Proof of Basic Properties of the Mean-Square Riemann Integral

Fix versions of all mean-square Riemann integrals appearing; they exist by Existence and Uniqueness of the Mean-Square Riemann Integral for Mean-Square Continuous Families. For a mean-square continuous family $(K_t)_{t\in[s,t']}$ and $n\ge1$, write $S_n(K;[s,t'])$ for the mean-square Riemann sum over the partition of $[s,t']$ into $n$ equal intervals with left endpoints as tags, and abbreviate $S_n(K)=S_n(K;[a,b])$. By Mean-Square Riemann Integral of a Family of Random Variables, $\lVert S_n(K;[s,t'])-\int_s^{t'}K_u\,du\rVert_2\to0$ for every version of the integral. We use throughout the triangle inequality and Cauchy-Schwarz inequality of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm, and uniqueness of mean-square limits: if $\lVert Z_n-V\rVert_2\to0$ and $\lVert Z_n-V'\rVert_2\to0$ then $\lVert V-V'\rVert_2=0$, so $V=V'$ almost surely by Square-Integrable Random Variables and the Mean-Square Inner Product. Thus, to prove each asserted identity it suffices to show that the mean-square Riemann sums converge in mean square to the right-hand side; the version convention of Mean-Square Riemann Integral of a Family of Random Variables is then satisfied because every version is also a limit of the same sums.

We also record: if $\varphi:[s,t']\to\mathbb{R}$ is continuous ($s<t'$), then the ordinary Riemann sums $R_n(\varphi;[s,t'])$ over the same left-tagged uniform partitions converge to the Riemann integral $\int_s^{t'}\varphi(u)\,du$; abbreviate $R_n(\varphi)=R_n(\varphi;[a,b])$. Indeed $\varphi$ is Riemann integrable by Continuous Functions on a Closed Interval are Riemann Integrable, and Riemann Integrability on a Closed Interval states precisely that for every $\varepsilon>0$ there is $\delta>0$ such that every Riemann sum on a tagged partition of mesh less than $\delta$ is within $\varepsilon$ of the integral; the mesh $(t'-s)/n$ tends to $0$. Linearity and monotonicity of the Riemann integral on continuous integrands follow from Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval together with Linearity and Monotonicity of the Lebesgue Integral.

1. Mean-square continuity of $(\alpha H_t+\beta G_t)$: $\lVert(\alpha H_t+\beta G_t)-(\alpha H_s+\beta G_s)\rVert_2\le|\alpha|\lVert H_t-H_s\rVert_2+|\beta|\lVert G_t-G_s\rVert_2$. The sums satisfy $S_n(\alpha H+\beta G)=\alpha S_n(H)+\beta S_n(G)$ exactly, and

2. Product family. For $s,t\in[a,b]$,

The function $|c|$ is bounded on $[a,b]$ by Extreme Value Theorem on a Compact Interval, and $\sup_{s\in[a,b]}\lVert H_s\rVert_2$ is finite because $s\mapsto\mathbb{E}[H_s^2]$ is bounded by Uniform Mean-Square Continuity on a Compact Interval; mean-square continuity of $(c(t)H_t)$ at each point follows from continuity of $c$ and mean-square continuity of $(H_t)$. Constant random factor. $\lVert c(t)Z-c(s)Z\rVert_2=|c(t)-c(s)|\,\lVert Z\rVert_2$, so $(c(t)Z)$ is mean-square continuous, and $S_n(cZ)=R_n(c)\,Z$, with $\lVert R_n(c)Z-(\int_a^b c)Z\rVert_2=|R_n(c)-\int_a^b c|\,\lVert Z\rVert_2\to0$ by the recorded fact.

