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(b) Express \({\bf1}_{A\cup B}(x) \) in terms of the indicator functions in (a).

(a) Need to check for all \(x\in\Omega\). There are four possibilities (make the old Venn diagram): \(x\) is in \(A\) but not in \(B\); \(x\) is in \(B\) but not in \(A\); \(x\) is in \(A\) and in \(B\); \(x\) is neither in \(A\) nor in \(B\). By considering each of those cases one sees that:
$$
{\bf1}_A(x){\bf1}_B(x)=1\quad \iff \quad{\bf1}_A(x)={\bf1}_B(x)=1\quad \iff \quad x\in A\ \hbox{and}\ x\in B.
$$
(b) From the Venn diagram, again checking all the possibilities, one sees that
$$
{\bf1}_{A\cup B}(x) + {\bf1}_{A\cap B}(x)={\bf1}_A(x)+{\bf1}_B(x).
$$

\(\{\omega\in(0,1)\,|\, X(\omega)\le 2\}\ =\ \{\omega\in(0,1)\,|\, -\log\omega\le 2\}\ =\ \{\omega\in(0,1)\,|\, \omega\ge e^{-2}\}\ =\ [e^{-2},1)\).
\(\{\omega\in(0,1)\,|\, Y(\omega)>.5\}\ =\ \{\omega\in(0,1)\,|\, 1-\omega>.5\}\ =\ (0,.5)\).
Then \(A\cap B=[e^{-2},.5)\).

If \(A\) is in \(\mathcal G\) then \(A^c=\Omega-A\) is necessarily in \(\mathcal G\) too, and the same applies to \(A\cup B\), \(A\cap B\) for any \(A,B\) that are subsets of \(\Omega\). So \(\mathcal G\) is a field.
It is also a \(\sigma\)-field because the union of any number of subsets of \(\Omega\) is also a subset of \(\Omega\). There is no difference between a field and a \(\sigma\)-field when the number of elements in \(\mathcal G\) is finite, as is the case here.

The field (or \(\sigma\)-field) generated by this set of subsets of \(\Omega\) must be closed for complementation, so \(\{\triangle, \bigcirc\}\) must also be in it. It must be closed for unions, so \(\Omega\) must be in it too. The complement of \(\Omega\) is \(\emptyset\). Hence, a tentative answer is:
$$
\mathcal F\ =\ \{\emptyset,\Omega, \{\square\}, \{\triangle, \bigcirc\}\}.
$$
One can check that the conditions required for a set of sets to be a field are satisfied by \(\mathcal F\).

We know that \(A^c \in\mathcal F\), since \(\mathcal F\) is a \(\sigma\)-field. Therefore the intersection of \(B\) and \(A^c\) is also in the \(\sigma\)-field \(\mathcal F\).

\({\mathcal F}_0\) is closed under unions, intersections and complements.

(a) The function \(\log(1+\omega)\) is increasing, so the minimum value of \(X\) is \(\log 1=0\) and its maximum value is \(\log2\). For \(x\) in between 0 and \(\log2\) we have
$$\eqalign{
\pro(X\le x)\ &=\ \pro(\{\omega\,|\, \log(1+\omega)\le x\})\cr
& =\ \pro(\{\omega\,|\, \omega\le e^x-1\})\cr
& =\ \pro((0,e^x-1))\ =\ e^x-1.
}
$$
Density is \(e^{x}{\bf1}_{(0,\log2)}(x)\).
(b) For \(x\in(0,1)\), \(\{\omega\,|\, \omega\le\sqrt x\}=(0,\sqrt x)\), so DF is \(\sqrt x\) for those \(x\); PDF is \({1\over 2\sqrt x}{\bf1}_{(0,1)}(x)\).