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If $0<\theta <1$ and nonnegative measurable functions satisfy $h((1-\theta )x+\theta y)\ge f(x{)}^{1-\theta }g(y{)}^{\theta }$ for all $x,y\in {\mathbb{R}}^n$, then their Lebesgue integrals obeyThe quantile derivative identity and monotone substitution inequality prove the one-dimensional case; Tonelli theorem extends it by induction. It implies the Brunn–Minkowski inequality.

For $1<p\le \infty$, the Hardy-Littlewood maximal function is bounded on $L^p({\mathbb{R}}^n)$. A direct proof for finite $p$ applies the weak $(1,1)$ estimate to $f{1}_{\{|f|>s/2\}}$, integrates the resulting distribution-function bound against $p{s}^{p-1}$, and uses Tonelli theorem.

Start traps at a unit-intensity Poisson random measure in Euclidean space and attach mutually independent Brownian motions, independent of the initial measure. For a deterministic target path $f$, independently thin initial positions according to whether their marked path comes within radius one of the target by time $t$. The retention probability $p_t(x)$ integrates, by the Tonelli theorem, to the expected volume of the sausage along $f-B$. Symmetry of Brownian motion makes this the expected volume of the sausage along $f+B$. The Poisson zero-count probability proves the displayed survival formula. This derives the formula directly from independent thinning, without assuming a separate marking or displacement theorem.

The ultimate survival probability satisfies $\phi (u)=\phi (0)+(\lambda /c){\int }_0^u\phi (u-x)(1-F(x))dx$. Condition on the first arrival of the Poisson process, differentiate the resulting exponentially weighted integral, and integrate the convolution derivative equation from zero. Tonelli theorem converts the claim-density convolution to the tail convolution. The kernel mass is $\lambda \mu /c<1$, so this is a defective renewal equation.

For $q>0$ and a random variable $X$ with finite absolute $q$th moment, Tonelli theorem gives

For a nonnegative random variable $X$, Tonelli theorem givesConsequently, for each $a>0$, $\mathbb{E}[X]=\infty$ exactly when ${\sum }_{n\ge 1}\mathbb{P}(X>an)=\infty$; compare the integral on successive intervals of length $a$.