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Proof of Mean-Square Completeness of Square-Integrable Random Variables (Riesz-Fischer)

lemmalem:mean-square-completeness-2026a
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Reason: Proof of mean-square completeness via a geometrically spaced subsequence, Markov and Borel-Cantelli, an everywhere-defined G-measurable limit on the G-measurable Cauchy set, and Fatou. Approved by Aaron.

Proof

Throughout, expectations are handled with Linearity and Monotonicity of the Lebesgue Integral, mean-square norms and distances with Square-Integrable Random Variables and the Mean-Square Inner Product, and the triangle inequality with Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm. The measurability preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product (sums, differences, products, squares of measurable functions are measurable) apply verbatim with G\mathcal{G} in place of F\mathcal{F}, since their arguments use only preimages of the functions involved; we use them for G\mathcal{G} without further comment.

Step 1 (a subsequence with geometrically small increments). Using the Cauchy hypothesis with ε=2j\varepsilon=2^{-j}, choose recursively indices n1<n2<n_1<n_2<\dots such that XnXm2<2j\lVert X_n-X_m\rVert_{2}<2^{-j} for all n,mnjn,m\ge n_j. Set Dj=Xnj+1XnjD_j=X_{n_{j+1}}-X_{n_j}; then DjD_j is G\mathcal{G}-measurable and square-integrable with Dj2<2j\lVert D_j\rVert_{2}<2^{-j}, and more generally XniXnj2<2j\lVert X_{n_i}-X_{n_j}\rVert_{2}<2^{-j} for all iji\ge j.

Step 2 (almost sure absolute convergence along the subsequence). Let Aj={Dj(3/4)j}={Dj2(9/16)j}A_j=\{|D_j|\ge(3/4)^{j}\}=\{D_j^{2}\ge(9/16)^{j}\} (the two events coincide because squaring preserves the order of nonnegative reals). By Markov's inequality applied to the nonnegative random variable Dj2D_j^{2},

P(Aj)(169)jE[Dj2](169)j(14)j=(49)j.P(A_j)\le\Bigl(\frac{16}{9}\Bigr)^{j}\,\mathbb{E}[D_j^{2}]\le\Bigl(\frac{16}{9}\Bigr)^{j}\Bigl(\frac{1}{4}\Bigr)^{j}=\Bigl(\frac{4}{9}\Bigr)^{j}.

The series j(4/9)j\sum_j(4/9)^{j} converges: by the telescoping identity (1r)(r+r2++rm)=rrm+1(1-r)(r+r^{2}+\dots+r^{m})=r-r^{m+1} with r=4/9r=4/9, its partial sums are nondecreasing and bounded by r/(1r)=4/5r/(1-r)=4/5, hence converge to their least upper bound, which exists by the least upper bound property (for any ε>0\varepsilon>0 some partial sum exceeds the supremum minus ε\varepsilon, and all later partial sums lie between it and the supremum, which is the definition of the limit). By comparison (partial sums of jP(Aj)\sum_j P(A_j) are nondecreasing and bounded by those of j(4/9)j\sum_j(4/9)^{j}, hence by 4/54/5), jP(Aj)\sum_j P(A_j) converges. By the first Borel-Cantelli lemma, the event A=JNjJAjA=\bigcap_{J\in\mathbb{N}}\bigcup_{j\ge J}A_j (occurrence of infinitely many AjA_j) has P(A)=0P(A)=0.

Fix ωA\omega\notin A. There is J0J_0 with Dj(ω)<(3/4)j|D_j(\omega)|<(3/4)^{j} for all jJ0j\ge J_0. For J0j<jJ_0\le j<j', the telescoping bound gives

Xnj(ω)Xnj(ω)i=jj1Di(ω)i=jj1(34)i4(34)j,\bigl|X_{n_{j'}}(\omega)-X_{n_j}(\omega)\bigr|\le\sum_{i=j}^{j'-1}|D_i(\omega)|\le\sum_{i=j}^{j'-1}\Bigl(\frac{3}{4}\Bigr)^{i}\le4\Bigl(\frac{3}{4}\Bigr)^{j},

using the same telescoping identity with r=3/4r=3/4. Since 0<3/4<10<3/4<1, the sequence ((3/4)j)j((3/4)^{j})_j has limit 00: writing 4/3=1+x4/3=1+x with x=1/3x=1/3, Bernoulli's inequality gives (4/3)j1+jx(4/3)^{j}\ge1+jx, which exceeds any prescribed bound for large jj by the Archimedean property. Hence (Xnj(ω))j(X_{n_j}(\omega))_{j} is a Cauchy sequence of real numbers, and it converges by Every Cauchy Sequence of Real Numbers Converges.

