Random variable problems (a) Given $\omega = \left\{a,b,c,d\right\}$, $\mathscr{F} = \sigma(\left\{a,b\right\})$ and $H(a) = H(b) = -1, \ H(c) = 1, \ H(d) = 2$. Prove or disprove that $H$ is a random variable.
(b) Given 2 events $B_1, B_2$ and $\mathscr{F}$-measurable, define $X(w) = -1$ if $w\in B_1$, $= 1$ if $w\in B_1^{c}B_2$ and $= 0$ if $w\in B_1^{c}B_2^{c}$ 
My attempt Since $\mathscr{F}$ does not contain singleton sets, so for any $B\in \mathscr{B}(R)$ such that $ B = {1}$, then $f^{-1}(B) = {c}$, which is not in $\mathscr{F}$. Thus, $H$ is NOT a random variable.
Similarly, for any $B\in \mathscr{B}(R)$, $X^{-1}(B) = B_1$ if $(-1\in B) + (0, 1\not\in B)$, $= B_1^{c}B_2$ if $(1\in B$) + (0,-1\not\in B)$, = B_1^{c}B_2^{c}$  if $\ (0\in B) + (-1, 1\not\in B)$, $ = B_1\cup B_2\ $ if $\ (-1, 1\in B, 0\not\in B)$, $ = B_1^{c}B_2\cup (B_1^{c}B_2^{c}) $ if $\ (1, 0\in B, -1\not\in B)$, etc. Since in all of these cases, the result are all elements contained in $\mathscr{F}$, $X$ is a random variable.
My question Could someone please help review my proof above to see if they are correct? If not, please help me correct the mistake.
 A: Your proof looks correct to me.  Your argument is correct.  You could clean it up a bit in places if you want, but the changes are mostly aesthetic.
For example, your argument in part (a) could be rewritten: Since $\mathscr{F}$ doesn't contain singleton sets, we have $\{c\} = H^{-1}(\{1\}) \not\in \mathscr{F}$.  So $H$ isn't a random variable.
In part (b), your argument is again correct.  It might be worth noting that, if you want to check whether $X$ is a random variable, it suffices to check that $X^{-1}(B)$ is measurable (i.e. in $\mathscr{F}$) for all $B$ in some generating set for $\mathscr{B}(\mathbb{R}) \cap X(\Omega)$ instead of checking all possible $B$ (as you've done).  So, in this case, since $\mathscr{B}(\mathbb{R}) \cap X(\Omega) = \sigma(\{-1\}, \{0\}, \{1\}) \cap X(\Omega)$, it suffices to check that $X^{-1}(\{-1\}), X^{-1}(\{0\}),$ and $X^{-1}(\{1\})$ are measurable instead of checking all combinations (as you've done).
As a side note, you probably mean $B_1^c \cap B_2$ instead of $B_1^cB_2$?  I haven't seen the latter notation used.