3. By Cauchy-Schwarz, $|\mathbb{E}[ZH_t]-\mathbb{E}[ZH_s]|=|\mathbb{E}[Z(H_t-H_s)]|\le\lVert Z\rVert_2\lVert H_t-H_s\rVert_2$, giving continuity of $t\mapsto\mathbb{E}[ZH_t]$ (expectations are linear by Linearity and Monotonicity of the Lebesgue Integral). Next, $\mathbb{E}[Z\,S_n(H)]=\sum_i\mathbb{E}[ZH_{\tau_i}](x_i-x_{i-1})=R_n(\mathbb{E}[ZH_\cdot])\to\int_a^b\mathbb{E}[ZH_t]\,dt$ by the recorded fact, while

and uniqueness of real limits gives the first identity. With $Z=1$ (the constant random variable $1$, square-integrable since $\mathbb{E}[1^2]=1$, and $\mathbb{E}[1\cdot H_t]=\mathbb{E}[H_t]$) this yields $\mathbb{E}[\int_a^bH_t\,dt]=\int_a^b\mathbb{E}[H_t]\,dt$ and continuity of $t\mapsto\mathbb{E}[H_t]$. For the covariance version, by Covariance of Square-Integrable Random Variables, $\operatorname{Cov}(Z,X)=\mathbb{E}[ZX]-\mathbb{E}[Z]\mathbb{E}[X]$; hence $t\mapsto\operatorname{Cov}(Z,H_t)$ is continuous, and using the two identities just proved and linearity of the Riemann integral on continuous integrands,

4. By the triangle inequality, $\lVert H_t\rVert_2\le\lVert H_t-H_s\rVert_2+\lVert H_s\rVert_2$ and symmetrically, so $|\lVert H_t\rVert_2-\lVert H_s\rVert_2|\le\lVert H_t-H_s\rVert_2$ and $t\mapsto\lVert H_t\rVert_2$ is continuous. Also $\lVert S_n(H)\rVert_2\le\sum_i\lVert H_{\tau_i}\rVert_2(x_i-x_{i-1})=R_n(\lVert H_\cdot\rVert_2)\to\int_a^b\lVert H_t\rVert_2\,dt$ by the recorded fact, and $\lVert\int_a^bH_t\,dt\rVert_2\le\lVert\int_a^bH_t\,dt-S_n(H)\rVert_2+\lVert S_n(H)\rVert_2$; letting $n\to\infty$ gives the bound.

5. If $r=a$ or $r=b$ the identity is the convention $\int_s^s=0$ of Mean-Square Riemann Integral of a Family of Random Variables. Otherwise, for $n\ge1$ consider the tagged partition $Q_n$ of $[a,b]$ formed by the division points and tags of the two left-tagged uniform partitions of $[a,r]$ and $[r,b]$ with $n$ intervals each; its mesh is at most $\max(r-a,b-r)/n$. The associated sum splits exactly as $S(Q_n)=S_n(H;[a,r])+S_n(H;[r,b])$, whose mean-square limit is $\int_a^rH_t\,dt+\int_r^bH_t\,dt$ by the triangle inequality. On the other hand, given $\varepsilon>0$, take $\delta>0$ from Mean-Square Riemann Integral of a Family of Random Variables for the integral over $[a,b]$; for all $n$ with $\max(r-a,b-r)/n<\delta$ the sum $S(Q_n)$ is within $\varepsilon$ of $\int_a^bH_t\,dt$, so $S(Q_n)\to\int_a^bH_t\,dt$ in mean square. Uniqueness of mean-square limits concludes.

6. For $s<t$, claim 4 applied to the restricted family on $[s,t]$ gives $\lVert\int_s^tH_u\,du\rVert_2\le\int_s^t\lVert H_u\rVert_2\,du$; the second inequality holds since $\lVert H_u\rVert_2\le M:=\max_{u\in[a,b]}\lVert H_u\rVert_2$ (the maximum exists by claim 4 and Extreme Value Theorem on a Compact Interval) and the Riemann integral is monotone on continuous integrands. For $s=t$ all quantities vanish by the degenerate-interval conventions. Mean-square continuity of the indefinite integral: let $s\le t$ in $[a,b]$; if $s=t$ there is nothing to show, and the case $t=a$ forces $s=t=a$. For $a<t$ and $s<t$, applying claim 5 to the restricted family on $[a,t]$ with intermediate point $s$ (the convention covering $s=a$) gives $\int_a^tH_u\,du=\int_a^sH_u\,du+\int_s^tH_u\,du$ almost surely, whence

which tends to $0$ as $t-s\to0$; mean-square continuity follows. $\blacksquare$