put $D_t=Z_t-p(U_{t-1}-b_t)$. The unseen-count recursion givesHere $M_t$ is a martingale. Up to $T_0=\lceil 3\epsilon n\rceil$, martingale-difference orthogonality gives $\mathbb{E}{M}_{T_0}^2\le C\epsilon n$, since each conditional variance is at most $np<2$ and $q^{-2T_0}$ is bounded.

Take $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$. Then $h\to 0$, $\epsilon =o(h)$ and $h^{2}n{\epsilon }^3\to \infty$. The Doob L2 maximal inequality impliesThe Taylor expansion, uniformly for $t\le T_0$, gives $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$. Thus $F(t)\ge h{\epsilon }^{2}n/3$ throughout the integer interval $[\lceil h\epsilon n\rceil ,\lfloor (2-h)\epsilon n\rfloor ]$ for large $n$. The new-root sum is nonnegative, so on the preceding event $A_t>0$ throughout this interval. No graph component finishes there: all these explored vertices belong to one graph component, of order at least $(2-2h)\epsilon n-O(1)$, with high probability. Its expected value is therefore at least $(2-o(1))\epsilon n$. Combining both inequalities provesThe barely-supercritical largest-component expectation uses both a positive exploration window and a finite-progeny bound; a bare convergence-in-probability assertion would not by itself justify this expectation.

Write ${\Delta }_{i}M=M_{t_i}-M_{t_{i-1}}$ and $Q={\sum }_i({\Delta }_{i}M{)}^2$. The identityshows, using the martingale-difference orthogonality, thatAlso $\mathbb{E}[Q]=\mathbb{E}[M_t^2]-\mathbb{E}[M_0^2]\le C^2$. Hence $\mathbb{E}[Q^2]\le 10C^4$, which in particular proves the requested bound