Proof of The Integrals of min(My², M|y|³/(2√n)) Tend to Zero under a Finite Second Moment

Step 1 (Borel functions). Fix $n\in\mathbb{N}$. Read in $\mathbb{R}$, $n\ge1$ (by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (xii) at $x=-1$, as $0^{n}=0$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii)); since $\sqrt{n}\ge0$ and $(\sqrt{n})^{2}=n>0$ (The k-th Root and the Square Root of a Nonnegative Real Number (i)), $\sqrt{n}>0$. Let $P,Q:\mathbb{R}\to\mathbb{R}$ be the functions

$$P(y)=M\,y\cdot y=My^{2},\qquad Q(y)=\frac{M}{2\sqrt{n}}\,(|y|\cdot|y|)\cdot|y|=\frac{M|y|^{3}}{2\sqrt{n}}$$

(for $y^{2}=y\cdot y$ and $|y|^{3}=(|y|\cdot|y|)\cdot|y|$ see Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (i) and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii)). With $(P-Q)^{+}$ the positive part of $P-Q$,

$$g_{n}=P-(P-Q)^{+},$$

because $\min\{s,t\}=s-\max\{s-t,0\}$ for all $s,t\in\mathbb{R}$: if $s\le t$ both sides equal $s$, and if $t\le s$ both equal $t$ (The Maximum and Minimum of Two Elements of a Total Order (ii)). The identity $y\mapsto y$ of $\mathbb{R}$ lies in $\mathcal{L}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii)). Applying Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i) on the measurable space $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ step by step, to products, absolute values, constant multiples, a difference and a positive part, the functions $y\mapsto y\cdot y$, $y\mapsto|y|$, $y\mapsto(|y|\cdot|y|)\cdot|y|$, $P$, $Q$, $P-Q$, $(P-Q)^{+}$ and finally $g_{n}$ are maps $\mathbb{R}\to\mathbb{R}$ lying in $\mathcal{L}$. So $(g_{n})_{n\in\mathbb{N}}$ is a sequence in $\mathcal{L}$ (the maps $n\mapsto M/(2\sqrt{n})$, $\mathbb{N}\to\mathbb{R}$, and $n\mapsto g_{n}$, $\mathbb{N}\to\mathcal{L}$, exist by Maps and Relations Given by Formulas (i), $\mathcal{L}$ being a set by Measurable Map, and the Set of Measurable Maps (ii); that lemma is used only for these sequences, each $g_{n}$ being a map already by Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)). This proves (i).

Step 2 (bounds). Let $n\in\mathbb{N}$ and $y\in\mathbb{R}$; then $P(y)\ge0$ and $Q(y)\ge0$. The value $r=(P-Q)^{+}(y)=\max\{P(y)-Q(y),0\}$ satisfies $r\ge0$ (Positive Part, Negative Part and Absolute Value of a Real-Valued Function (i)) and $r\ge P(y)-Q(y)$ (The Maximum and Minimum of Two Elements of a Total Order (iii)), so $g_{n}(y)=P(y)-r$ is at most $P(y)$ and at most $Q(y)$; and $r\le P(y)$, since $P(y)-Q(y)\le P(y)$ and $0\le P(y)$ (The Maximum and Minimum of Two Elements of a Total Order (iv)), so $g_{n}(y)\ge0$. Thus

$$0\le g_{n}(y)\le My^{2},\qquad 0\le g_{n}(y)\le\frac{M|y|^{3}}{2\sqrt{n}}.$$

Step 3 (pointwise limit). Fix $y\in\mathbb{R}$, let $K=M|y|^{3}/2\ge0$, and let a real $\varepsilon>0$ be given. Choose $N\in\mathbb{N}$ such that $n>K^{2}/\varepsilon^{2}$ for every $n\in\mathbb{N}$ with $n\ge N$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (vii)), and let $n\ge N$. Then $(\sqrt{n}\,\varepsilon)^{2}=n\varepsilon^{2}>K^{2}$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii)), and as $K$ and $\sqrt{n}\,\varepsilon$ are nonnegative, $K<\sqrt{n}\,\varepsilon$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (viii)), that is, $K/\sqrt{n}<\varepsilon$. By Step 2,

$$|g_{n}(y)-0|=g_{n}(y)\le\frac{K}{\sqrt{n}}<\varepsilon .$$

So for every $y$ and every $\varepsilon>0$ there is $N$ with $|g_{n}(y)-0|<\varepsilon$ for all $n\ge N$, which is

$$g_{n}(y)\to0\qquad(y\in\mathbb{R})$$

(Convergent Sequences of Real Numbers (i)).

Step 4 (integrability and the limit). Let $\mu$ be as in (ii) and (iii), and let $q\in\mathcal{L}^{1}(\mu)$ be the function $q(y)=y^{2}$. By Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii) (with $a=M$, $b=0$ and $f=g=q$) the function $Mq$, $y\mapsto My^{2}$, lies in $\mathcal{L}^{1}(\mu)$. By Step 1, $(g_{n})$ is a sequence in $\mathcal{L}$; by Step 3 it converges pointwise to the zero function $0:\mathbb{R}\to\mathbb{R}$; and by Step 2, $|g_{n}(y)|=g_{n}(y)\le My^{2}$ for every $n$ and every $y$. So the dominated convergence theorem applies on the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu)$ to $(g_{n})$, with limit function $0$ and dominating function $Mq$: by Dominated Convergence Theorem (i), $g_{n}\in\mathcal{L}^{1}(\mu)$ for every $n$, which is (ii), and $0\in\mathcal{L}^{1}(\mu)$. The integrals form a sequence $n\mapsto\int_{\mathbb{R}}g_{n}\,d\mu$ in $\mathbb{R}$ (the composite of the sequence $(g_{n})$ in $\mathcal{L}^{1}(\mu)$ with the integral on $\mathcal{L}^{1}(\mu)$, Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)), and by Dominated Convergence Theorem (iii),

$$\int_{\mathbb{R}}g_{n}\,d\mu\to\int_{\mathbb{R}}0\,d\mu .$$

Finally the zero function is $0\cdot q+0\cdot q$, so $\int_{\mathbb{R}}0\,d\mu=0\cdot\int_{\mathbb{R}}q\,d\mu+0\cdot\int_{\mathbb{R}}q\,d\mu=0$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)). This proves (iii).