Convergence rate of $\log(1-X_n)-\log(1-c)$ Suppose that $\{X_n\}_{n\ge1}$ are random variables such that $X_n-c=O_p(a_n)$ as $n\to\infty$, where $0<c<1$ and $\{a_n\}_{n\ge1}$ is a positive sequence such that $a_n\to0$ as $n\to\infty$. I suppose that $\log(1-X_n)-\log(1-c)=O_p(a_n)$ as $n\to\infty$ as well and I am looking for a nice argument to prove this.
By the mean value theorem,
$$
\log(1-X_n(\omega))-\log(1-c)=\frac{1}{\xi_n(\omega)-1}(X_n(\omega)-c)
$$
for each $n\ge1$ and $\omega\in\Omega$, where $c\wedge X_n(\omega)<\xi_n(\omega)<c\vee X_n(\omega)$. We have that
$$
|\xi_n-c|<|X_n-c|
$$
and $\xi_n\xrightarrow{p} c$ as $n\to\infty$. Since $1/(\xi_n-1)=O_p(1)$ and $X_n-c=O_p(a_n)$ as $n\to\infty$, we conclude that $\log(1-X_n)-\log(1-c)=O_p(a_n)$ as $n\to\infty$.

Is this correct? Is this rigorous? Is $\xi_n$ a random variable, i.e. is $\xi_n$ measurable (we defined $\xi_n(\omega)$ for each $n\ge1$ and $\omega\in\Omega$)? Is there a simpler way to show that $\log(1-X_n)-\log(1-c)=O_p(a_n)$ as $n\to\infty$?

Any help is much appreaciated!
 A: I think your proof is correct. To avoid measurability issues with $\xi_n$ you can use the integral representation for the remainder term in Taylor's formula:
$$\log(1-x) - \log(1-c) =  (x-c) \int_0^1 \frac{1}{(c+r(x-c))-1} \, dr$$
and so
$$\log(1-X_n(\omega))-\log(1-c) = (X_n(\omega)-c) \left( \int_0^1 \frac{1}{(c+r(X_n(\omega)-c))-1} \, dr \right); \tag{1}$$
the term in the brackets corresponds to $1/(\xi_n(\omega)-1)$; in particular you can use this representation to prove the measurability of $\xi_n$.
Regarding the question about an easier proof: I would write it up a bit differently, but essentially the idea is the same. For any $\epsilon>0$ we can choose $R>0$ and $N \in \mathbb{N}$ such that $$A_n := \left\{\frac{|X_n-c|}{a_n} \leq R \right\}$$ satisfies $\mathbb{P}(A_n) \geq 1-\epsilon$ for all $n \geq N$. As $a_n \to 0$, we may assume without loss of generality that $N$ is sufficently large such that $$a_n R \leq \frac{c}{2}  \quad \text{for all $n \geq N$};$$ this implies
$$|c+r(X_n(\omega))-c)| \geq c - r |X_n(\omega)-c| \geq \frac{c}{2}$$
for any $\omega \in A_n$ and $r \in [0,1]$. It now follows from $(1)$ that
$$|\log(1-X_n(\omega))-\log(1-c)| \stackrel{(1)}{\leq} |X_n(\omega)-c| \frac{1}{1-c/2} \stackrel{\omega \in A_n}{\leq} \frac{R a_n}{1-c/2} $$
for any $\omega \in A_n$ and $n \geq N$. Hence,
$$1- \epsilon \leq \mathbb{P}(A_n) \leq \mathbb{P} \left( \frac{|\log(1-X_n))-\log(1-c)|}{a_n} \leq \frac{R}{1-c/2} \right).$$
As $\epsilon>0$ is arbitrary, this means that $\log(1-X_n)-\log(1-c) = O_P(a_n)$ by the very definition of boundedness in probability.
