Completeness of Probability Distribution as a Measure. Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a complete probability space
and $X:\Omega \to \mathbb{R}^n$ be a $\mathcal{F}$-measurable function, i.e., it is a random variable.
Then, in the book 
"Bernt Øksendal, stochastic differential equations: an introduction with applications, Springer-Verlag, 2003,"
it is said that $X$ induces the probability space $(\mathbb{R}^n, \mathcal{B}, \mu_X)$, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}^n$, 
and $\mu_X: \mathcal{B} \to [0, 1]$ is the probability measure called the distribution given by
\begin{equation}
   \mu_X(B) \doteq \mathbb{P}(X^{-1}(B)) 
\end{equation}
for any $B \in \mathcal{B}$. My question is that is $\mu_X$ a complete measure? Since $\mathbb{P}$ is assumed complete, I think $\mu_X$ would be also complete, but I stuck with its proof. Perhaps, one need to show that 
\begin{equation}
   \mu_X(B) = 0 \textrm{ and } C \subset B \;\; \Longrightarrow \;\; 
   C \in \mathcal{B},
\end{equation}
which obviously implies $\mu_X(C) = 0$ by monotonicy.
 A: No, the image measure of a random variable with respect to a complete measure is, in general, not complete.

Consider $((0,1),\mathcal{B}(0,1))$ endowed with the Lebesgue measure $\lambda$ and denote by $((0,1),\bar{\mathcal{B}}(0,1),\bar{\lambda})$ its completion. Define
$$X: ((0,1),\bar{\mathcal{B}}(0,1),\bar{\lambda}) \to ((0,1),\mathcal{B}(0,1),\lambda), x \mapsto x.$$
Obviously, $X$ is measurable and the underlying probability space is, by construction, complete. However, the image measure
$$\mu_X(B) = \mathbb{P}(X^{-1}(B)) = \lambda(B), \qquad B \in \mathcal{B}(0,1)$$
is not complete (because the space $((0,1),\mathcal{B}(0,1),\lambda)$ is not complete).

More generally: Consider any random variable $$X:(\Omega,\mathcal{A},\mathbb{P}) \to (X,\mathcal{B},\mu_X)$$ such that $(X,\mathcal{B},\mu_X)$ is not complete. Then $$X: (\Omega,\bar{\mathcal{A}},\bar{\mathbb{P}}) \to (X,\mathcal{B},\mu_X)$$ is obviously still a random variable; the underlying probability space is complete, but $(X,\mathcal{B},\mu_X)$ is not complete. 
