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

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 $\mathcal{G}$ in place of $\mathcal{F}$, since their arguments use only preimages of the functions involved; we use them for $\mathcal{G}$ without further comment.

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

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

$$P(A_j)\le\Bigl(\frac{16}{9}\Bigr)^{j}\,\mathbb{E}[D_j^{2}]\le\Bigl(\frac{16}{9}\Bigr)^{j}\Bigl(\frac{1}{4}\Bigr)^{j}=\Bigl(\frac{4}{9}\Bigr)^{j}.$$

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

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

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

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

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

$$\Omega^{*}=\bigl\{\omega\in\Omega:(X_{n_j}(\omega))_{j\in\mathbb{N}}\ \text{is a Cauchy sequence}\bigr\}=\bigcap_{k\in\mathbb{N}}\ \bigcup_{J\in\mathbb{N}}\ \bigcap_{j,j'\ge J}\Bigl\{\bigl|X_{n_j}-X_{n_{j'}}\bigr|\le\tfrac1k\Bigr\},$$

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

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

the limit existing on $\Omega^{*}$ by Every Cauchy Sequence of Real Numbers Converges. Put $Z_j=X_{n_j}\mathbf{1}_{\Omega^{*}}$, where $\mathbf{1}_{\Omega^{*}}$ is $1$ on $\Omega^{*}$ and $0$ off it; $\mathbf{1}_{\Omega^{*}}$ is $\mathcal{G}$-measurable because $\Omega^{*}\in\mathcal{G}$, so each $Z_j$ is $\mathcal{G}$-measurable, and $Z_j(\omega)\to X(\omega)$ for every $\omega\in\Omega$ (on $\Omega^{*}$ this is the definition of $X$; off $\Omega^{*}$ both sides are $0$). For every real $a$,

$$\{X>a\}=\bigcup_{k\in\mathbb{N}}\ \bigcup_{J\in\mathbb{N}}\ \bigcap_{j\ge J}\Bigl\{Z_j>a+\tfrac1k\Bigr\}.$$

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

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

Fix $j$. For $i\ge j$ consider the nonnegative random variables $f_i=(X_{n_j}-X_{n_i})^{2}\,\mathbf{1}_{\Omega^{*}}$. For every $\omega$, $f_i(\omega)\to(X_{n_j}(\omega)-X(\omega))^{2}\,\mathbf{1}_{\Omega^{*}}(\omega)$ as $i\to\infty$ (on $\Omega^{*}$ by the algebra of limits applied to the square of a convergent sequence; off $\Omega^{*}$ both sides are $0$). By Fatou's lemma and monotonicity ($f_i\le(X_{n_j}-X_{n_i})^{2}$ pointwise), together with Step 1,

$$\mathbb{E}\bigl[(X_{n_j}-X)^{2}\mathbf{1}_{\Omega^{*}}\bigr]\le\liminf_{i\to\infty}\mathbb{E}[f_i]\le\liminf_{i\to\infty}\ \mathbb{E}\bigl[(X_{n_j}-X_{n_i})^{2}\bigr]\le4^{-j}.$$

The function $(X_{n_j}-X)^{2}\mathbf{1}_{\Omega\setminus\Omega^{*}}$ is nonnegative and vanishes on $\Omega^{*}$, an event of probability $1$, so its expectation is $0$ by the null-support principle. Adding the two pieces (pointwise $(X_{n_j}-X)^{2}=(X_{n_j}-X)^{2}\mathbf{1}_{\Omega^{*}}+(X_{n_j}-X)^{2}\mathbf{1}_{\Omega\setminus\Omega^{*}}$, and expectation is additive on nonnegative random variables by Linearity and Monotonicity of the Lebesgue Integral),

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

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

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

$$\lVert X_n-X\rVert_{2}\le\lVert X_n-X_{n_j}\rVert_{2}+\lVert X_{n_j}-X\rVert_{2}<\frac{\varepsilon}{2}+2^{-j}<\varepsilon.$$

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

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

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

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