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If $k(x)=\varphi (|x|)\ge 0$ is decreasing in radius and has integral one, then $|k_t\ast f(x)|\le Mf(x)$, where $k_t(x)=t^{-d}k(x/t)$ and $M$ is the centered Hardy-Littlewood maximal function of $|f|dx$. The layer cake representation expresses $k_t$ as a positive mixture of ball indicators; each corresponding integral is bounded by its ball volume times $Mf(x)$. Integrating these volumes gives exactly one. This applies to the Poisson kernel for the upper half-space and supplies a weak type (1,1) domination for its boundary convergence.

A simple function is a measurable function whose image is a finite set. Its value sets $A_j=\{x:s(x)=a_j\}$ form a finite measurable partition. For nonnegative $a_j$, its Lebesgue integral is ${\sum }_j{a}_j\mu (A_j)$, with the convention $0\cdot \infty =0$. The pieces need not be intervals, or even connected sets. See MIT lecture 10, Definitions 100 and 106.

For example, on $[0,1]$ let $s$ be $2$ on $[0,1/4)$, $1$ on $[1/4,3/4)$, and $0$ on $[3/4,1]$. Its Lebesgue integral is $2(1/4)+1(1/2)=1$. Its superlevel sets have Lebesgue measure $3/4$ for $0\le t<1$, $1/4$ for $1\le t<2$, and $0$ thereafter. The layer cake representation gives the same answer: $(3/4)(1)+(1/4)(1)=1$. These two calculations are the building blocks of the general Lebesgue integral.

On the unit sphere, write $x=A(V)$ and $s=4\pi$. The expression equals $|xA(D\setminus V)-(s-x)A(D\cap V)|$, at most $x(s-x)$. A constant-version spherical isoperimetric inequality therefore gives the displayed bound, uniformly in nonempty open $D$. A smooth positive area density $\theta$ gives a second metric; integrating this bound over $D_t=\{\theta >t\}$ by the layer cake representation bounds the discrepancy between the two normalized area fractions by $\kappa (\max \theta -\min \theta )L^2/(s\tilde{A}(S))$.

Let $S$ be a sublinear operator with positive constants and $\mu \{Sf>a\}\le C_1\Vert f{\Vert }_1/a$ and $\Vert Sf{\Vert }_{\infty }\le C_{\infty }\Vert f{\Vert }_{\infty }$. Split $f$ into its part above $a/(2C_{\infty })$ and its bounded remainder. The latter contributes at most $a/2$, so the weak bound controls the superlevel measure by $2C_1{a}^{-1}{\int }_{|f|>a/(2C_{\infty })}|f|$. Integrating this estimate with layer cake representation proves the displayed bound for $1<p<\infty$ and $q=p/(p-1)$. In particular it gives the strong $L^p$ estimate for the Hardy-Littlewood maximal function.

For $1<q<\infty$, the weak Lq space consists of measurable functions for which $A_q(f)={\sup }_{t>0}t|\{|f|>t\}{|}^{1/q}$ is finite. This distribution-function expression is a quasi-norm. The equivalent norm $N_q(f)={\sup }_{0<|E|<\infty }|E{|}^{-1/q^{\prime }}{\int }_E|f|$ satisfies $A_q\le N_q\le q^{\prime }{A}_q$, with $q^{\prime }=q/(q-1)$. The layer cake representation bounds the integral by ${\int }_0^{\infty }\min (|E|,(A_q/t{)}^q)dt=q^{\prime }{A}_q|E{|}^{1/q^{\prime }}$. The lower bound uses finite-measure subsets of a superlevel set.