Hi this is an exercise from other old exam. I'd like to know if my attempt is correct. I'd really appreciate your help or your suggestions to improve my arguments and also recommendation to similar problems.
Question: Let $F: \mathbb{R} \to \mathbb{R}$ defined by
$$F(x)=\begin{cases}0& \text{ if $\,-\infty<x<-1$}\\ 1& \text{ if $\,-1\le x<1$}\\ 2& \text{ if $\,1\le x<+\infty$}\\ \end{cases}$$
1) We set for each $\,-\infty<a\le b <+\infty$, $$\,\mu_F((a,b])=F(b)-F(a)$$ Find the value of $\mu_F$ for each interval $(a,b]$
2) We can extent $\mu_F$ in the Borel $\sigma$-algebra $\mathscr{B}(\mathbb{R})$ uniquely?
3) Let $g: \mathbb{R} \to \mathbb{R}$ a Borel function and let $h: \mathbb{R}\to \mathbb{R}$ a continuous function. Prove that $h\circ g$ is $\mu$-integrable. What is the value of $\int_{\mathbb{R}}h\circ g d\mu$?
Answer: It is easy to show that $$\mu_F ((a,b])=\delta_{\{-1\}}((a,b])+\delta_{\{1\}}((a,b])$$
where $\delta_{\{x\}}((a,b])=\chi_{(a,b]}(x)$. Since $F$ is bounded, non-decreasing, right continuous and satisfies $\lim_{x\to -\infty} F(x)=0$, thus there is a finite measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ such that $\mu ((-\infty, x])=F(x)$ holds for each $x$ in $\mathbb{R}$.
To show uniqueness of $\mu$. Suppose there is a measure $\nu$ defined on the class $\mathscr{C}=\{(a,b] \mid a\le b \in \mathbb{R} \}$ such that $\nu ((a,b])= F(b)-F(a)$, that is, such that $\nu$ agrees with $\mu_F$ in the class $\mathscr{C}$. It is clear that $\mathscr{C}$ is a $\pi$-system. We will show that
$$\mathscr{D}=\{A\in \mathscr{B}(\mathbb{R}) \mid \mu(A)=\nu(A)\}$$
is a Dynkin Class. Let $A_n=(-n,n]$, for $n$ a natural number, so $A_n\uparrow \mathbb{R}$. Then
$$\nu(\mathbb{R})=\lim_n \nu (A_n)=\lim_n\big(F(n)-F(-n)\big)=\lim_n \mu(A_n)=\mu(\mathbb{R})$$
Then $\mathbb{R}$ belongs to $\mathscr{D}$ and also follows that $\nu$ is a finite measure. Suppose that $A\subset B$ and $A$ and $B$ belong to $\mathscr{D}$, so $$\nu(B-A)=\nu(B)-\nu(A)=\mu(B)-\mu(A)=\mu(B-A)$$
Since $\mu$ and $\nu$ are finite measures. So $A-B$ is in $\mathscr{D}$. If $\{A_n\}$ is non-decreasing sequence of elements in $\mathscr{D}$, then $A_n\uparrow A=\bigcup_n A_n$ and by the monotonicity of the measure, we have $$\mu(A)=\lim_n \mu(A_n)=\lim_n \nu(A_n)=\nu(A)$$
Thus $A$ belongs to $\mathscr{D}$. It follows that $\mathscr{D}$ is a Dynkin class. Now since $\mathscr{C}$ is clearly included in $\mathscr{D}$, it follows that $\mathscr{B}(\mathbb{R})=\sigma(\mathscr{C})\subset \mathscr{D}$. Hence $\nu$ agrees with $\mu$ on $\mathscr{B}(\mathbb{R})$
For (3). Since $h$ is continuous so is measurable and $h\circ g$ is also a Borel function. Let $A=\mathbb{R}-\{\pm 1\}$, we will show that $A$ is $\mu$-negligible. Since $A_n=(-n,-1-1/n]\uparrow (-\infty, -1)$ , thus $$\mu((-\infty, -1)) =\lim_n\mu(A_n)=\lim_n \bigg(F(-1-1/n)-F(-n)\bigg)=\lim_{x\,\to 1^{-}} F(x)=0$$
Let $(-1,1-1/n]\uparrow (-1,1)$ and $(1,n]\uparrow (1,+\infty)$ similarly $\mu((-1,1))=0=\mu((1,+\infty))$. Hence $$\mu(A)=\mu( (-\infty, -1))+\mu( (-1,1))+\mu((1,+\infty))=0$$
as desired. It follow that $$(h\circ g) (x) = h(g(1)) \chi_{\{1\}}(x)+h(g(-1)) \chi_{\{-1\}}(x)$$ $\mu$- a.e., hence the integral exists as the integral of simple function exists. Moreover
\begin{align}\int_{\mathbb{R}}(h\circ g) d\mu =\int_{\{\pm 1\}}(h\circ g) d\mu &= h(g(1)) \mu({\{1\}})+h(g(-1)) \mu(({\{-1\}})\\ & = h(g(1))+h(g(-1)) \end{align}
Any reference for exercises of this kind?
The proofs seem fine, although what is 'fine' in an exam depends heavily on the background of the students and the material covered/emphasised by the lecturer. Some points-
For similar exercises, you might want to find other uses of Dynkin's $π-\lambda$ theorem and prove them without referring to the proof. An easy suggestion: Lebesgue on $ℝ^d$ is the unique measure up to scaling that is translation invariant on rectangles, and assigns the unit box $[0,1]^d$ finite measure.