Step 3 (a G\mathcal{G}-measurable limit defined everywhere). Let

Ω={ωΩ:(Xnj(ω))jN is a Cauchy sequence}=kN JN j,jJ{XnjXnj1k},\Omega^{*}=\bigl\{\omega\in\Omega:(X_{n_j}(\omega))_{j\in\mathbb{N}}\ \text{is a Cauchy sequence}\bigr\}=\bigcap_{k\in\mathbb{N}}\ \bigcup_{J\in\mathbb{N}}\ \bigcap_{j,j'\ge J}\Bigl\{\bigl|X_{n_j}-X_{n_{j'}}\bigr|\le\tfrac1k\Bigr\},

a set built from the G\mathcal{G}-measurable random variables XnjXnjX_{n_j}-X_{n_{j'}} by countable intersections and unions, hence ΩG\Omega^{*}\in\mathcal{G}. By Step 2, ΩΩA\Omega\setminus\Omega^{*}\subseteq A, so P(Ω)=1P(\Omega^{*})=1 by monotonicity and complementation. Define

X(ω)=limjXnj(ω)  (ωΩ),X(ω)=0  (ωΩ),X(\omega)=\lim_{j\to\infty}X_{n_j}(\omega)\ \ (\omega\in\Omega^{*}),\qquad X(\omega)=0\ \ (\omega\notin\Omega^{*}),

the limit existing on Ω\Omega^{*} by Every Cauchy Sequence of Real Numbers Converges. Put Zj=Xnj1ΩZ_j=X_{n_j}\mathbf{1}_{\Omega^{*}}, where 1Ω\mathbf{1}_{\Omega^{*}} is 11 on Ω\Omega^{*} and 00 off it; 1Ω\mathbf{1}_{\Omega^{*}} is G\mathcal{G}-measurable because ΩG\Omega^{*}\in\mathcal{G}, so each ZjZ_j is G\mathcal{G}-measurable, and Zj(ω)X(ω)Z_j(\omega)\to X(\omega) for every ωΩ\omega\in\Omega (on Ω\Omega^{*} this is the definition of XX; off Ω\Omega^{*} both sides are 00). For every real aa,

{X>a}=kN JN jJ{Zj>a+1k}.\{X>a\}=\bigcup_{k\in\mathbb{N}}\ \bigcup_{J\in\mathbb{N}}\ \bigcap_{j\ge J}\Bigl\{Z_j>a+\tfrac1k\Bigr\}.

Indeed, if X(ω)>aX(\omega)>a, choose kk with X(ω)>a+2/kX(\omega)>a+2/k; convergence gives JJ with Zj(ω)>a+1/kZ_j(\omega)>a+1/k for all jJj\ge J. Conversely, if Zj(ω)>a+1/kZ_j(\omega)>a+1/k for all jJj\ge J, then the limit satisfies X(ω)a+1/k>aX(\omega)\ge a+1/k>a by the order properties of limits. Hence {X>a}G\{X>a\}\in\mathcal{G} for every aa, and by the generator criterion of Measurable Function and Real-Valued Measurable Function, XX is G\mathcal{G}-measurable; in particular XX is a random variable, as GF\mathcal{G}\subseteq\mathcal{F}.

Step 4 (square-integrability and mean-square convergence). A nonnegative random variable WW with W=0W=0 on the complement of an event of probability 00 has E[W]=0\mathbb{E}[W]=0: every nonnegative simple function sWs\le W has each of its nonzero values attained on a subset of that null event, so its integral is 00 by monotonicity of the measure PP, and the integral of WW is the supremum of these. We refer to this as the null-support principle.

