Proof of Layer-Cake Formula for the Second Moment

Step 1: $Y^2$ is a nonnegative random variable. For real $a<0$ we have $\{Y^2>a\}=\Omega$. For $a\ge0$, since $Y\ge0$ pointwise, $Y(\omega)^2>a$ holds if and only if $Y(\omega)>\sqrt{a}$, where $\sqrt{a}$ is the nonnegative square root of $a$: indeed, for $y\ge0$, $y>\sqrt{a}$ implies $y^2>\sqrt{a}\,y\ge\sqrt{a}\sqrt{a}=a$ when $y>0$ (and $a<y^2$ trivially if $\sqrt{a}=0<y$), while $y\le\sqrt{a}$ implies $y^2\le\sqrt{a}^2=a$. Hence $\{Y^2>a\}=\{Y>\sqrt{a}\}\in\mathcal{F}$, and $Y^2$ is measurable by the half-line criterion of Measurable Function and Real-Valued Measurable Function. Its integral $\int_\Omega Y^2\,dP$ is defined in $[0,\infty]$ by Lebesgue Integral of a Nonnegative Measurable Function.

Step 2: the product space. The measure $P$ is finite and $m$ is $\sigma$-finite: $\mathbb{R}=\bigcup_{n\in\mathbb{N}}[-n,n]$ and $m([-n,n])=2n<\infty$ by the interval property of Existence of Lebesgue Measure on the Real Line. By Product Sigma-Algebra and Existence and Uniqueness of the Product Measure the product measure $P\otimes m$ exists on the product $\sigma$-algebra $\mathcal{F}\otimes\mathcal{B}(\mathbb{R})$ of $\Omega\times\mathbb{R}$.

Step 3: a measurable region and integrand. Let

Then $E=\bigcup_{q}\bigl(\{Y>q\}\times(0,q)\bigr)$, the union over all positive rational $q$: if $0<u<Y(\omega)$, the density of the rationals provides a rational $q$ with $u<q<Y(\omega)$, so $(\omega,u)\in\{Y>q\}\times(0,q)$; conversely $u<q<Y(\omega)$ implies $0<u<Y(\omega)$. Each set $\{Y>q\}\times(0,q)$ is a measurable rectangle in $\mathcal{F}\otimes\mathcal{B}(\mathbb{R})$ (Product Sigma-Algebra), the rationals form a countable family, and a $\sigma$-algebra is closed under countable unions; hence $E\in\mathcal{F}\otimes\mathcal{B}(\mathbb{R})$.

Define $f:\Omega\times\mathbb{R}\to\mathbb{R}$ by $f(\omega,u)=2u$ for $(\omega,u)\in E$ and $f(\omega,u)=0$ otherwise; $f\ge0$ everywhere since $u>0$ on $E$. For $a<0$, $\{f>a\}=\Omega\times\mathbb{R}$; for $a\ge0$,

an intersection of members of $\mathcal{F}\otimes\mathcal{B}(\mathbb{R})$. By the half-line criterion of Measurable Function and Real-Valued Measurable Function, $f$ is measurable.

Step 4: the two iterated integrals. By the Tonelli part of Tonelli and Fubini Theorems applied to the nonnegative measurable function $f$, both iterated integrals are defined, the slice-integral functions are measurable, and both iterated integrals equal $\int f\,d(P\otimes m)$.

The $\omega$-slices. Fix $\omega$ and put $c=Y(\omega)\ge0$. The slice $f(\omega,\cdot)$ is the function $u\mapsto2u$ on $(0,c)$ and $0$ elsewhere. If $c=0$ the slice is identically $0$ and its integral is $0=c^2$. If $c>0$: the function $h(u)=2u$ is continuous on $[0,c]$ (for $u_0\in[0,c]$ and $\varepsilon>0$ take $\delta=\varepsilon/2$ in Continuity at a Point), and by Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval its zero extension $\tilde{h}=2u\,\mathbf{1}_{[0,c]}$ is integrable with

the Riemann integral. The function $F(u)=u^2$ on $\mathbb{R}$ is an antiderivative of $h$ extended to $\mathbb{R}$: every real point is an interior point of the interval $\mathbb{R}$, the identity function has difference quotients constantly $1$ and hence derivative $1$, and the product rule (claim 3 of Sum and Product Rules for One-Dimensional Derivatives and Continuity) gives $F'(u)=1\cdot u+u\cdot1=2u$. By Fundamental Theorem of Calculus, Part II in One Dimension, $\int_0^c2u\,du=F(c)-F(0)=c^2$. Finally $f(\omega,\cdot)=\tilde{h}-2c\,\mathbf{1}_{\{c\}}$ pointwise (the two sides agree off $\{0,c\}$, at $0$ both vanish, and at $c$ both equal $0$), the simple function $2c\,\mathbf{1}_{\{c\}}$ has integral $2c\,m(\{c\})=0$ since $m(\{c\})\le m((c-\varepsilon,c])=\varepsilon$ for every $\varepsilon>0$ by monotonicity and the interval property of Existence of Lebesgue Measure on the Real Line, and by the linearity of the integral (Linearity and Monotonicity of the Lebesgue Integral)

Hence the first iterated integral is $\int_{\Omega}Y^{2}\,dP$.

The $u$-slices. Fix $u\in\mathbb{R}$. If $u\le0$ the slice $f(\cdot,u)$ is identically $0$ with integral $0=\varphi(u)$. If $u>0$, the slice is the simple function $2u\,\mathbf{1}_{\{Y>u\}}$ with integral $2u\,P(Y>u)=\varphi(u)$. Hence the second iterated integral is $\int_{\mathbb{R}}\varphi\,dm$, and Tonelli asserts that $\varphi$, being the slice-integral function, is measurable — proving claim 1 — and that

proving claim 2.

Step 5: claim 3. By Square-Integrable Random Variables and the Mean-Square Inner Product, $Y$ is square-integrable exactly when $\int_\Omega Y^2\,dP<\infty$, which by claim 2 is equivalent to $\int_{\mathbb{R}}\varphi\,dm<\infty$; in that case the second moment $\mathbb{E}[Y^2]$ of Expectation, Variance, and Moments is the integral $\int_\Omega Y^2\,dP$, and the identity of claim 2 gives claim 3. $\blacksquare$