Proving independence of random variables 
If $X$ and $Y$ are independent exponential random variables with parameter $\lambda$ and $\mu$. Let $Z=\min(X,Y)$, prove that $Z$ and $\mathbf 1_{\{X<Y\}}$ are independent.

I don't know, how to begin, to show is 
$$\Pr(Z\in A, \mathbf 1_{\{X<Y\}}\in B)=\Pr(Z\in A)\Pr(\mathbf 1_{\{X<Y\}}\in B)$$
for any Borel $A,B$
but this doesn't seem very useful, or do I have to find first the distribution function of $Z$, can you help please.
EDIT: maybe the parameters are not significant for this part of the exercise, there are also other questions, where the parameters must be used. I tried also;
$\Pr(Z\in A, \mathbf 1_{\{X<Y\}}\in B)$
$=\Pr([Z\in A, \mathbf 1_{\{X<Y\}}\in B]\mathbf 1_{\{X<Y\}})+\Pr([Z\in A, \mathbf 1_{\{X<Y\}}\in B]\mathbf 1_{\{X>Y\}})$
$=\Pr(X\in A, 1\in B)+\Pr(Y\in A, 0\in B)\overset!=\Pr(X\in A)\Pr(1\in B)+\Pr(Y\in A)\Pr(0\in B)$
$=\Pr(Z\in A)\Pr(\mathbf 1_{\{X<Y\}}\in B)$
the second last step (!) is allowed since, $1$ and $0$ are constant random variables and therefore independent of $X$ and $Y$ ?
 A: Consider independent Poisson processes $N_t$, $M_t$ with rates $\lambda$, $\mu$, such that $X$ is the time until the first occurrence of $N_t$ and $Y$ is the time until the first occurrence of $M_t$.  Then $N_t + M_t$ is a Poisson process  with rate $\lambda + \mu$, and $Z$ is the time until its first occurrence.  One way to
realize this is to start with a Poisson process of rate $\lambda + \mu$, and assign each occurrence (independently) to process $N_t$ with probability $\lambda/(\lambda + \mu)$, otherwise to $M_t$.  Then $X<Y$ is the event that the first occurrence is assigned to $N_t$, and by construction that is independent
of $Z$.
A: Maybe an easier answer is to follow the initial approach and forget about the exponential law because it doesn't matter. Intuitively, just knowing the maximum of the two iid rv tells you nothing about which of the two is larger than the other. Formally, you are supposed to show
\begin{align*}
\mathbb{P}[Z\in A,\mathbf{1}_{X<Y} \in B] &=\mathbb{P}[Z\in A] \mathbb{P}[\mathbf{1}_{X<Y} \in B]
\end{align*}
for all $A,B \in \mathcal{B}(R)$. Borel sets are not as bad as Lebesgue sets, but still, they are very complicated, so showing something for all Borel sets seems hard. Until you realize that there are not so many values that $\mathbf{1}_{X<Y}$ can take. If $B$ contains neither 0 nor 1, the equality should be obvious to you (is it?). By symmetry, you only have to consider $B$ such that $0\in B$ and $1\notin B$. Then you have
\begin{align}
\mathbb{P}[Z\in A,\mathbf{1}_{X<Y}\in B]&=\mathbb{P}[Z\in A,X<Y].
\end{align}
Can you finish it? 
Hints: Use iid and the continuity of the law to find:
1) $\mathbb{P}[X<Y]=?$.  
2) $\mathbb{P}[Z\in A,X<Y]=\mathbb{P}[Z\in A,Y<X]$
Finish it off with $\mathbb{P}[Z\in A]=\mathbb{P}[Z\in A,X<Y]+\mathbb{P}[Z\in A,Y<X]$.
