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Letbe the event that the partial sums first cross level $t$ at time $k$. On $A_k$, write $R_k=S_n-S_k$. Conditional on $X_1,\dots ,X_k$, the random variable $R_k$ is independent and symmetric. Sinceat least one of the two norms on the right exceeds $t$. Symmetry of $R_k$ consequently giveson $A_k$. The events $A_k$ are disjoint, so summation proves the Lévy maximal inequality

For the Gaussian series, put $S_m={\sum }_{j=1}^m{g}_j{u}_j$. Apply the inequality to the symmetric independent increments from $n+1$ through $m$, followed by Markov inequality in squared norm:Here the cross terms vanish by orthogonality of independent centered Hilbert-space random variables. Letting $m\to \infty$ and then $n\to \infty$ showsThe convergence criterion in the question now shows that ${\sum }_j{g}_j{u}_j$ converges almost surely in $V$.

In a dense graph with many common neighbours for every pair of vertices, choose disjoint random small sets. A set is good if few vertices have no neighbour in it. The Markov inequality controls the number of bad sets, and random sampling makes almost every pair of good sets adjacent. Add unused common neighbours to connect each set and to repair the remaining missing adjacencies. If the number of used vertices stays below every common-neighbour count, this constructs disjoint branch sets of a graph minor representing a large complete graph.

Suppose $|Z(u)-Z(v)|\le L\Vert u-v\Vert$ with a random finite $L$. On $L\le \ell$, a finite net with radius $\eta$ bounds the full supremum by the grid maximum plus $\ell \eta$. The Markov inequality controls the exceptional event using $\mathbb{E}L$, and a union bound controls the grid maximum. Choosing the net spacing balances these two costs.

If $\kappa >8$, then for every $0<\alpha <(1-8/\kappa )/2$ and fixed $T$, there is an almost surely finite random $C$ such thatfor $|x|\le 1$ and $0<y\le 1$. The proof combines the derivative martingale, Markov inequality, a dyadic lattice, the Borel-Cantelli lemmas, and the Koebe distortion theorem.

For a fixed mean-zero variance-one real entry law, centered truncation of a Wigner matrix satisfies $\mathbb{E}[N^{-1}\mathrm{Tr}(X-\hat{X}{)}^2]\le t(C)$, where $t(C)=\mathbb{E}[Y^2{1}_{|Y|\ge C}]\to 0$. The spectral Lipschitz bound from Frobenius distance and Markov inequality give probability at most $t(C)/{\epsilon }^2$ for a spectral test-function discrepancy exceeding $\epsilon$. Thus $t(C)<{\epsilon }^3$ works uniformly in matrix size. A varying entry-law family requires uniform second-moment tail decay instead.

Suppose a graph with $n$ vertices and $m>0$ edges has a vector coloring with adjacent inner products at most $-1/2$. Take $r$ independent copies of random hyperplane rounding and use the $r$ signs as a color. Each edge remains monochromatic with probability at most $3^{-r}$, so the expected number $B$ of monochromatic edges is at most $m{3}^{-r}$ by linearity of expectation.

Choose $r=\max \{0,\lceil {\log }_3(4m/n)\rceil \}$. Then Markov inequality gives $\mathbb{P}(B>n/2)\le 1/2$. On success, delete one endpoint of every monochromatic edge. This random alteration method leaves at least $n/2$ vertices properly colored withThe removed vertices can retain their original colors, as the semicoloring only requires the retained induced subgraph to be proper. A failed draw can be detected and repeated, with at most two trials on average. The case $m=0$ needs just one color. This is the elementary independent-hyperplane construction in Karger, Motwani and Sudan's paper on approximate graph coloring.

For fixed $0<\lambda <1$ in $G(n,\lambda /n)$, the largest component of a graph has the displayed order with high probability. A dominating Galton-Watson process gives the tail bound $\mathbb{P}(T\ge j)\le e^{-(\lambda -1-\log \lambda )(j-1)}$ by the exponential Markov inequality. For the lower bound, the tree-component expectation in the Erdős-Rényi model diverges at $j=\lfloor (1-\eta )\log n/(\lambda -1-\log \lambda )\rfloor$. The disjoint-set covariance factor and the exclusion of overlaps give relative variance tending to zero, so the second moment method supplies a tree component of that order.

For a finite set of primes choose $0<g(p)<1$, put $k(p)=g(p)/(1-g(p))$, and let $W={\prod }_p(1+k(p))$. Give each squarefree integer $d$ on these primes probability $k(d)/W$. Inclusion of each prime is represented by a Bernoulli random variable with parameter $g(p)$; these are independent random variables, so $\mathbb{E}[\log d]={\sum }_{p}g(p)\log p$. If this is at most $\theta \log L$ with $0<\theta <1$, the Markov inequality givesThis turns a full Euler product into a lower bound for the truncated normalizing sum of a Selberg sieve.

If ${\sup }_n\mathbb{E}[X_n^2]<\infty$, then $(X_n)$ is uniformly integrable. Indeed, the Cauchy-Schwarz inequality and Markov inequality give

Stopping a uniformly integrable discrete-time martingale at any stopping time preserves uniform integrability, including when the stopping time can be infinite. For $C={\sup }_n\mathbb{E}|X_n|$, the optional stopping theorem and the Markov inequality giveChoose $R$ first, then $K$, to make this uniformly small.

For one integrable random variable $Y$, the family $Z_{\mathcal{G}}=\mathbb{E}[Y\mid \mathcal{G}]$, over arbitrary sigma-algebras $\mathcal{G}$, has uniform integrability. On $A=\{|Z_{\mathcal{G}}|>K\}$, the defining property of conditional expectation gives $\mathbb{E}[|Z_{\mathcal{G}}|1_A]\le \mathbb{E}[|Y|1_A]$, while $\mathbb{P}(A)\le \mathbb{E}|Y|/K$ by the Markov inequality. The integrable random variable $Y$ has uniformly small absolute integrals over events of sufficiently small probability.