Uniform Distribution Problem on $X, Y, Z$ Problem:

Let $X \sim \text{Uniform}(0,1)$. Let $0 < a < b < 1$. Let
$$
Y = \begin{cases} 1 & 0 < X < b \\
                  0 & \text{otherwise}
    \end{cases}
$$
and let
$$
Z = \begin{cases} 1 & a < X < 1 \\
                  0 & \text{otherwise}
    \end{cases}
$$

Question 1: How does one determine if $Y$ and $Z$ are independent? 
I can see that $Y$ and $Z$ are independent for all $x \notin [0,1]$. Things are less clear otherwise.
Question 2: How does one compute $\mathbb{E}(Y \mid Z = z)$?
Of course, I know that if $Z$, $Y$ are independent then $\mathbb{E}(Y \mid Z = z) = \mathbb{E}(Y)$
 A: It's obvious that $Y$ and $Z$ cannot be independent.  If you observe $Y = 1$, then you know $X \in (0,b)$, and intuitively, this information affects the probability of observing $Z = 1$.  Formally, $$\Pr[Z = 1 \mid Y = 1] = \frac{\Pr[(Z = 1) \cap (Y = 1)]}{\Pr[Y = 1]} = \frac{b-a}{b},$$ but $$\Pr[Z = 1] = 1-a \ne 1 - \frac{a}{b}$$ if $b \ne 1$.
A: Define events A and B s.t.
$$Y = 1_{(X < b)} := 1_{B}$$
$$Z = 1_{(a < X)} := 1_{A}$$
Observe that Y and Z are independent iff A and B are independent.
$$P(A \cap B) = \int_a^b 1 dx = b-a$$
$$P(B) = \int_0^b 1 dx = b$$
$$P(A) = \int_a^1 1 dx = 1-a$$
So for $x \in [0,1]$,
A and B are independent iff
$$b-a=b(1-a) \iff b-a=b-ba \iff -a=-ba \iff a = ba \iff b=1 \ \text{or} \ a=0$$
Recall that events whose probabilities are 0 or 1 are independent of any other event, including itself.

Definition:
$$E[X|A] = \frac{E[X1_A]}{P(A)}$$
Thus, we have
$$E[Y | Z=z] = \frac{E[Y1_{Z=z}]}{P(Z=z)}$$
$$ = \frac{E[1_B1_{1_A=z}]}{P(1_A=z)}$$
Thus, $E[Y | Z=z]$ is undefined if $z \ne 1$
For $z=1$, we have
$$ = \frac{E[1_B1_{1_A=1}]}{P(1_A=1)}$$
$$ = \frac{E[1_B1_A]}{P(A)}$$
$$ = \frac{E[1_{A \cap B}]}{P(A)}$$
$$ = \frac{P(A \cap B)}{P(A)}$$
$$ = P(B|A)$$
Can you take it from here?
