Layer cake representation

In mathematics, the layer cake representation of a non-negative, real-valued measurable function ${\displaystyle f}$ defined on a measure space ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ is the formula

${\displaystyle f(x)=\int _{0}^{\infty }1_{L(f,t)}(x)\,\mathrm {d} t,}$

Layer cake representation.

for all ${\displaystyle x\in \Omega }$, where ${\displaystyle 1_{E}}$ denotes the indicator function of a subset ${\displaystyle E\subseteq \Omega }$ and ${\displaystyle L(f,t)}$ denotes the (${\displaystyle \color {red}{\text{strict}}}$) super-level set:

${\displaystyle L(f,t)=\{y\in \Omega \mid f(y)\geq t\}\;\;\;{\color {red}{\text{or}}\;L(f,t)=\{y\in \Omega \mid f(y)>t\}}.}$

The layer cake representation follows easily from observing that

${\displaystyle 1_{L(f,t)}(x)=1_{[0,f(x)]}(t)\;\;\;{\color {red}{\text{or}}\;1_{L(f,t)}(x)=1_{[0,f(x))}(t)}}$

where either integrand gives the same integral:

${\displaystyle f(x)=\int _{0}^{f(x)}\,\mathrm {d} t.}$

The layer cake representation takes its name from the representation of the value ${\displaystyle f(x)}$ as the sum of contributions from the "layers" ${\displaystyle L(f,t)}$: "layers"/values ${\displaystyle t}$ below ${\displaystyle f(x)}$ contribute to the integral, while values ${\displaystyle t}$ above ${\displaystyle f(x)}$ do not. It is a generalization of Cavalieri's principle and is also known under this name.^{[1]: cor. 2.2.34}

Applications

The layer cake representation can be used to rewrite the Lebesgue integral as an improper Riemann integral. For the measure space, ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$, let ${\displaystyle S\subseteq \Omega }$, be a measureable subset (${\displaystyle S\in {\mathcal {A}})}$ and ${\displaystyle f}$ a non-negative measureable function. By starting with the Lebesgue integral, then expanding ${\displaystyle f(x)}$, then exchanging integration order (see Fubini-Tonelli theorem) and simplifying in terms of the Lebesgue integral of an indicator function, we get the Riemann integral:

${\displaystyle {\begin{aligned}\int _{S}f(x)\,{\text{d}}\mu (x)&=\int _{S}\int _{0}^{\infty }1_{\{x\in \Omega \mid f(x)>t\}}(x)\,{\text{d}}t\,{\text{d}}\mu (x)\\&=\int _{0}^{\infty }\!\!\int _{S}1_{\{x\in \Omega \mid f(x)>t\}}(x)\,{\text{d}}\mu (x)\,{\text{d}}t\\&=\int _{0}^{\infty }\!\!\int _{\Omega }1_{\{x\in S\mid f(x)>t\}}(x)\,{\text{d}}\mu (x)\,{\text{d}}t\\&=\int _{0}^{\infty }\mu (\{x\in S\mid f(x)>t\})\,{\text{d}}t.\end{aligned}}}$

This can be used in turn, to rewrite the integral for the L^p-space p-norm, for ${\displaystyle 1\leq p<+\infty }$:

${\displaystyle \int _{\Omega }|f(x)|^{p}\,\mathrm {d} \mu (x)=p\int _{0}^{\infty }s^{p-1}\mu (\{x\in \Omega$ :|f(x)|>s\})\mathrm {d} s,}

which follows immediately from the change of variables ${\displaystyle t=s^{p}}$ in the layer cake representation of ${\displaystyle |f(x)|^{p}}$. This representation can be used to prove Markov's inequality and Chebyshev's inequality.

See also

• Symmetric decreasing rearrangement