Mean Value Theorem for stochastic processes In Lemma 3.1 and Theorem 3.2 from the article Numerical solution of random differential equations: A mean square approach, it is stated and proved a Mean Value Theorem for stochastic integrals and derivatives in the mean square sense. See Chapter 4 in the book Random differential equations in science and engineering for an introduction to mean square theory.
My question is about the proof of Lemma 3.1.
Statement of Lemma 3.1: Let $Y(t)$ be a mean square continuous process with finite second order moments on $T=[t_0,t_1]$. Then, there exists $\xi\in [t_0,t_1]$ such that $\int_{t_0}^t Y(s)ds=Y(\xi)(t-t_0)$, $t_0<t<t_1$.
Question: The first thing that surprises me is the fact that $\xi$ is the same for every $t_0<t<t_1$. Is this correct?
Proof of Lemma 3.1 in the article: By page 90 from the book, $\Gamma_Y(r,s)=E[Y(r)Y(s)]$ is continuous on $T\times T$. Then $\Gamma_Y(r,\cdot)$ is continuous on $T$, for each $r\in T$. By the Mean Value Theorem for Riemann integrals, $\int_{t_0}^t \Gamma_Y(r,s)ds=\Gamma_Y(r,\xi)(t-t_0)$, $\xi\in[t_0,t]$. From here, $\xi$ is considered constant. 
Question: I think $\xi$ depends on $r$ and $t$, so this invalidates the proof. Is this correct?
Question: In case Lemma 3.1 is correct, for Theorem 3.2 I think it misses the hypothesis $\dot{X}(t)$ be mean square integrable, by page 104 from the book. Is this correct?
 A: Of course, $\xi$ in formula (3.2) must in general depend on $r$ and $t$. Also, there is no reason for the last sentence in the "proof" of Lemma 3.1 in the paper to be true, and I think one can easily construct a relevant counterexample. 
It is also easy to construct examples showing that $\xi$ in the equality $\int_{t_0}^t Y(s)ds=Y(\xi)(t-t_0)$ in Lemma 3.1 in the paper must in general depend, not only on $t$, but also on the path/realization of the process $Y(\cdot)$. 
For instance, for $t_0=0$ consider a process $Y(\cdot)$ on $[t_0,t_1]=[0,1]$ with only two possible paths, say $s\mapsto y_1(s):=s$ and $s\mapsto y_2(s):=s^2$, each of them having the same probability, $1/2$. Then for real $t>0$
$$\text{on the event $\{Y=y_1\}$ one has $\int_{t_0}^t Y(s)ds=\frac{t^2}2=Y\Big(\frac t2\Big)(t-t_0)$ and hence $\xi=\frac t2$}$$ 
and 
$$\text{on the event $\{Y=y_2\}$ one has $\int_{t_0}^t Y(s)ds=\frac{t^3}3=Y\Big(\frac t{\sqrt3}\Big)(t-t_0)$ and hence $\xi=\frac t{\sqrt3}$.}$$ 
So, we see that $\xi$ depends on the path and on $t$. 
Thus, Lemma 3.1 in the paper is not correct if $\xi$ is thought of as not depending on $t$ and/or the path. The proof of that lemma is incorrect anyway. 
However, if $\xi$ is allowed to depend on $t$ and on the path and if $Y(\cdot)$ is, not only mean-square continuous, but also path-wise continuous, then the path-wise mean value theorem obviously holds, as it immediately follows from the standard, "non-random" mean value theorem of calculus. 
On the other hand, if $Y(\cdot)$ is only assumed to be mean-square continuous but not path-wise continuous, then $\xi$ may not exist at all. E.g., let $Y(\cdot)$ be the stochastic process on the interval $[0,1]$ defined by the formula $Y(t)=I\{t>U\}$, where $U$ is a random variable uniformly distributed on $[0,1]$ and $I$ denotes the indicator. Then $Y(\cdot)$ is bounded and hence has finite second order moments. Moreover, $Y(\cdot)$ is mean square continuous, since $E(Y(t)-Y(s))^2=t-s$ if $0\le s\le t\le1$. However, for any $t\in(0,1]$, on the event $\{0<U<t\}$ we have 
$$\int_0^t Y(s)ds=\int_U^t Y(s)ds=t-U\ne Y(\xi)t$$ 
for any $\xi\in[0,1]$, since $Y(\cdot)$ can only take values $0$ and $1$. 
