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The triangle count of a finite simple graph is the number of unordered triples of vertices inducing a triangle in a graph. In a tripartite graph with parts $X,Y,Z$ and adjacency indicator functions $g_{XY},g_{YZ},g_{XZ}$, it equals $|X||Y||Z|{\mathbb{E}}_{x,y,z}{g}_{XY}(x,y)g_{YZ}(y,z)g_{XZ}(x,z)$, using uniform expectations on nonempty parts.

Let a tripartite graph have nonempty parts $X,Y,Z$, pair edge density of a bipartite graph values $\alpha ,\beta ,\gamma$ on $XY,YZ,XZ$, and $\Vert G(X,Y)-\alpha {\Vert }_{\square }\le c$. If every $z\in Z$ has exactly $\beta |Y|$ neighbors in $Y$, its normalized triangle count $\tau$ obeysThe constant-degree assumption makes the contribution of the constant $\alpha$ exactly $\alpha \beta \gamma$. For each fixed $z$, apply the bilinear correlation bound for the box norm to the indicator functions of its two vertex neighbourhoods. Their squared $L^2$ norms are $\beta$ and the relative $X$-degree of $z$. Average over $z$ and use the Cauchy-Schwarz inequality to bound the mean square root of that degree by $\sqrt{\gamma }$.

On $(0,1)$, for $k\ge 0$ and $0\le j<2^k$, the displayed indicator functions form a sequence that converges to zero in every finite Lp norm while failing to converge pointwise anywhere. Indeed, for $1\le p<\infty$,Every point belongs to exactly one interval in each dyadic block, and for $k\ge 1$ it misses at least one other interval in that block. Thus every point sees infinitely many ones and infinitely many zeros. Half-open intervals specify the behaviour at dyadic endpoints. This separates norm convergence in an Lp space from pointwise convergence and convergence almost everywhere.

In a one-period complete market with deterministic initial holdings, two integrable positive pricing variables assigning the same prices to every traded asset are equal almost surely. Replicate indicator functions to obtain equality of their integrals on every event, then apply this to the event where one density exceeds the other.

Wald's equation turns a random sum into a product of two expected values: the average contribution per step and the average number of steps. Its surprising feature is that the number of steps may depend on what has already happened. A procedure can stop on its first successful Bernoulli trial, for example, while the identity still accounts for its total cost. The crucial restriction is that the decision to take the next step cannot use that step's unobserved outcome.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, let $\mu =\mathbb{E}[X_1]$, and put $S_0=0$ and $S_n={\sum }_{i=1}^n{X}_i$. Suppose $N$ is a nonnegative integer-valued stopping time for the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$, with $\mathbb{E}[N]<\infty$. Then $S_N$ is an integrable random variable andThese hypotheses allow statistical dependence between $N$ and the increments; they do not allow access to future increments. The finite-expected value requirement is stronger than stopping almost surely.^[ref]

For the proof idea, count step $i$ only on $\{N\ge i\}$. This event is determined by the first $i-1$ observations, so its indicator function is independent of $X_i$. The Proof of Wald's equation by stopping indicators combines this observation with the tail-sum formula for expectation. The same calculation with $|X_i|$ gives $\mathbb{E}{\sum }_{i=1}^N|X_i|=\mathbb{E}|X_1|\mathbb{E}[N]<\infty$, which justifies interchanging the expected value and the infinite series. This is why the result survives a data-dependent end point without assuming an independent count.

The identity is useful in randomized algorithms and repeated sampling: one can translate an average amount of work into an average number of attempts, or conversely, without first finding the full probability distribution of the total. The examples below separate a permissible stopping rule from a rule that peeks ahead and from a valid rule whose average waiting time is infinite.

For finite subsets $A_i$ of a finite ambient set $X$, the indicator function of their complement is ${\prod }_i(1-1_{A_i})$. Expanding this finite product and summing it against any weight $f:X\to \mathbb{R}$ proves the weighted formula. The empty intersection is $X$. Negative weights are allowed; finiteness removes every convergence issue.

A window function is a function that vanishes outside a chosen finite interval; multiplying an input by it selects and weights that part of the history before a Fourier transform. The rectangular window $w_T$ is the indicator function of $[0,T]$. Applied to the weighted causal extension by zero of an input, it produces the finite-history integral ${\int }_0^T{e}^{-st}f(t)dt$ used to approximate a Laplace transform.