Fix jj. For iji\ge j consider the nonnegative random variables fi=(XnjXni)21Ωf_i=(X_{n_j}-X_{n_i})^{2}\,\mathbf{1}_{\Omega^{*}}. For every ω\omega, fi(ω)(Xnj(ω)X(ω))21Ω(ω)f_i(\omega)\to(X_{n_j}(\omega)-X(\omega))^{2}\,\mathbf{1}_{\Omega^{*}}(\omega) as ii\to\infty (on Ω\Omega^{*} by the algebra of limits applied to the square of a convergent sequence; off Ω\Omega^{*} both sides are 00). By Fatou's lemma and monotonicity (fi(XnjXni)2f_i\le(X_{n_j}-X_{n_i})^{2} pointwise), together with Step 1,

E[(XnjX)21Ω]lim infiE[fi]lim infi E[(XnjXni)2]4j.\mathbb{E}\bigl[(X_{n_j}-X)^{2}\mathbf{1}_{\Omega^{*}}\bigr]\le\liminf_{i\to\infty}\mathbb{E}[f_i]\le\liminf_{i\to\infty}\ \mathbb{E}\bigl[(X_{n_j}-X_{n_i})^{2}\bigr]\le4^{-j}.

The function (XnjX)21ΩΩ(X_{n_j}-X)^{2}\mathbf{1}_{\Omega\setminus\Omega^{*}} is nonnegative and vanishes on Ω\Omega^{*}, an event of probability 11, so its expectation is 00 by the null-support principle. Adding the two pieces (pointwise (XnjX)2=(XnjX)21Ω+(XnjX)21ΩΩ(X_{n_j}-X)^{2}=(X_{n_j}-X)^{2}\mathbf{1}_{\Omega^{*}}+(X_{n_j}-X)^{2}\mathbf{1}_{\Omega\setminus\Omega^{*}}, and expectation is additive on nonnegative random variables by Linearity and Monotonicity of the Lebesgue Integral),

XnjX22=E[(XnjX)2]4j.\lVert X_{n_j}-X\rVert_{2}^{2}=\mathbb{E}\bigl[(X_{n_j}-X)^{2}\bigr]\le4^{-j}.

In particular XnjXX_{n_j}-X is square-integrable, hence so is X=Xnj(XnjX)X=X_{n_j}-(X_{n_j}-X) by the closure properties of Square-Integrable Random Variables and the Mean-Square Inner Product.

Finally, let ε>0\varepsilon>0. Choose NN with XnXm2<ε/2\lVert X_n-X_m\rVert_{2}<\varepsilon/2 for n,mNn,m\ge N, and choose jj with njNn_j\ge N and 2j<ε/22^{-j}<\varepsilon/2 (possible since 2j02^{-j}\to0 as in Step 2). For every nNn\ge N, the triangle inequality of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm applied to XnX=(XnXnj)+(XnjX)X_n-X=(X_n-X_{n_j})+(X_{n_j}-X) gives

XnX2XnXnj2+XnjX2<ε2+2j<ε.\lVert X_n-X\rVert_{2}\le\lVert X_n-X_{n_j}\rVert_{2}+\lVert X_{n_j}-X\rVert_{2}<\frac{\varepsilon}{2}+2^{-j}<\varepsilon.

Hence XnX20\lVert X_n-X\rVert_{2}\to0 in the sense of Limit of a Sequence of Real Numbers.

Step 5 (uniqueness up to almost-sure equality). Let XX' be square-integrable with XnX20\lVert X_n-X'\rVert_{2}\to0. For every nn, the triangle inequality gives

0XX2XXn2+XnX2,0\le\lVert X-X'\rVert_{2}\le\lVert X-X_n\rVert_{2}+\lVert X_n-X'\rVert_{2},

and the right side has limit 00, so the constant XX2\lVert X-X'\rVert_{2} is ε\le\varepsilon for every ε>0\varepsilon>0, hence equals 00. By Square-Integrable Random Variables and the Mean-Square Inner Product, P(X=X)=1P(X=X')=1. \blacksquare

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